{"id": "Rudin|exercise_1_1b", "formal_statement": "theorem exercise_1_1b\n(x : \u211d)\n(y : \u211a)\n(h : y \u2260 0)\n: ( irrational x ) -> irrational ( x * y ) :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "If $r$ is rational $(r \\neq 0)$ and $x$ is irrational, prove that $rx$ is irrational.\n", "nl_proof": "\\begin{proof}\n\n    If $r x$ were rational, then $x=\\frac{r x}{r}$ would also be rational.\n\n\\end{proof}"}
{"id": "Rudin|exercise_1_4", "formal_statement": "theorem exercise_1_4\n(\u03b1 : Type*) [partial_order \u03b1]\n(s : set \u03b1)\n(x y : \u03b1)\n(h\u2080 : set.nonempty s)\n(h\u2081 : x \u2208 lower_bounds s)\n(h\u2082 : y \u2208 upper_bounds s)\n: x \u2264 y :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "Let $E$ be a nonempty subset of an ordered set; suppose $\\alpha$ is a lower bound of $E$ and $\\beta$ is an upper bound of $E$. Prove that $\\alpha \\leq \\beta$.\n", "nl_proof": "\\begin{proof}\n\nSince $E$ is nonempty, there exists $x \\in E$. Then by definition of lower and upper bounds we have $\\alpha \\leq x \\leq \\beta$, and hence by property $i i$ in the definition of an ordering, we have $\\alpha<\\beta$ unless $\\alpha=x=\\beta$.\n\n\\end{proof}"}
{"id": "Rudin|exercise_1_8", "formal_statement": "theorem exercise_1_8 : \u00ac \u2203 (r : \u2102 \u2192 \u2102 \u2192 Prop), is_linear_order \u2102 r :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "Prove that no order can be defined in the complex field that turns it into an ordered field.\n", "nl_proof": "\\begin{proof}\n\n    By Part (a) of Proposition $1.18$, either $i$ or $-i$ must be positive. Hence $-1=i^2=(-i)^2$ must be positive. But then $1=(-1)^2$, must also be positive, and this contradicts Part $(a)$ of Proposition 1.18, since 1 and $-1$ cannot both be positive.\n\n\\end{proof}"}
{"id": "Rudin|exercise_1_12", "formal_statement": "theorem exercise_1_12 (n : \u2115) (f : \u2115 \u2192 \u2102) : \n  abs (\u2211 i in finset.range n, f i) \u2264 \u2211 i in finset.range n, abs (f i) :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "If $z_1, \\ldots, z_n$ are complex, prove that $|z_1 + z_2 + \\ldots + z_n| \\leq |z_1| + |z_2| + \\cdots + |z_n|$.\n", "nl_proof": "\\begin{proof}\n\n    We can apply the case $n=2$ and induction on $n$ to get\n\n$$\n\n\\begin{aligned}\n\n\\left|z_1+z_2+\\cdots z_n\\right| &=\\left|\\left(z_1+z_2+\\cdots+z_{n-1}\\right)+z_n\\right| \\\\\n\n& \\leq\\left|z_1+z_2+\\cdots+z_{n-1}\\right|+\\left|z_n\\right| \\\\\n\n& \\leq\\left|z_1\\right|+\\left|z_2\\right|+\\cdots+\\left|z_{n-1}\\right|+\\left|z_n\\right|\n\n\\end{aligned}\n\n$$\n\n\\end{proof}"}
{"id": "Rudin|exercise_1_14", "formal_statement": "theorem exercise_1_14\n  (z : \u2102) (h : abs z = 1)\n  : (abs (1 + z)) ^ 2 + (abs (1 - z)) ^ 2 = 4 :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "If $z$ is a complex number such that $|z|=1$, that is, such that $z \\bar{z}=1$, compute $|1+z|^{2}+|1-z|^{2}$.\n", "nl_proof": "\\begin{proof}\n\n    $|1+z|^2=(1+z)(1+\\bar{z})=1+\\bar{z}+z+z \\bar{z}=2+z+\\bar{z}$. Similarly $|1-z|^2=(1-z)(1-\\bar{z})=1-z-\\bar{z}+z \\bar{z}=2-z-\\bar{z}$. Hence\n\n$$\n\n|1+z|^2+|1-z|^2=4 \\text {. }\n\n$$\n\n\\end{proof}"}
{"id": "Rudin|exercise_1_17", "formal_statement": "theorem exercise_1_17\n  (n : \u2115)\n  (x y : euclidean_space \u211d (fin n)) -- R^n\n  : \u2016x + y\u2016^2 + \u2016x - y\u2016^2 = 2*\u2016x\u2016^2 + 2*\u2016y\u2016^2 :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "Prove that $|\\mathbf{x}+\\mathbf{y}|^{2}+|\\mathbf{x}-\\mathbf{y}|^{2}=2|\\mathbf{x}|^{2}+2|\\mathbf{y}|^{2}$ if $\\mathbf{x} \\in R^{k}$ and $\\mathbf{y} \\in R^{k}$.\n", "nl_proof": "\\begin{proof}\n\n    The proof is a routine computation, using the relation\n\n$$\n\n|x \\pm y|^2=(x \\pm y) \\cdot(x \\pm y)=|x|^2 \\pm 2 x \\cdot y+|y|^2 .\n\n$$\n\nIf $\\mathrm{x}$ and $\\mathrm{y}$ are the sides of a parallelogram, then $\\mathrm{x}+\\mathrm{y}$ and $\\mathbf{x}-\\mathrm{y}$ are its diagonals. Hence this result says that the sum of the squares on the diagonals of a parallelogram equals the sum of the squares on the sides.\n\n\\end{proof}"}
{"id": "Rudin|exercise_1_18b", "formal_statement": "theorem exercise_1_18b\n  : \u00ac \u2200 (x : \u211d), \u2203 (y : \u211d), y \u2260 0 \u2227 x * y = 0 :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "If $k = 1$ and $\\mathbf{x} \\in R^{k}$, prove that there does not exist $\\mathbf{y} \\in R^{k}$ such that $\\mathbf{y} \\neq 0$ but $\\mathbf{x} \\cdot \\mathbf{y}=0$\n", "nl_proof": "\\begin{proof}\n\n    Not true when $k=1$, since the product of two nonzero real numbers is nonzero.\n\n\\end{proof}"}
{"id": "Rudin|exercise_2_19a", "formal_statement": "theorem exercise_2_19a {X : Type*} [metric_space X]\n  (A B : set X) (hA : is_closed A) (hB : is_closed B) (hAB : disjoint A B) :\n  separated_nhds A B :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "If $A$ and $B$ are disjoint closed sets in some metric space $X$, prove that they are separated.\n", "nl_proof": "\\begin{proof}\n\n    We are given that $A \\cap B=\\varnothing$. Since $A$ and $B$ are closed, this means $A \\cap \\bar{B}=\\varnothing=\\bar{A} \\cap B$, which says that $A$ and $B$ are separated.\n\n\\end{proof}"}
{"id": "Rudin|exercise_2_25", "formal_statement": "theorem exercise_2_25 {K : Type*} [metric_space K] [compact_space K] :\n  \u2203 (B : set (set K)), set.countable B \u2227 is_topological_basis B :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "Prove that every compact metric space $K$ has a countable base.\n", "nl_proof": "\\begin{proof}\n\n    $K$ can be covered by a finite union of neighborhoods of radius $1 / n$, and this shows that this implies that $K$ is separable.\n\n\n\nIt is not entirely obvious that a metric space with a countable base is separable. To prove this, let $\\left\\{V_n\\right\\}_{n=1}^{\\infty}$ be a countable base, and let $x_n \\in V_n$. The points $V_n$ must be dense in $X$. For if $G$ is any non-empty open set, then $G$ contains $V_n$ for some $n$, and hence $x_n \\in G$. (Thus for a metric space, having a countable base and being separable are equivalent.)\n\n\\end{proof}"}
{"id": "Rudin|exercise_2_27b", "formal_statement": "theorem exercise_2_27b (k : \u2115) (E P : set (euclidean_space \u211d (fin k)))\n  (hE : E.nonempty \u2227 \u00ac set.countable E)\n  (hP : P = {x | \u2200 U \u2208 \ud835\udcdd x, (P \u2229 E).nonempty \u2227 \u00ac set.countable (P \u2229 E)}) :\n  set.countable (E \\ P) :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "Suppose $E\\subset\\mathbb{R}^k$ is uncountable, and let $P$ be the set of condensation points of $E$. Prove that at most countably many points of $E$ are not in $P$.\n", "nl_proof": "\\begin{proof}\n\n    If $x \\in W^c$, and $O$ is any neighborhood of $x$, then $x \\in V_n \\subseteq O$ for some n. Since $x \\notin W, V_n \\cap E$ is uncountable. Hence $O$ contains uncountably many points of $E$, and so $x$ is a condensation point of $E$. Thus $x \\in P$, i.e., $W^c \\subseteq P$.\n\nConversely if $x \\in W$, then $x \\in V_n$ for some $V_n$ such that $V_n \\cap E$ is countable. Hence $x$ has a neighborhood (any neighborhood contained in $V_n$ ) containing at most a countable set of points of $E$, and so $x \\notin P$, i.e., $W \\subseteq P^c$. Hence $P=W^c$.\n\nIt is clear that $P$ is closed (since its complement $W$ is open), so that we need only show that $P \\subseteq P^{\\prime}$. Hence suppose $x \\in P$, and $O$ is any neighborhood of $x$. (By definition of $P$ this means $O \\cap E$ is uncountable.) We need to show that there is a point $y \\in P \\cap(O \\backslash\\{x\\})$. If this is not the case, i.e., if every point $y$ in $O \\backslash\\{x\\}$ is in $P^c$, then for each such point $y$ there is a set $V_n$ containing $y$ such that $V_n \\cap E$ is at most countable. That would mean that $y \\in W$, i.e., that $O \\backslash\\{x\\}$ is contained in $W$. It would follow that $O \\cap E \\subseteq\\{x\\} \\cup(W \\cap E)$, and so $O \\cap E$ contains at most a countable set of points, contrary to the hypothesis that $x \\in P$. Hence $O$ contains a point of $P$ different from $x$, and so $P \\subseteq P^{\\prime}$. Thus $P$ is perfect.\n\n\\end{proof}"}
{"id": "Rudin|exercise_2_29", "formal_statement": "theorem exercise_2_29 (U : set \u211d) (hU : is_open U) :\n  \u2203 (f : \u2115 \u2192 set \u211d), (\u2200 n, \u2203 a b : \u211d, f n = {x | a < x \u2227 x < b}) \u2227 (\u2200 n, f n \u2286 U) \u2227\n  (\u2200 n m, n \u2260 m \u2192 f n \u2229 f m = \u2205) \u2227\n  U = \u22c3 n, f n :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "Prove that every open set in $\\mathbb{R}$ is the union of an at most countable collection of disjoint segments.\n", "nl_proof": "\\begin{proof}\n\n    Let $O$ be open. For each pair of points $x \\in O, y \\in O$, we define an equivalence relation $x \\sim y$ by saying $x \\sim y$ if and only if $[\\min (x, y), \\max (x, y)] \\subset$ 0 . This is an equivalence relation, since $x \\sim x([x, x] \\subset O$ if $x \\in O)$; if $x \\sim y$, then $y \\sim x$ (since $\\min (x, y)=\\min (y, x)$ and $\\max (x, y)=\\max (y, x))$; and if $x \\sim y$ and $y \\sim z$, then $x \\sim z([\\min (x, z), \\max (x, z)] \\subseteq[\\min (x, y), \\max (x, y)] \\cup$ $[\\min (y, z), \\max (y, z)] \\subseteq O)$. In fact it is easy to prove that\n\n$$\n\n\\min (x, z) \\geq \\min (\\min (x, y), \\min (y, z))\n\n$$\n\nand\n\n$$\n\n\\max (x, z) \\leq \\max (\\max (x, y), \\max (y, z))\n\n$$\n\nIt follows that $O$ can be written as a disjoint union of pairwise disjoint equivalence classes. We claim that each equivalence class is an open interval.\n\n\n\nTo show this, for each $x \\in O$; let $A=\\{z:[z, x] \\subseteq O\\}$ and $B=\\{z:[x, z] \\subseteq$ $O\\}$, and let $a=\\inf A, b=\\sup B$. We claim that $(a, b) \\subset O$. Indeed if $a<z<b$, there exists $c \\in A$ with $c<z$ and $d \\in B$ with $d>z$. Then $z \\in[c, x] \\cup[x, d] \\subseteq O$. We now claim that $(a, b)$ is the equivalence class containing $x$. It is clear that each element of $(a, b)$ is equivalent to $x$ by the way in which $a$ and $b$ were chosen. We need to show that if $z \\notin(a, b)$, then $z$ is not equivalent to $x$. Suppose that $z<a$. If $z$ were equivalent to $x$, then $[z, x]$ would be contained in $O$, and so we would have $z \\in A$. Hence $a$ would not be a lower bound for $A$. Similarly if $z>b$ and $z \\sim x$, then $b$ could not be an upper bound for $B$.\n\n\n\nWe have now established that $O$ is a union of pairwise disjoint open intervals. Such a union must be at most countable, since each open interval contains a rational number not in any other interval.\n\n\\end{proof}"}
{"id": "Rudin|exercise_3_2a", "formal_statement": "theorem exercise_3_2a\n  : tendsto (\u03bb (n : \u211d), (sqrt (n^2 + n) - n)) at_top (\ud835\udcdd (1/2)) :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "Prove that $\\lim_{n \\rightarrow \\infty}\\sqrt{n^2 + n} -n = 1/2$.\n", "nl_proof": "\\begin{proof}\n\n    Multiplying and dividing by $\\sqrt{n^2+n}+n$ yields\n\n$$\n\n\\sqrt{n^2+n}-n=\\frac{n}{\\sqrt{n^2+n}+n}=\\frac{1}{\\sqrt{1+\\frac{1}{n}}+1} .\n\n$$\n\nIt follows that the limit is $\\frac{1}{2}$.\n\n\\end{proof}"}
{"id": "Rudin|exercise_3_5", "formal_statement": "theorem exercise_3_5 -- TODO fix\n  (a b : \u2115 \u2192 \u211d)\n  (h : limsup a + limsup b \u2260 0) :\n  limsup (\u03bb n, a n + b n) \u2264 limsup a + limsup b :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "For any two real sequences $\\left\\{a_{n}\\right\\},\\left\\{b_{n}\\right\\}$, prove that $\\limsup _{n \\rightarrow \\infty}\\left(a_{n}+b_{n}\\right) \\leq \\limsup _{n \\rightarrow \\infty} a_{n}+\\limsup _{n \\rightarrow \\infty} b_{n},$ provided the sum on the right is not of the form $\\infty-\\infty$.\n", "nl_proof": "\\begin{proof}\n\n    Since the case when $\\limsup _{n \\rightarrow \\infty} a_n=+\\infty$ and $\\limsup _{n \\rightarrow \\infty} b_n=-\\infty$ has been excluded from consideration, we note that the inequality is obvious if $\\limsup _{n \\rightarrow \\infty} a_n=+\\infty$. Hence we shall assume that $\\left\\{a_n\\right\\}$ is bounded above.\n\n\n\nLet $\\left\\{n_k\\right\\}$ be a subsequence of the positive integers such that $\\lim _{k \\rightarrow \\infty}\\left(a_{n_k}+\\right.$ $\\left.b_{n_k}\\right)=\\limsup _{n \\rightarrow \\infty}\\left(a_n+b_n\\right)$. Then choose a subsequence of the positive integers $\\left\\{k_m\\right\\}$ such that\n\n$$\n\n\\lim _{m \\rightarrow \\infty} a_{n_{k_m}}=\\limsup _{k \\rightarrow \\infty} a_{n_k} .\n\n$$\n\nThe subsequence $a_{n_{k_m}}+b_{n_{k_m}}$ still converges to the same limit as $a_{n_k}+b_{n_k}$, i.e., to $\\limsup _{n \\rightarrow \\infty}\\left(a_n+b_n\\right)$. Hence, since $a_{n_k}$ is bounded above (so that $\\limsup _{k \\rightarrow \\infty} a_{n_k}$ is finite), it follows that $b_{n_{k_m}}$ converges to the difference\n\n$$\n\n\\lim _{m \\rightarrow \\infty} b_{n_{k_m}}=\\lim _{m \\rightarrow \\infty}\\left(a_{n_{k_m}}+b_{n_{k_m}}\\right)-\\lim _{m \\rightarrow \\infty} a_{n_{k_m}} .\n\n$$\n\nThus we have proved that there exist subsequences $\\left\\{a_{n_{k_m}}\\right\\}$ and $\\left\\{b_{n_{k_m}}\\right\\}$ which converge to limits $a$ and $b$ respectively such that $a+b=\\limsup _{n \\rightarrow \\infty}\\left(a_n+b_n^*\\right)$. Since $a$ is the limit of a subsequence of $\\left\\{a_n\\right\\}$ and $b$ is the limit of a subsequence of $\\left\\{b_n\\right\\}$, it follows that $a \\leq \\limsup _{n \\rightarrow \\infty} a_n$ and $b \\leq \\limsup _{n \\rightarrow \\infty} b_n$, from which the desired inequality follows.\n\n\\end{proof}"}
{"id": "Rudin|exercise_3_7", "formal_statement": "theorem exercise_3_7\n  (a : \u2115 \u2192 \u211d)\n  (h : \u2203 y, (tendsto (\u03bb n, (\u2211 i in (finset.range n), a i)) at_top (\ud835\udcdd y))) :\n  \u2203 y, tendsto (\u03bb n, (\u2211 i in (finset.range n), sqrt (a i) / n)) at_top (\ud835\udcdd y) :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "Prove that the convergence of $\\Sigma a_{n}$ implies the convergence of $\\sum \\frac{\\sqrt{a_{n}}}{n}$ if $a_n\\geq 0$.\n", "nl_proof": "\\begin{proof}\n\n    Since $\\left(\\sqrt{a_n}-\\frac{1}{n}\\right)^2 \\geq 0$, it follows that\n\n$$\n\n\\frac{\\sqrt{a_n}}{n} \\leq \\frac{1}{2}\\left(a_n^2+\\frac{1}{n^2}\\right) .\n\n$$\n\nNow $\\Sigma a_n^2$ converges by comparison with $\\Sigma a_n$ (since $\\Sigma a_n$ converges, we have $a_n<1$ for large $n$, and hence $\\left.a_n^2<a_n\\right)$. Since $\\Sigma \\frac{1}{n^2}$ also converges ($p$ series, $p=2$ ), it follows that $\\Sigma \\frac{\\sqrt{a_n}}{n}$ converges.\n\n\\end{proof}"}
{"id": "Rudin|exercise_3_13", "formal_statement": "theorem exercise_3_13\n  (a b : \u2115 \u2192 \u211d)\n  (ha : \u2203 y, (tendsto (\u03bb n, (\u2211 i in (finset.range n), |a i|)) at_top (\ud835\udcdd y)))\n  (hb : \u2203 y, (tendsto (\u03bb n, (\u2211 i in (finset.range n), |b i|)) at_top (\ud835\udcdd y))) :\n  \u2203 y, (tendsto (\u03bb n, (\u2211 i in (finset.range n),\n  \u03bb i, (\u2211 j in finset.range (i + 1), a j * b (i - j)))) at_top (\ud835\udcdd y)) :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "Prove that the Cauchy product of two absolutely convergent series converges absolutely.\n", "nl_proof": "\\begin{proof}\n\n    Since both the hypothesis and conclusion refer to absolute convergence, we may assume both series consist of nonnegative terms. We let $S_n=\\sum_{k=0}^n a_n, T_n=\\sum_{k=0}^n b_n$, and $U_n=\\sum_{k=0}^n \\sum_{l=0}^k a_l b_{k-l}$. We need to show that $U_n$ remains bounded, given that $S_n$ and $T_n$ are bounded. To do this we make the convention that $a_{-1}=T_{-1}=0$, in order to save ourselves from having to separate off the first and last terms when we sum by parts. We then have\n\n$$\n\n\\begin{aligned}\n\nU_n &=\\sum_{k=0}^n \\sum_{l=0}^k a_l b_{k-l} \\\\\n\n&=\\sum_{k=0}^n \\sum_{l=0}^k a_l\\left(T_{k-l}-T_{k-l-1}\\right) \\\\\n\n&=\\sum_{k=0}^n \\sum_{j=0}^k a_{k-j}\\left(T_j-T_{j-1}\\right) \\\\\n\n&=\\sum_{k=0}^n \\sum_{j=0}^k\\left(a_{k-j}-a_{k-j-1}\\right) T_j \\\\\n\n&=\\sum_{j=0}^n \\sum_{k=j}^n\\left(a_{k-j}-a_{k-j-1}\\right) T_j\n\n&=\\sum_{j=0}^n a_{n-j} T_j \\\\\n\n&\\leq T \\sum_{m=0}^n a_m \\\\\n\n&=T S_n \\\\\n\n&\\leq S T .\n\n\\end{aligned}\n\n$$\n\nThus $U_n$ is bounded, and hence approaches a finite limit.\n\n\n\n\\end{proof}"}
{"id": "Rudin|exercise_3_21", "formal_statement": "theorem exercise_3_21\n  {X : Type*} [metric_space X] [complete_space X]\n  (E : \u2115 \u2192 set X)\n  (hE : \u2200 n, E n \u2283 E (n + 1))\n  (hE' : tendsto (\u03bb n, metric.diam (E n)) at_top (\ud835\udcdd 0)) :\n  \u2203 a, set.Inter E = {a} :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "If $\\left\\{E_{n}\\right\\}$ is a sequence of closed nonempty and bounded sets in a complete metric space $X$, if $E_{n} \\supset E_{n+1}$, and if $\\lim _{n \\rightarrow \\infty} \\operatorname{diam} E_{n}=0,$ then $\\bigcap_{1}^{\\infty} E_{n}$ consists of exactly one point.\n", "nl_proof": "\\begin{proof}\n\n    Choose $x_n \\in E_n$. (We use the axiom of choice here.) The sequence $\\left\\{x_n\\right\\}$ is a Cauchy sequence, since the diameter of $E_n$ tends to zero as $n$ tends to infinity and $E_n$ contains $E_{n+1}$. Since the metric space $X$ is complete, the sequence $x_n$ converges to a point $x$, which must belong to $E_n$ for all $n$, since $E_n$ is closed and contains $x_m$ for all $m \\geq n$. There cannot be a second point $y$ in all of the $E_n$, since for any point $y \\neq x$ the diameter of $E_n$ is less $\\operatorname{than} d(x, y)$ for large $n$.\n\n\\end{proof}"}
{"id": "Rudin|exercise_4_1a", "formal_statement": "theorem exercise_4_1a\n  : \u2203 (f : \u211d \u2192 \u211d), (\u2200 (x : \u211d), tendsto (\u03bb y, f(x + y) - f(x - y)) (\ud835\udcdd 0) (\ud835\udcdd 0)) \u2227 \u00ac continuous f :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "Suppose $f$ is a real function defined on $\\mathbb{R}$ which satisfies $\\lim_{h \\rightarrow 0} f(x + h) - f(x - h) = 0$ for every $x \\in \\mathbb{R}$. Show that $f$ does not need to be continuous.\n", "nl_proof": "\\begin{proof}\n\n    $$\n\nf(x)= \\begin{cases}1 & \\text { if } x \\text { is an integer } \\\\ 0 & \\text { if } x \\text { is not an integer. }\\end{cases}\n\n$$\n\n(If $x$ is an integer, then $f(x+h)-f(x-h) \\equiv 0$ for all $h$; while if $x$ is not an integer, $f(x+h)-f(x-h)=0$ for $|h|<\\min (x-[x], 1+[x]-x)$.\n\n\\end{proof}"}
{"id": "Rudin|exercise_4_3", "formal_statement": "theorem exercise_4_3\n  {\u03b1 : Type} [metric_space \u03b1]\n  (f : \u03b1 \u2192 \u211d) (h : continuous f) (z : set \u03b1) (g : z = f\u207b\u00b9' {0})\n  : is_closed z :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "Let $f$ be a continuous real function on a metric space $X$. Let $Z(f)$ (the zero set of $f$ ) be the set of all $p \\in X$ at which $f(p)=0$. Prove that $Z(f)$ is closed.\n", "nl_proof": "\\begin{proof}\n\n    $Z(f)=f^{-1}(\\{0\\})$, which is the inverse image of a closed set. Hence $Z(f)$ is closed.\n\n\\end{proof}"}
{"id": "Rudin|exercise_4_4b", "formal_statement": "theorem exercise_4_4b\n  {\u03b1 : Type} [metric_space \u03b1]\n  {\u03b2 : Type} [metric_space \u03b2]\n  (f g : \u03b1 \u2192 \u03b2)\n  (s : set \u03b1)\n  (h\u2081 : continuous f)\n  (h\u2082 : continuous g)\n  (h\u2083 : dense s)\n  (h\u2084 : \u2200 x \u2208 s, f x = g x)\n  : f = g :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "Let $f$ and $g$ be continuous mappings of a metric space $X$ into a metric space $Y$, and let $E$ be a dense subset of $X$. Prove that if $g(p) = f(p)$ for all $p \\in P$ then $g(p) = f(p)$ for all $p \\in X$.\n", "nl_proof": "\\begin{proof}\n\n    The function $\\varphi: X \\rightarrow R^1$ given by\n\n$$\n\n\\varphi(p)=d_Y(f(p), g(p))\n\n$$\n\nis continuous, since\n\n$$\n\n\\left|d_Y(f(p), g(p))-d_Y(f(q), g(q))\\right| \\leq d_Y(f(p), f(q))+d_Y(g(p), g(q))\n\n$$\n\n(This inequality follows from the triangle inequality, since\n\n$$\n\nd_Y(f(p), g(p)) \\leq d_Y(f(p), f(q))+d_Y(f(q), g(q))+d_Y(g(q), g(p)),\n\n$$\n\nand the same inequality holds with $p$ and $q$ interchanged. The absolute value $\\left|d_Y(f(p), g(p))-d_Y(f(q), g(q))\\right|$ must be either $d_Y(f(p), g(p))-d_Y(f(q), g(q))$ or $d_Y(f(q), g(q))-d_Y(f(p), g(p))$, and the triangle inequality shows that both of these numbers are at most $d_Y(f(p), f(q))+d_Y(g(p), g(q))$.)\n\nBy the previous problem, the zero set of $\\varphi$ is closed. But by definition\n\n$$\n\nZ(\\varphi)=\\{p: f(p)=g(p)\\} .\n\n$$\n\nHence the set of $p$ for which $f(p)=g(p)$ is closed. Since by hypothesis it is dense, it must be $X$.\n\n\\end{proof}"}
{"id": "Rudin|exercise_4_5b", "formal_statement": "theorem exercise_4_5b\n  : \u2203 (E : set \u211d) (f : \u211d \u2192 \u211d), (continuous_on f E) \u2227\n  (\u00ac \u2203 (g : \u211d \u2192 \u211d), continuous g \u2227 \u2200 x \u2208 E, f x = g x) :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "Show that there exist a set $E \\subset \\mathbb{R}$ and a real continuous function $f$ defined on $E$, such that there does not exist a continuous real function $g$ on $\\mathbb{R}$ such that $g(x)=f(x)$ for all $x \\in E$.\n", "nl_proof": "\\begin{proof}\n\n    Let $E:=(0,1)$, and define $f(x):=1 / x$ for all $x \\in E$. If $f$ has a continuous extension $g$ to $\\mathbb{R}$, then $g$ is continuous on $[-1,1]$, and therefore bounded by the intermediate value theorem. However, $f$ is not bounded in any neighborhood of $x=0$, so therefore $g$ is not bounded either, a contradiction.\n\n\\end{proof}"}
{"id": "Rudin|exercise_4_8a", "formal_statement": "theorem exercise_4_8a\n  (E : set \u211d) (f : \u211d \u2192 \u211d) (hf : uniform_continuous_on f E)\n  (hE : metric.bounded E) : metric.bounded (set.image f E) :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "Let $f$ be a real uniformly continuous function on the bounded set $E$ in $R^{1}$. Prove that $f$ is bounded on $E$.\n", "nl_proof": "\\begin{proof}\n\n    Let $a=\\inf E$ and $b=\\sup E$, and let $\\delta>0$ be such that $|f(x)-f(y)|<1$ if $x, y \\in E$ and $|x-y|<\\delta$. Now choose a positive integer $N$ larger than $(b-a) / \\delta$, and consider the $N$ intervals $I_k=\\left[a+\\frac{k-1}{b-a}, a+\\frac{k}{b-a}\\right], k=1,2, \\ldots, N$. For each $k$ such that $I_k \\cap E \\neq \\varnothing$ let $x_k \\in E \\cap I_k$. Then let $M=1+\\max \\left\\{\\left|f\\left(x_k\\right)\\right|\\right\\}$. If $x \\in E$, we have $\\left|x-x_k\\right|<\\delta$ for some $k$, and hence $|f(x)|<M$.\n\n\\end{proof}"}
{"id": "Rudin|exercise_4_11a", "formal_statement": "theorem exercise_4_11a\n  {X : Type*} [metric_space X]\n  {Y : Type*} [metric_space Y]\n  (f : X \u2192 Y) (hf : uniform_continuous f)\n  (x : \u2115 \u2192 X) (hx : cauchy_seq x) :\n  cauchy_seq (\u03bb n, f (x n)) :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "Suppose $f$ is a uniformly continuous mapping of a metric space $X$ into a metric space $Y$ and prove that $\\left\\{f\\left(x_{n}\\right)\\right\\}$ is a Cauchy sequence in $Y$ for every Cauchy sequence $\\{x_n\\}$ in $X$.\n", "nl_proof": "\\begin{proof}\n\n    Suppose $\\left\\{x_n\\right\\}$ is a Cauchy sequence in $X$. Let $\\varepsilon>0$ be given. Let $\\delta>0$ be such that $d_Y(f(x), f(u))<\\varepsilon$ if $d_X(x, u)<\\delta$. Then choose $N$ so that $d_X\\left(x_n, x_m\\right)<\\delta$ if $n, m>N$. Obviously $d_Y\\left(f\\left(x_n\\right), f\\left(x_m\\right)\\right)<\\varepsilon$ if $m, n>N$, showing that $\\left\\{f\\left(x_n\\right)\\right\\}$ is a Cauchy sequence.\n\n\\end{proof}"}
{"id": "Rudin|exercise_4_15", "formal_statement": "theorem exercise_4_15 {f : \u211d \u2192 \u211d}\n  (hf : continuous f) (hof : is_open_map f) :\n  monotone f :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "Prove that every continuous open mapping of $R^{1}$ into $R^{1}$ is monotonic.\n", "nl_proof": "\\begin{proof}\n\n    Suppose $f$ is continuous and not monotonic, say there exist points $a<b<c$ with $f(a)<f(b)$, and $f(c)<f(b)$. Then the maximum value of $f$ on the closed interval $[a, c]$ is assumed at a point $u$ in the open interval $(a, c)$. If there is also a point $v$ in the open interval $(a, c)$ where $f$ assumes its minimum value on $[a, c]$, then $f(a, c)=[f(v), f(u)]$. If no such point $v$ exists, then $f(a, c)=(d, f(u)]$, where $d=\\min (f(a), f(c))$. In either case, the image of $(a, c)$ is not open.\n\n\\end{proof}"}
{"id": "Rudin|exercise_4_21a", "formal_statement": "theorem exercise_4_21a {X : Type*} [metric_space X]\n  (K F : set X) (hK : is_compact K) (hF : is_closed F) (hKF : disjoint K F) :\n  \u2203 (\u03b4 : \u211d), \u03b4 > 0 \u2227 \u2200 (p q : X), p \u2208 K \u2192 q \u2208 F \u2192 dist p q \u2265 \u03b4 :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "Suppose $K$ and $F$ are disjoint sets in a metric space $X, K$ is compact, $F$ is closed. Prove that there exists $\\delta>0$ such that $d(p, q)>\\delta$ if $p \\in K, q \\in F$.\n", "nl_proof": "\\begin{proof}\n\nFollowing the hint, we observe that $\\rho_F(x)$ must attain its minimum value on $K$, i.e., there is some point $r \\in K$ such that\n\n$$\n\n\\rho_F(r)=\\min _{q \\in K} \\rho_F(q) .\n\n$$\n\nSince $F$ is closed and $r \\notin F$, it follows from Exercise $4.20$ that $\\rho_F(r)>0$. Let $\\delta$ be any positive number smaller than $\\rho_F(r)$. Then for any $p \\in F, q \\in K$, we have\n\n$$\n\nd(p, q) \\geq \\rho_F(q) \\geq \\rho_F(r)>\\delta .\n\n$$\n\nThis proves the positive assertion.\n\n\\end{proof}"}
{"id": "Rudin|exercise_5_1", "formal_statement": "theorem exercise_5_1\n  {f : \u211d \u2192 \u211d} (hf : \u2200 x y : \u211d, | (f x - f y) | \u2264 (x - y) ^ 2) :\n  \u2203 c, f = \u03bb x, c :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "Let $f$ be defined for all real $x$, and suppose that $|f(x)-f(y)| \\leq (x-y)^{2}$ for all real $x$ and $y$. Prove that $f$ is constant.\n", "nl_proof": "\\begin{proof}\n\n    Dividing by $x-y$, and letting $x \\rightarrow y$, we find that $f^{\\prime}(y)=0$ for all $y$. Hence $f$ is constant.\n\n\\end{proof}"}
{"id": "Rudin|exercise_5_3", "formal_statement": "theorem exercise_5_3 {g : \u211d \u2192 \u211d} (hg : continuous g)\n  (hg' : \u2203 M : \u211d, \u2200 x : \u211d, | deriv g x | \u2264 M) :\n  \u2203 N, \u2200 \u03b5 > 0, \u03b5 < N \u2192 function.injective (\u03bb x : \u211d, x + \u03b5 * g x) :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "Suppose $g$ is a real function on $R^{1}$, with bounded derivative (say $\\left|g^{\\prime}\\right| \\leq M$ ). Fix $\\varepsilon>0$, and define $f(x)=x+\\varepsilon g(x)$. Prove that $f$ is one-to-one if $\\varepsilon$ is small enough.\n", "nl_proof": "\\begin{proof}\n\n    If $0<\\varepsilon<\\frac{1}{M}$, we certainly have\n\n$$\n\nf^{\\prime}(x) \\geq 1-\\varepsilon M>0,\n\n$$\n\nand this implies that $f(x)$ is one-to-one, by the preceding problem.\n\n\\end{proof}"}
{"id": "Rudin|exercise_5_5", "formal_statement": "theorem exercise_5_5\n  {f : \u211d \u2192 \u211d}\n  (hfd : differentiable \u211d f)\n  (hf : tendsto (deriv f) at_top (\ud835\udcdd 0)) :\n  tendsto (\u03bb x, f (x + 1) - f x) at_top at_top :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "Suppose $f$ is defined and differentiable for every $x>0$, and $f^{\\prime}(x) \\rightarrow 0$ as $x \\rightarrow+\\infty$. Put $g(x)=f(x+1)-f(x)$. Prove that $g(x) \\rightarrow 0$ as $x \\rightarrow+\\infty$.\n", "nl_proof": "\\begin{proof}\n\n    Let $\\varepsilon>0$. Choose $x_0$ such that $\\left|f^{\\prime}(x)\\right|<\\varepsilon$ if $x>x_0$. Then for any $x \\geq x_0$ there exists $x_1 \\in(x, x+1)$ such that\n\n$$\n\nf(x+1)-f(x)=f^{\\prime}\\left(x_1\\right) .\n\n$$\n\nSince $\\left|f^{\\prime}\\left(x_1\\right)\\right|<\\varepsilon$, it follows that $|f(x+1)-f(x)|<\\varepsilon$, as required.\n\n\\end{proof}"}
{"id": "Rudin|exercise_5_7", "formal_statement": "theorem exercise_5_7\n  {f g : \u211d \u2192 \u211d} {x : \u211d}\n  (hf' : differentiable_at \u211d f 0)\n  (hg' : differentiable_at \u211d g 0)\n  (hg'_ne_0 : deriv g 0 \u2260 0)\n  (f0 : f 0 = 0) (g0 : g 0 = 0) :\n  tendsto (\u03bb x, f x / g x) (\ud835\udcdd x) (\ud835\udcdd (deriv f x / deriv g x)) :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "Suppose $f^{\\prime}(x), g^{\\prime}(x)$ exist, $g^{\\prime}(x) \\neq 0$, and $f(x)=g(x)=0$. Prove that $\\lim _{t \\rightarrow x} \\frac{f(t)}{g(t)}=\\frac{f^{\\prime}(x)}{g^{\\prime}(x)}.$\n", "nl_proof": "\\begin{proof}\n\n    Since $f(x)=g(x)=0$, we have\n\n$$\n\n\\begin{aligned}\n\n\\lim _{t \\rightarrow x} \\frac{f(t)}{g(t)} &=\\lim _{t \\rightarrow x} \\frac{\\frac{f(t)-f(x)}{t-x}}{\\frac{g(t)-g(x)}{t-x}} \\\\\n\n&=\\frac{\\lim _{t \\rightarrow x} \\frac{f(t)-f(x)}{t-x}}{\\lim _{t \\rightarrow x} \\frac{g(t)-g(x)}{t-x}} \\\\\n\n&=\\frac{f^{\\prime}(x)}{g^{\\prime}(x)}\n\n\\end{aligned}\n\n$$\n\n\\end{proof}"}
{"id": "Rudin|exercise_5_17", "formal_statement": "theorem exercise_5_17\n  {f : \u211d \u2192 \u211d}\n  (hf' : differentiable_on \u211d f (set.Icc (-1) 1))\n  (hf'' : differentiable_on \u211d (deriv f) (set.Icc 1 1))\n  (hf''' : differentiable_on \u211d (deriv (deriv f)) (set.Icc 1 1))\n  (hf0 : f (-1) = 0)\n  (hf1 : f 0 = 0)\n  (hf2 : f 1 = 1)\n  (hf3 : deriv f 0 = 0) :\n  \u2203 x, x \u2208 set.Ioo (-1 : \u211d) 1 \u2227 deriv (deriv (deriv f)) x \u2265 3 :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "Suppose $f$ is a real, three times differentiable function on $[-1,1]$, such that $f(-1)=0, \\quad f(0)=0, \\quad f(1)=1, \\quad f^{\\prime}(0)=0 .$ Prove that $f^{(3)}(x) \\geq 3$ for some $x \\in(-1,1)$.\n", "nl_proof": "\\begin{proof}\n\n    Following the hint, we observe that Theorem $5.15$ (Taylor's formula with remainder) implies that\n\n$$\n\n\\begin{aligned}\n\nf(1) &=f(0)+f^{\\prime}(0)+\\frac{1}{2} f^{\\prime \\prime}(0)+\\frac{1}{6} f^{(3)}(s) \\\\\n\nf(-1) &=f(0)-f^{\\prime}(0)+\\frac{1}{2} f^{\\prime \\prime}(0)-\\frac{1}{6} f^{(3)}(t)\n\n\\end{aligned}\n\n$$\n\nfor some $s \\in(0,1), t \\in(-1,0)$. By subtracting the second equation from the first and using the given values of $f(1), f(-1)$, and $f^{\\prime}(0)$, we obtain\n\n$$\n\n1=\\frac{1}{6}\\left(f^{(3)}(s)+f^{(3)}(t)\\right),\n\n$$\n\nwhich is the desired result. Note that we made no use of the hypothesis $f(0)=0$.\n\n\\end{proof}"}
{"id": "Munkres|exercise_13_3b", "formal_statement": "theorem exercise_13_3b : \u00ac \u2200 X : Type, \u2200s : set (set X),\n  (\u2200 t : set X, t \u2208 s \u2192 (set.infinite t\u1d9c \u2228 t = \u2205 \u2228 t = \u22a4)) \u2192 \n  (set.infinite (\u22c3\u2080 s)\u1d9c \u2228 (\u22c3\u2080 s) = \u2205 \u2228 (\u22c3\u2080 s) = \u22a4) :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Show that the collection $$\\mathcal{T}_\\infty = \\{U | X - U \\text{ is infinite or empty or all of X}\\}$$ does not need to be a topology on the set $X$.\n", "nl_proof": "\\begin{proof}\n\n    Let $X=\\mathbb{R}, U_1=(-\\infty, 0)$ and $U_2=(0, \\infty)$. Then $U_1$ and $U_2$ are in $\\mathcal{T}_{\\infty}$ but $U_1 \\cup U_2=\\mathbb{R} \\backslash\\{0\\}$ is not.\n\n\\end{proof}"}
{"id": "Munkres|exercise_13_4a2", "formal_statement": "theorem exercise_13_4a2 :\n  \u2203 (X I : Type*) (T : I \u2192 set (set X)),\n  (\u2200 i, is_topology X (T i)) \u2227 \u00ac  is_topology X (\u22c2 i : I, T i) :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "If $\\mathcal{T}_\\alpha$ is a family of topologies on $X$, show that $\\bigcup \\mathcal{T}_\\alpha$ does not need to be a topology on $X$.\n", "nl_proof": "\\begin{proof}\n\n    On the other hand, the union $\\bigcup_\\alpha \\mathcal{T}_\\alpha$ is in general not a topology on $X$. For instance, let $X=\\{a, b, c\\}$. Then $\\mathcal{T}_1=\\{\\emptyset, X,\\{a\\}\\}$ and $\\mathcal{T}_2=\\{\\emptyset, X,\\{b\\}\\}$ are topologies on $X$ but $\\mathcal{T}_1 \\cup \\mathcal{T}_2=$ $\\{\\emptyset, X,\\{a\\},\\{b\\}\\}$ is not.\n\n\\end{proof}"}
{"id": "Munkres|exercise_13_4b2", "formal_statement": "theorem exercise_13_4b2 (X I : Type*) (T : I \u2192 set (set X)) (h : \u2200 i, is_topology X (T i)) :\n  \u2203! T', is_topology X T' \u2227 (\u2200 i, T' \u2286 T i) \u2227\n  \u2200 T'', is_topology X T'' \u2192 (\u2200 i, T'' \u2286 T i) \u2192 T' \u2286 T'' :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Let $\\mathcal{T}_\\alpha$ be a family of topologies on $X$. Show that there is a unique largest topology on $X$ contained in all the collections $\\mathcal{T}_\\alpha$.\n", "nl_proof": "\\begin{proof}\n\n    Now we prove that there exists a unique largest topology contained in all $\\mathcal{T}_\\alpha$. Uniqueness of such topology is clear. Consider $\\mathcal{T}=\\bigcap_\\alpha \\mathcal{T}_\\alpha$. We already know that $\\mathcal{T}$ is a topology by, and clearly $\\mathcal{T} \\subset \\mathcal{T}_\\alpha$ for all $\\alpha$. If $\\mathcal{O}$ is another topology contained in all $\\mathcal{T}_\\alpha$, it must be contained in their intersection, so $\\mathcal{O} \\subset \\mathcal{T}$. I follows that $\\mathcal{T}$ is the unique largest topology contained in all $\\mathcal{T}_\\alpha$.\n\n\\end{proof}"}
{"id": "Munkres|exercise_13_5b", "formal_statement": "theorem exercise_13_5b {X : Type*}\n  [t : topological_space X] (A : set (set X)) (hA : t = generate_from A) :\n  generate_from A = generate_from (sInter {T | is_topology X T \u2227 A \u2286 T}) :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Show that if $\\mathcal{A}$ is a subbasis for a topology on $X$, then the topology generated by $\\mathcal{A}$ equals the intersection of all topologies on $X$ that contain $\\mathcal{A}$.\n", "nl_proof": "\\begin{proof}\n\n    If we now considered $\\mathcal{A}$ as a subbasis, then the elements of $\\mathcal{T}$ are union of finite intersections of elements of $\\mathcal{A}$. The inclusion $\\mathcal{O} \\subset \\mathcal{T}$ is again clear and $\\mathcal{T} \\subset \\mathcal{O}$ holds since every union of finite intersections of elements of $\\mathcal{A}$ belongs to $\\mathcal{O}$.\n\n\\end{proof}"}
{"id": "Munkres|exercise_13_8a", "formal_statement": "theorem exercise_13_8a :\n  topological_space.is_topological_basis {S : set \u211d | \u2203 a b : \u211a, a < b \u2227 S = Ioo a b} :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Show that the collection $\\{(a,b) \\mid a < b, a \\text{ and } b \\text{ rational}\\}$ is a basis that generates the standard topology on $\\mathbb{R}$.\n", "nl_proof": "\\begin{proof}\n\n    Exercise 13.8. (a) First note that $\\mathcal{B}$ is a basis for a topology on $\\mathbb{R}$. This follows from the fact that the union of its elements is all of $\\mathbb{R}$ and the intersection of two elements of $\\mathcal{B}$ is either empty or another element of $\\mathcal{B}$. Let $\\mathcal{T}$ be the standard topology on $\\mathbb{R}$. Clearly the topology generated by $\\mathcal{B}$ is coarser than $\\mathcal{T}$. Let $U \\in \\mathcal{T}$ and $x \\in U$. Then $U$ contains an open interval with centre $x$. Since the rationals are dense in $\\mathbb{R}$ with the standard topology, there exists $q \\in \\mathbb{Q}$ such that $x \\in(x-q, x+q) \\subset U$. This proves that $\\mathcal{T}$ is coarser than the topology generated by $\\mathcal{B}$. We conclude that $\\mathcal{B}$ generates the standard topology on $\\mathbb{R}$.\n\n\\end{proof}"}
{"id": "Munkres|exercise_16_1", "formal_statement": "theorem exercise_16_1 {X : Type*} [topological_space X]\n  (Y : set X)\n  (A : set Y) :\n  \u2200 U : set A, is_open U \u2194 is_open (subtype.val '' U) :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Show that if $Y$ is a subspace of $X$, and $A$ is a subset of $Y$, then the topology $A$ inherits as a subspace of $Y$ is the same as the topology it inherits as a subspace of $X$.\n", "nl_proof": "\\begin{proof}\n\n    Exercise 16.1. Let $\\mathcal{T}$ be the topology $A$ inherits as a subspace of $Y$, and $\\mathcal{O}$ be the topology it inherits as a subspace of $X$. A (standard) basis element for $\\mathcal{T}$ has the form $U \\cap A$ where $U$ is open in $Y$, so is of the form $(Y \\cap V) \\cap A=V \\cap A$ where $V$ is open in $X$. Therefore every basis element for $\\mathcal{T}$ is also a basis element for $\\mathcal{O}$. Conversely, a (standard) basis element for $\\mathcal{O}$ have the form $W \\cap A=W \\cap Y \\cap A$ where $W$ is open in $X$. Since $W \\cap Y$ is open in $Y$, this is a basis element for $\\mathcal{T}$, so every basis element for $\\mathcal{O}$ is a basis element for $\\mathcal{T}$. It follows that $\\mathcal{T}=\\mathcal{O}$.\n\n\\end{proof}"}
{"id": "Munkres|exercise_16_6", "formal_statement": "theorem exercise_16_6\n  (S : set (set (\u211d \u00d7 \u211d)))\n  (hS : \u2200 s, s \u2208 S \u2192 \u2203 a b c d, (rational a \u2227 rational b \u2227 rational c \u2227 rational d\n  \u2227 s = {x | \u2203 x\u2081 x\u2082, x = (x\u2081, x\u2082) \u2227 a < x\u2081 \u2227 x\u2081 < b \u2227 c < x\u2082 \u2227 x\u2082 < d})) :\n  is_topological_basis S :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Show that the countable collection \\[\\{(a, b) \\times (c, d) \\mid a < b \\text{ and } c < d, \\text{ and } a, b, c, d \\text{ are rational}\\}\\] is a basis for $\\mathbb{R}^2$.\n", "nl_proof": "\\begin{proof}\n\n    We know that $\\mathcal{B}=\\{(a,b)|a<b, a \\text{ and } b \\text rational\\}$ is a basis for $\\mathcal{R}$, therefore the set we are concerned with in the above question is a basis for $\\mathcal{R}^2$. \n\n\\end{proof}"}
{"id": "Munkres|exercise_18_8a", "formal_statement": "theorem exercise_18_8a {X Y : Type*} [topological_space X] [topological_space Y]\n  [linear_order Y] [order_topology Y] {f g : X \u2192 Y}\n  (hf : continuous f) (hg : continuous g) :\n  is_closed {x | f x \u2264 g x} :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Let $Y$ be an ordered set in the order topology. Let $f, g: X \\rightarrow Y$ be continuous. Show that the set $\\{x \\mid f(x) \\leq g(x)\\}$ is closed in $X$.\n", "nl_proof": "\\begin{proof}\n\n    We prove that $U=\\{x \\mid g(x)<f(x)\\}$ is open in $X$. Let $a \\in U$, so that $g(a)<f(a)$. If there is an element $c$ between $g(a)$ and $f(a)$, then $a \\in g^{-1}((-\\infty, c)) \\cap f^{-1}((c,+\\infty))$. If there are no elements between $g(a)$ and $f(a)$, then $a \\in g^{-1}\\left((-\\infty, f(a)) \\cap f^{-1}((g(a),+\\infty))\\right.$. Note that all these preimages are open since $f$ and $g$ are continuous. Thus $U=V \\cup W$, where\n\n$$\n\nV=\\bigcup_{c \\in X} g^{-1}((-\\infty, c)) \\cap f^{-1}((c,+\\infty)) \\quad \\text { and } \\quad W=\\bigcup_{\\substack{g(a)<f(a) \\\\(g(a), f(a))=\\emptyset}} g^{-1}\\left((-\\infty, f(a)) \\cap f^{-1}((g(a),+\\infty))\\right.\n\n$$\n\nare open in $X$. So $U$ is open in $X$ and therefore $X \\backslash U=\\{x \\mid f(x) \\leq g(x)\\}$ is closed in $X$.\n\n\\end{proof}"}
{"id": "Munkres|exercise_18_13", "formal_statement": "theorem exercise_18_13\n  {X : Type*} [topological_space X] {Y : Type*} [topological_space Y]\n  [t2_space Y] {A : set X} {f : A \u2192 Y} (hf : continuous f)\n  (g : closure A \u2192 Y)\n  (g_con : continuous g) :\n  \u2200 (g' : closure A \u2192 Y), continuous g' \u2192  (\u2200 (x : closure A), g x = g' x) :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Let $A \\subset X$; let $f: A \\rightarrow Y$ be continuous; let $Y$ be Hausdorff. Show that if $f$ may be extended to a continuous function $g: \\bar{A} \\rightarrow Y$, then $g$ is uniquely determined by $f$.\n", "nl_proof": "\\begin{proof}\n\n    Let $h, g: \\bar{A} \\rightarrow Y$ be continuous extensions of $f$. Suppose that there is a point $x \\in \\bar{A}$ such that $h(x) \\neq g(x)$. Since $h=g$ on $A$, we must have $x \\in A^{\\prime}$. Since $Y$ is Hausdorff, there is a neighbourhood $U$ of $h(x)$ and a neighbourhood $V$ of $g(x)$ such that $U \\cap V=\\emptyset$. Since $h$ and $g$ are continuous, $h^{-1}(U) \\cap g^{-1}(V)$ is a neighbourhood of $x$. Since $x \\in A^{\\prime}$, there is a point $y \\in h^{-1}(U) \\cap g^{-1}(V) \\cap A$ different from $x$. But $h=g$ on $A$, so $g^{-1}(V) \\cap A=h^{-1}(V) \\cap A$ and hence $y \\in h^{-1}(U) \\cap h^{-1}(V)=h^{-1}(U \\cap V)=\\emptyset$, a contradiction. It follows that $h=g$ on $\\bar{A}$.\n\n\\end{proof}"}
{"id": "Munkres|exercise_20_2", "formal_statement": "theorem exercise_20_2\n  [topological_space (\u211d \u00d7\u2097 \u211d)] [order_topology (\u211d \u00d7\u2097 \u211d)]\n  : metrizable_space (\u211d \u00d7\u2097 \u211d) :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Show that $\\mathbb{R} \\times \\mathbb{R}$ in the dictionary order topology is metrizable.\n", "nl_proof": "\\begin{proof}\n\n    The dictionary order topology on $\\mathbb{R} \\times \\mathbb{R}$ is the same as the product topology $\\mathbb{R}_d \\times \\mathbb{R}$, where $\\mathbb{R}_d$ denotes $\\mathbb{R}$ with the discrete topology. We know that $\\mathbb{R}_d$ and $\\mathbb{R}$ are metrisable. Thus, it suffices to show that the product of two metrisable spaces is metrisable.\n\nSo let $X$ and $Y$ be metrisable spaces, with metrics $d$ and $d^{\\prime}$ respectively. On $X \\times Y$, define\n\n$$\n\n\\rho(x \\times y, w \\times z)=\\max \\left\\{d(x, w), d^{\\prime}(y, z)\\right\\} .\n\n$$\n\nThen $\\rho$ is a metric on $X \\times Y$; it remains to prove that it induces the product topology on $X \\times Y$. If $B_d\\left(x, r_1\\right) \\times B_d\\left(y, r_2\\right)$ is a basis element for the product space $X \\times Y$, and $r=\\min \\left\\{r_1, r_2\\right\\}$, then $x \\times y \\in B_\\rho(x \\times y, r) \\subset B_d\\left(x, r_1\\right) \\times B_d\\left(y, r_2\\right)$, so the product topology is coarser than the $\\rho$-topology. Conversely, if $B_\\rho(x \\times y, \\delta)$ is a basis element for the $\\rho$-topology, then $x \\times y \\in B_d(x, \\delta) \\times B_{d^{\\prime}}(y, \\delta) \\subset$ $B_\\rho(x \\times y, \\delta)$, so the product topology is finer than the $\\rho$-topology. It follows that both topologies are equal, so the product space $X \\times Y$ is metrisable.\n\n\\end{proof}"}
{"id": "Munkres|exercise_21_6b", "formal_statement": "theorem exercise_21_6b\n  (f : \u2115 \u2192 I \u2192 \u211d )\n  (h : \u2200 x n, f n x = x ^ n) :\n  \u00ac \u2203 f\u2080, tendsto_uniformly f f\u2080 at_top :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Define $f_{n}:[0,1] \\rightarrow \\mathbb{R}$ by the equation $f_{n}(x)=x^{n}$. Show that the sequence $\\left(f_{n}\\right)$ does not converge uniformly.\n", "nl_proof": "\\begin{proof}\n\n    The sequence $\\left(f_n\\right)_n$ does not converge uniformly, since given $0<\\varepsilon<1$ and $N \\in \\mathbb{Z}_{+}$, for $x=\\varepsilon^{1 / N}$ we have $d\\left(f_N(x), f(x)\\right)=\\varepsilon$. We can also apply Theorem 21.6: the convergence is not uniform since $f$ is not continuous.\n\n\\end{proof}"}
{"id": "Munkres|exercise_22_2a", "formal_statement": "theorem exercise_22_2a {X Y : Type*} [topological_space X]\n  [topological_space Y] (p : X \u2192 Y) (h : continuous p) :\n  quotient_map p \u2194 \u2203 (f : Y \u2192 X), continuous f \u2227 p \u2218 f = id :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Let $p: X \\rightarrow Y$ be a continuous map. Show that if there is a continuous map $f: Y \\rightarrow X$ such that $p \\circ f$ equals the identity map of $Y$, then $p$ is a quotient map.\n", "nl_proof": "\\begin{proof}\n\nLet $1_Y: Y \\rightarrow Y$ be the identity map in $Y$. If $U$ is a subset of $Y$ and $p^{-1}(U)$ is open in $X$, then $f^{-1}\\left(p^{-1}(U)\\right)=1_Y^{-1}(U)=U$ is open in $Y$ by continuity of $f$. Thus $p$ is a quotient map.\n\n\\end{proof}"}
{"id": "Munkres|exercise_22_5", "formal_statement": "theorem exercise_22_5 {X Y : Type*} [topological_space X]\n  [topological_space Y] (p : X \u2192 Y) (hp : is_open_map p)\n  (A : set X) (hA : is_open A) : is_open_map (p \u2218 subtype.val : A \u2192 Y) :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Let $p \\colon X \\rightarrow Y$ be an open map. Show that if $A$ is open in $X$, then the map $q \\colon A \\rightarrow p(A)$ obtained by restricting $p$ is an open map.\n", "nl_proof": "\\begin{proof}\n\nLet $U$ be open in $A$. Since $A$ is open in $X, U$ is open in $X$ as well, so $p(U)$ is open in $Y$. Since $q(U)=p(U)=p(U) \\cap p(A)$, the set $q(U)$ is open in $p(A)$. Thus $q$ is an open map.\n\n\\end{proof}"}
{"id": "Munkres|exercise_23_3", "formal_statement": "theorem exercise_23_3 {X : Type*} [topological_space X]\n  [topological_space X] {A : \u2115 \u2192 set X}\n  (hAn : \u2200 n, is_connected (A n))\n  (A\u2080 : set X)\n  (hA : is_connected A\u2080)\n  (h : \u2200 n, A\u2080 \u2229 A n \u2260 \u2205) :\n  is_connected (A\u2080 \u222a (\u22c3 n, A n)) :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Let $\\left\\{A_{\\alpha}\\right\\}$ be a collection of connected subspaces of $X$; let $A$ be a connected subset of $X$. Show that if $A \\cap A_{\\alpha} \\neq \\varnothing$ for all $\\alpha$, then $A \\cup\\left(\\bigcup A_{\\alpha}\\right)$ is connected.\n", "nl_proof": "\\begin{proof}\n\n    For each $\\alpha$ we have $A \\cap A_\\alpha \\neq \\emptyset$, so each $A \\cup A_\\alpha$ is connected by Theorem 23.3. In turn $\\left\\{A \\cup A_\\alpha\\right\\}_\\alpha$ is a collection of connected spaces that have a point in common (namely any point in $A)$, so $\\bigcup_\\alpha\\left(A \\cup A_\\alpha\\right)=A \\cup\\left(\\bigcup_\\alpha A_\\alpha\\right)$ is connected. \n\n\\end{proof}"}
{"id": "Munkres|exercise_23_6", "formal_statement": "theorem exercise_23_6 {X : Type*}\n  [topological_space X] {A C : set X} (hc : is_connected C)\n  (hCA : C \u2229 A \u2260 \u2205) (hCXA : C \u2229 A\u1d9c \u2260 \u2205) :\n  C \u2229 (frontier A) \u2260 \u2205 :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Let $A \\subset X$. Show that if $C$ is a connected subspace of $X$ that intersects both $A$ and $X-A$, then $C$ intersects $\\operatorname{Bd} A$.\n", "nl_proof": "\\begin{proof}\n\n    Suppose that $C \\cap B d A=C \\cap \\bar{A} \\cap \\overline{X-A}=\\emptyset$. Then $C \\cap A$ and $C \\cap(X \\backslash A)$ are a pair of disjoint non-empty sets whose union is all of $C$, neither of which contains a limit point of the other. Indeed, if $C \\cap(X-A)$ contains a limit point $x$ of $C \\cap A$, then $x \\in C \\cap(X-A) \\cap A^{\\prime} \\subset C \\cap \\bar{A} \\cap \\overline{X-A}=\\emptyset$, a contradiction, and similarly $C \\cap A$ does not contain a limit point of $C \\cap(X-A)$. Then $C \\cap A$ and $C \\cap(X-A)$ constitute a separation of $C$, contradicting the fact that $C$ is connected (Lemma 23.1).\n\n\\end{proof}"}
{"id": "Munkres|exercise_23_11", "formal_statement": "theorem exercise_23_11 {X Y : Type*} [topological_space X] [topological_space Y]\n  (p : X \u2192 Y) (hq : quotient_map p)\n  (hY : connected_space Y) (hX : \u2200 y : Y, is_connected (p \u207b\u00b9' {y})) :\n  connected_space X :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Let $p: X \\rightarrow Y$ be a quotient map. Show that if each set $p^{-1}(\\{y\\})$ is connected, and if $Y$ is connected, then $X$ is connected.\n", "nl_proof": "\\begin{proof}\n\n    Suppose that $U$ and $V$ constitute a separation of $X$. If $y \\in p(U)$, then $y=p(x)$ for some $x \\in U$, so that $x \\in p^{-1}(\\{y\\})$. Since $p^{-1}(\\{y\\})$ is connected and $x \\in U \\cap p^{-1}(\\{y\\})$, we have $p^{-1}(\\{y\\}) \\subset U$. Thus $p^{-1}(\\{y\\}) \\subset U$ for all $y \\in p(U)$, so that $p^{-1}(p(U)) \\subset U$. The inclusion $U \\subset p^{-1}(p(U))$ if true for any subset and function, so we have the equality $U=p^{-1}(p(U))$ and therefore $U$ is saturated. Similarly, $V$ is saturated. Since $p$ is a quotient map, $p(U)$ and $p(V)$ are disjoint non-empty open sets in $Y$. But $p(U) \\cup p(V)=Y$ as $p$ is surjective, so $p(U)$ and $p(V)$ constitute a separation of $Y$, contradicting the fact that $Y$ is connected. We conclude that $X$ is connected.\n\n\\end{proof}"}
{"id": "Munkres|exercise_24_3a", "formal_statement": "theorem exercise_24_3a [topological_space I]\n  (f : I \u2192 I) (hf : continuous f) :\n  \u2203 (x : I), f x = x :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Let $f \\colon X \\rightarrow X$ be continuous. Show that if $X = [0, 1]$, there is a point $x$ such that $f(x) = x$. (The point $x$ is called a fixed point of $f$.)\n", "nl_proof": "\\begin{proof}\n\n    If $f(0)=0$ or $f(1)=1$ we are done, so suppose $f(0)>0$ and $f(1)<1$. Let $g:[0,1] \\rightarrow[0,1]$ be given by $g(x)=f(x)-x$. Then $g$ is continuous, $g(0)>0$ and $g(1)<0$. Since $[0,1]$ is connected and $g(1)<0<g(0)$, by the intermediate value theorem there exists $x \\in(0,1)$ such that $g(x)=0$, that is, such that $f(x)=x$.\n\n\\end{proof}"}
{"id": "Munkres|exercise_25_9", "formal_statement": "theorem exercise_25_9 {G : Type*} [topological_space G] [group G]\n  [topological_group G] (C : set G) (h : C = connected_component 1) :\n  is_normal_subgroup C :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Let $G$ be a topological group; let $C$ be the component of $G$ containing the identity element $e$. Show that $C$ is a normal subgroup of $G$.\n", "nl_proof": "\\begin{proof}\n\n    Given $x \\in G$, the maps $y \\mapsto x y$ and $y \\mapsto y x$ are homeomorphisms of $G$ onto itself. Since $C$ is a component, $x C$ and $C x$ are both components that contain $x$, so they are equal. Hence $x C=C x$ for all $x \\in G$, so $C$ is a normal subgroup of $G$.\n\n\\end{proof}"}
{"id": "Munkres|exercise_26_12", "formal_statement": "theorem exercise_26_12 {X Y : Type*} [topological_space X] [topological_space Y]\n  (p : X \u2192 Y) (h : function.surjective p) (hc : continuous p) (hp : \u2200 y, is_compact (p \u207b\u00b9' {y}))\n  (hY : compact_space Y) : compact_space X :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Let $p: X \\rightarrow Y$ be a closed continuous surjective map such that $p^{-1}(\\{y\\})$ is compact, for each $y \\in Y$. (Such a map is called a perfect map.) Show that if $Y$ is compact, then $X$ is compact.\n", "nl_proof": "\\begin{proof}\n\n    We first show that if $U$ is an open set containing $p^{-1}(\\{y\\})$, then there is a neighbourhood $W$ of $y$ such that $p^{-1}(W)$ is contained in $U$. Since $X-U$ is closed in $X$, $p(X-U)$ is closed in $Y$ and does not contain $y$, so $W=Y \\backslash p(X \\backslash U)$ is a neighbourhood of $y$. Moreover, since $X \\backslash U \\subset p^{-1}(p(X \\backslash U)$ ) (by elementary set theory), we have\n\n$$\n\np^{-1}(W)=p^{-1}(Y \\backslash p(X \\backslash U))=p^{-1}(Y) \\backslash p^{-1}(p(X \\backslash U)) \\subset X \\backslash(X \\backslash U)=U .\n\n$$\n\nNow let $\\mathcal{A}$ be an open covering of $X$. For each $y \\in Y$, let $\\mathcal{A}_y$ be a subcollection of $\\mathcal{A}$ such that\n\n$$\n\np^{-1}(\\{y\\}) \\subset \\bigcup_{A \\in \\mathcal{A}_y} A .\n\n$$\n\nSince $p^{-1}(\\{y\\})$ is compact, there exists a finite subcollection of $\\mathcal{A}_y$ that also covers $p^{-1}(\\{y\\})$, say $\\left\\{A_y^1, \\ldots, A_y^{n_y}\\right\\}$. Thus $\\bigcup_{i=1}^{n_y} A_y^i$ is open and contains $p^{-1}(\\{y\\})$, so there exists a neighbourhood $W_y$ of $y$ such that $p^{-1}\\left(W_y\\right)$ is contained in $\\bigcup_{i=1}^{n_y} A_y^i$. Then $\\left\\{W_y\\right\\}_{y \\in Y}$ is an open covering of $Y$, so there exist $y_1, \\ldots, y_k \\in Y$ such that $\\left\\{W_{y_j}\\right\\}_{j=1}^k$ also covers $Y$. Then\n\n$$\n\nX=p^{-1}(Y) \\subset p^{-1}\\left(\\bigcup_{j=1}^k W_{y_j}\\right)=\\bigcup_{j=1}^k p^{-1}\\left(W_{y_j}\\right) \\subset \\bigcup_{j=1}^k\\left(\\bigcup_{i=1}^{n_{y_j}} A_{y_j}^i\\right)\n\n$$\n\nso\n\n$$\n\n\\left\\{A_{y_j}^i\\right\\}_{\\substack{j=1, \\ldots, k . \\\\ i=1, \\ldots, n_{y_j}}}\n\n$$\n\nis a finite subcollection of $\\mathcal{A}$ that also covers $X$. Therefore, $X$ is compact.\n\n\\end{proof}"}
{"id": "Munkres|exercise_28_4", "formal_statement": "theorem exercise_28_4 {X : Type*}\n  [topological_space X] (hT1 : t1_space X) :\n  countably_compact X \u2194 limit_point_compact X :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "A space $X$ is said to be countably compact if every countable open covering of $X$ contains a finite subcollection that covers $X$. Show that for a $T_1$ space $X$, countable compactness is equivalent to limit point compactness.\n", "nl_proof": "\\begin{proof}\n\n    First let $X$ be a countable compact space. Note that if $Y$ is a closed subset of $X$, then $Y$ is countable compact as well, for if $\\left\\{U_n\\right\\}_{n \\in \\mathbb{Z}_{+}}$is a countable open covering of $Y$, then $\\left\\{U_n\\right\\}_{n \\in \\mathbb{Z}_{+}} \\cup(X \\backslash Y)$ is a countable open covering of $X$; there is a finite subcovering of $X$, hence a finite subcovering of $Y$. Now let $A$ be an infinite subset. We show that $A$ has a limit point. Let $B$ be a countable infinite subset of $A$. Suppose that $B$ has no limit point, so that $B$ is closed in $X$. Then $B$ is countable compact. Since $B$ has no limit point, for each $b \\in B$ there is a neighbourhood $U_b$ of $b$ that intersects $B$ in the point $b$ alone. Then $\\left\\{U_b\\right\\}_{b \\in B}$ is an open covering of $B$ with no finite subcovering, contradicting the fact that $B$ is countable compact. Hence $B$ has a limit point, so that $A$ has a limit point as well. Since $A$ was arbitrary, we deduce that $X$ is limit point compact. (Note that the $T_1$ property is not necessary in this direction.)\n\n\n\nNow assume that $X$ is a limit point compact $T_1$ space. We show that $X$ is countable compact. Suppose, on the contrary, that $\\left\\{U_n\\right\\}_{n \\in \\mathbb{Z}_{+}}$is a countable open covering of $X$ with no finite subcovering. For each $n$, take a point $x_n$ in $X$ not in $U_1 \\cup \\cdots \\cup U_n$. By assumption, the infinite set $A=\\left\\{x_n \\mid n \\in \\mathbb{Z}_{+}\\right\\}$has a limit point $y \\in X$. Since $\\left\\{U_n\\right\\}_{n \\in \\mathbb{Z}_{+}}$covers $X$, there exists $N \\in \\mathbb{Z}_{+}$such that $y \\in U_1 \\cup \\cdots \\cup U_N$. Now $X$ is $T_1$, so for each $i=1, \\ldots, N$ there exists a neighbourhood $V_i$ of $y$ that does not contain $x_i$. Then\n\n$$\n\nV=\\left(V_1 \\cap \\cdots \\cap V_N\\right) \\cap\\left(U_1 \\cup \\cdots \\cup U_N\\right)\n\n$$\n\nis a neighbourhood of $y$ that does not contain any of the points $x_i$, contradicting the fact that $y$ is a limit point of $A$. It follows that every countable open covering of $X$ must have a finite subcovering, so $X$ is countable compact.\n\n\\end{proof}"}
{"id": "Munkres|exercise_28_6", "formal_statement": "theorem exercise_28_6 {X : Type*} [metric_space X]\n  [compact_space X] {f : X \u2192 X} (hf : isometry f) :\n  function.bijective f :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Let $(X, d)$ be a metric space. If $f: X \\rightarrow X$ satisfies the condition $d(f(x), f(y))=d(x, y)$ for all $x, y \\in X$, then $f$ is called an isometry of $X$. Show that if $f$ is an isometry and $X$ is compact, then $f$ is bijective and hence a homeomorphism.\n", "nl_proof": "\\begin{proof}\n\n    Note that $f$ is an imbedding. It remains to prove that $f$ is surjective. Suppose it is not, and let $a \\in f(X)$. Since $X$ is compact, $f(X)$ is compact and hence closed (every metric space is Hausdorff). Thus, there exists $\\varepsilon>0$ such that the $\\varepsilon^{-}$ neighbourhood of $a$ is contained in $X \\backslash f(X)$. Set $x_1=a$, and inductively $x_{n+1}=f\\left(x_n\\right)$ for $n \\in \\mathbb{Z}_{+}$. We show that $d\\left(x_n, x_m\\right) \\geq \\varepsilon$ for $n \\neq m$. Indeed, we may assume $n<m$. If $n \\geq 1$, then $d\\left(x_n, x_m\\right)=$ $d\\left(f^{-1}\\left(x_n\\right), f^{-1}\\left(x_m\\right)\\right)=d\\left(x_{n-1}, x_{m-1}\\right)$. By induction it follows that $d\\left(x_n, x_m\\right)=d\\left(x_{n-i}, x_{m-i}\\right)$ for all $i \\geq 1$, and hence $d\\left(x_n, x_m\\right)=d\\left(a, x_{m-n}\\right)=d\\left(a, f\\left(x_{m-n-1}\\right)\\right)$. Since $f\\left(x_{m-n-1}\\right) \\in f(X)$ and $B(a, \\varepsilon) \\cap f(X)=\\emptyset$, we have $d\\left(x_n, x_m\\right) \\geq \\varepsilon$, as claimed. Thus $\\left\\{x_n\\right\\}_{n \\in \\mathbb{Z}_{+}}$is a sequence with no convergent subsequence, so $X$ is not sequentially compact. This contradicts the fact that $X$ is compact. Therefore $f$ is surjective and hence a homeomorphism.\n\n\\end{proof}"}
{"id": "Munkres|exercise_29_4", "formal_statement": "theorem exercise_29_4 [topological_space (\u2115 \u2192 I)] :\n  \u00ac locally_compact_space (\u2115 \u2192 I) :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Show that $[0, 1]^\\omega$ is not locally compact in the uniform topology.\n", "nl_proof": "\\begin{proof}\n\n    Consider $\\mathbf{0} \\in[0,1]^\\omega$ and suppose that $[0,1]^\\omega$ is locally compact at $\\mathbf{0}$. Then there exists a compact $C$ containing an open ball $B=B_\\rho(\\mathbf{0}, \\varepsilon) \\subset[0,1]^\\omega$. Note that $\\bar{B}=[0, \\varepsilon]^\\omega$. Then $[0, \\varepsilon]^\\omega$ is closed and contained in the compact $C$, so it is compact. But $[0, \\varepsilon]^\\omega$ is homeomorphic to $[0,1]^\\omega$, which is not compact by Exercise 28.1. This contradiction proves that $[0,1]^\\omega$ is not locally compact in the uniform topology.\n\n\\end{proof}"}
{"id": "Munkres|exercise_30_10", "formal_statement": "theorem exercise_30_10\n  {X : \u2115 \u2192 Type*} [\u2200 i, topological_space (X i)]\n  (h : \u2200 i, \u2203 (s : set (X i)), countable s \u2227 dense s) :\n  \u2203 (s : set (\u03a0 i, X i)), countable s \u2227 dense s :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Show that if $X$ is a countable product of spaces having countable dense subsets, then $X$ has a countable dense subset.\n", "nl_proof": "\\begin{proof}\n\n    Let $\\left(X_n\\right)$ be spaces having countable dense subsets $\\left(A_n\\right)$. For each $n$, fix an arbitrary $x_n \\in X_n$. Consider the subset $A$ of $X$ defined by\n\n$$\n\nA=\\bigcup\\left\\{\\prod U_n: U_n=A_n \\text { for finitely many } n \\text { and is }\\left\\{x_n\\right\\} \\text { otherwise }\\right\\} .\n\n$$\n\nThis set is countable because the set of finite subsets of $\\mathbb{N}$ is countable and each of the inner sets is countable. Now, let $x \\in X$ and $V=\\prod V_n$ be a basis element containing $x$ such that each $V_n$ is open in $X_n$ and $V_n=X_n$ for all but finitely many $n$. For each $n$, if $V_n \\neq X_n$, choose a $y_n \\in\\left(A_n \\cap V_n\\right)$ (such a $y_n$ exists since $A_n$ is dense in $\\left.X_n\\right)$. Otherwise, let $y_n=x_n$. Then $\\left(y_n\\right) \\in(A \\cap V)$, proving that $A$ is dense in $X$.\n\n\\end{proof}"}
{"id": "Munkres|exercise_31_1", "formal_statement": "theorem exercise_31_1 {X : Type*} [topological_space X]\n  (hX : regular_space X) (x y : X) :\n  \u2203 (U V : set X), is_open U \u2227 is_open V \u2227 x \u2208 U \u2227 y \u2208 V \u2227 closure U \u2229 closure V = \u2205 :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Show that if $X$ is regular, every pair of points of $X$ have neighborhoods whose closures are disjoint.\n", "nl_proof": "\\begin{proof}\n\n    Let $x, y \\in X$ be two points such that $x \\neq y$. Since $X$ is regular (and thus Hausdorff), there exist disjoint open sets $U, V \\subseteq X$ such that $x \\in U$ and $y \\in V$.\n\nNote that $y \\notin \\bar{U}$. Otherwise $V$ must intersect $U$ in a point different from $y$ since $V$ is an open neighborhood of $y$, which is a contradiction since $U$ and $V$ are disjoint.\n\nSince $X$ is regular and $\\bar{U}$ is closed, there exist disjoint open sets $U^{\\prime}, V^{\\prime} \\subseteq X$ such that $\\bar{U} \\subseteq U^{\\prime}$ and $y \\in V^{\\prime}$.\n\n\n\nAnd now $U$ and $V^{\\prime}$ are neighborhoods of $x$ and $y$ whose closures are disjoint. If $\\bar{U} \\cap \\overline{V^{\\prime}} \\neq \\emptyset$, then it follows that $U^{\\prime} \\supseteq U$ intersects $\\overline{V^{\\prime}}$. Since $U^{\\prime}$ is open, it follows that $U^{\\prime}$ intersects $V^{\\prime}$, which is a contradiction.\n\n\\end{proof}"}
{"id": "Munkres|exercise_31_3", "formal_statement": "theorem exercise_31_3 {\u03b1 : Type*} [partial_order \u03b1]\n  [topological_space \u03b1] (h : order_topology \u03b1) : regular_space \u03b1 :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Show that every order topology is regular.\n", "nl_proof": "\\begin{proof}\n\n    Let $X$ be an ordered set.\n\nFirst we show that $X$ is a $T_1$-space. For $x \\in X$ we have that\n\n$$\n\nX \\backslash\\{x\\}=\\langle-\\infty, x\\rangle \\cup\\langle x,+\\infty\\rangle\n\n$$\n\nwhich is an open set as an union of two open intervals. Therefore, the set $\\{x\\}$ is closed.\n\nStep 2\n\n2 of 3\n\nNow to prove that $X$ is regular we use Lemma $\\mathbf{3 1 . 1 .}$\n\nLet $x \\in X$ be any point and $U \\subseteq X$ any open neighborhood of $x$. Then there exist $a, b \\in X$ such that $x \\in\\langle a, b\\rangle \\subseteq U$. Now we have four possibilities.\n\n1. If there exist $x_1, x_2 \\in U$ such that $a<x_1<x<x_2<b$, then\n\n$$\n\nx \\in\\left\\langle x_1, x_2\\right\\rangle \\subseteq \\overline{\\left\\langle x_1, x_2\\right\\rangle} \\subseteq\\left[x_1, x_2\\right] \\subseteq\\langle a, b\\rangle \\subseteq U\n\n$$\n\n2. If there exists $x_1 \\in U$ such that $a<x_1<x$, but there's no $x_2 \\in U$ such that $x<x_2<b$, then\n\n$$\n\nx \\in\\left\\langle x_1, b\\right\\rangle=\\left(x_1, x\\right] \\subseteq \\overline{\\left(x_1, x\\right]} \\subseteq\\left[x_1, x\\right] \\subseteq\\langle a, b\\rangle \\subseteq U\n\n$$\n\n3. If there exists $x_2 \\in U$ such that $x<x_2<b$, but there's no $x_1 \\in U$ such that $a<x_1<x$, then\n\n$$\n\nx \\in\\left\\langle a, x_2\\right\\rangle=\\left[x, x_2\\right) \\subseteq \\overline{\\left[x, x_2\\right)} \\subseteq\\left[x, x_2\\right] \\subseteq\\langle a, b\\rangle \\subseteq U\n\n$$\n\n4. If there's no $x_1 \\in U$ such that $a<x_1<x$ and no $x_2 \\in U$ such that $x<x_2<b$, then\n\n$$\n\nx \\in\\langle a, b\\rangle=\\{x\\}=\\overline{\\{x\\}}=\\{x\\} \\subseteq U\n\n$$\n\nWe have that $\\overline{\\{x\\}}=\\{x\\}$ because $X$ is a $T_1$-space.\n\nIn all four cases we proved that there exists an open interval $V$ such that $x \\in V \\subseteq \\bar{V} \\subseteq U$, so $X$ is regular.\n\n\\end{proof}"}
{"id": "Munkres|exercise_32_2a", "formal_statement": "theorem exercise_32_2a\n  {\u03b9 : Type*} {X : \u03b9 \u2192 Type*} [\u2200 i, topological_space (X i)]\n  (h : \u2200 i, nonempty (X i)) (h2 : t2_space (\u03a0 i, X i)) :\n  \u2200 i, t2_space (X i) :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Show that if $\\prod X_\\alpha$ is Hausdorff, then so is $X_\\alpha$. Assume that each $X_\\alpha$ is nonempty.\n", "nl_proof": "\\begin{proof}\n\n    Suppose that $X=\\prod_\\beta X_\\beta$ is Hausdorff and let $\\alpha$ be any index.\n\nLet $x, y \\in X_\\alpha$ be any points such that $x \\neq y$. Since all $X_\\beta$ are nonempty, there exist points $\\mathbf{x}, \\mathbf{y} \\in X$ such that $x_\\beta=y_\\beta$ for every $\\beta \\neq \\alpha$ and $x_\\alpha=x, y_\\alpha=y$.\n\nSince $x \\neq y$, it follows that $\\mathbf{x} \\neq \\mathbf{y}$. Since $X$ is Hausdorff, there exist open disjoint sets $U, V \\subseteq X$ such that $\\mathbf{x} \\in U$ and $\\mathbf{y} \\in V$.\n\nFor $\\beta \\neq \\alpha$ we have that $x_\\beta=y_\\beta \\in \\pi_\\beta(U) \\cap \\pi_\\beta(V)$, hence $\\pi_\\beta(U)$ and $\\pi_\\beta(V)$ are not disjoint.\n\nSince $U$ and $V$ are disjoint, it follows that $\\pi_\\alpha(U) \\cap \\pi_\\alpha(V)=\\emptyset$.\n\nWe also have that $x \\in \\pi_\\alpha(U)$ and $y \\in \\pi_\\alpha(V)$ and since the projections are open maps, it follows that the sets $\\pi_\\alpha(U)$ and $\\pi_\\alpha(V)$ are open.\n\nThis proves that $x$ and $y$ can be separated by open sets, so $X_\\alpha$ is Hausdorff.\n\n\\end{proof}"}
{"id": "Munkres|exercise_32_2c", "formal_statement": "theorem exercise_32_2c\n  {\u03b9 : Type*} {X : \u03b9 \u2192 Type*} [\u2200 i, topological_space (X i)]\n  (h : \u2200 i, nonempty (X i)) (h2 : normal_space (\u03a0 i, X i)) :\n  \u2200 i, normal_space (X i) :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Show that if $\\prod X_\\alpha$ is normal, then so is $X_\\alpha$. Assume that each $X_\\alpha$ is nonempty.\n", "nl_proof": "\\begin{proof}\n\n    Suppose that $X=\\prod_\\beta X_\\beta$ is normal and let $\\alpha$ be any index.\n\nSince $X$ is normal, it follows that $X$ is Hausdorff (or regular), which then implies that $X_\\alpha$ is Hausdorff (or regular). This imples that $X_\\alpha$ satisfies the $T_1$ axiom.\n\nNow the proof that $X_\\alpha$ satisfies the $T_4$ axiom is the same as for regular spaces.\n\nIf $F, G \\subseteq X_\\alpha$ are disjoint closed sets, then $\\prod_\\beta F_\\beta$ and $\\prod_\\beta G_\\beta$, where $F_\\alpha=F, G_\\alpha=G$ and $F_\\beta=G_\\beta=X_\\beta$ for $\\beta \\neq \\alpha$, are disjoint closed sets in $X$.\n\nSince $X$ is normal (and therefore satisfies the $T_4$ axiom), there exist disjoint open sets $U, V \\subseteq X$ such that $\\prod_\\beta F_\\beta \\subseteq U$ and $\\prod_\\beta G_\\beta \\subseteq V$\n\nThen $\\pi_\\alpha(U)$ and $\\pi_\\alpha(V)$ are disjoint open sets in $X_\\alpha$ such that $F \\subseteq \\pi_\\alpha(U)$ and $G \\subseteq \\pi_\\alpha(V)$.\n\n\\end{proof}"}
{"id": "Munkres|exercise_33_7", "formal_statement": "theorem exercise_33_7 {X : Type*} [topological_space X]\n  (hX : locally_compact_space X) (hX' : t2_space X) :\n  \u2200 x A, is_closed A \u2227 \u00ac x \u2208 A \u2192\n  \u2203 (f : X \u2192 I), continuous f \u2227 f x = 1 \u2227 f '' A = {0}\n  :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Show that every locally compact Hausdorff space is completely regular.\n", "nl_proof": "\\begin{proof}\n\n    $X$ is a subspace of a compact Hausdorff space $Y$, its one-point compactification. $Y$ is normal, and so by the Urysohn lemma $Y$ is completely regular. Therefore by corollary $X$ is completely regular.\n\n\\end{proof}"}
{"id": "Munkres|exercise_34_9", "formal_statement": "theorem exercise_34_9\n  (X : Type*) [topological_space X] [compact_space X]\n  (X1 X2 : set X) (hX1 : is_closed X1) (hX2 : is_closed X2)\n  (hX : X1 \u222a X2 = univ) (hX1m : metrizable_space X1)\n  (hX2m : metrizable_space X2) : metrizable_space X :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Let $X$ be a compact Hausdorff space that is the union of the closed subspaces $X_1$ and $X_2$. If $X_1$ and $X_2$ are metrizable, show that $X$ is metrizable.\n", "nl_proof": "\\begin{proof}\n\n    Both $X_1$ and $X_2$ are compact, Hausdorff and metrizable, so by exercise 3 they are second countable, i.e. there are countable bases $\\left\\{U_{i, n} \\subset X_i \\mid n \\in \\mathbb{N}\\right\\}$ for $i \\in\\{1,2\\}$. By the same exercise it is enough to show that $X$ is second countable. If $X_1 \\cap X_2=\\emptyset$ both $X_1$ and $X_2$ are open and the union $\\left\\{U_{i, n} \\mid i \\in\\{1,2\\} ; n \\in \\mathbb{N}\\right\\}$ of their countable bases form a countable base for $X$.\n\n\n\nSuppose now $X_1 \\cap X_2 \\neq \\emptyset$. Let $x \\in X$ and $U \\subset X$ be an open neighborhood of $x$. If $x \\in X_i-X_j=X-$ $X_j$ then $U \\cap X_i$ is open in $X_i$ and there is a basis neighborhood $U_{i, n}$ of $x$ such that $x \\in U_{i, n} \\cap X-X_j$ is an open neighborhood of $x$ in the open subset $X-X_j$, so $U_{i, n} \\cap X-X_j$ is also open in $X$.\n\nSuppose now that $x \\in X_1 \\cap X_2$. We have that $U \\cap X_i$ is open in $X_i$ so there is a basis neighborhood $U_{i, n_i}$ contained in $U \\cap X_i$. By definition of sub-space topology there is some open subset $V_{i, n_i} \\subset X$ such that $U_{i, n_i}=$ $X_i \\cap V_{i, n_i}$. Then\n\n$$\n\nx \\in V_{1, n_1} \\cap V_{2, n_2}=\\left(V_{1, n_1} \\cap V_{2, n_2} \\cap X_1\\right) \\cup\\left(V_{1, n_1} \\cap V_{2, n_2} \\cap X_2\\right)=\\left(U_{1, n_1} \\cap V_{2, n_2}\\right) \\cup\\left(V_{1, n_1} \\cap U_{2, n_2}\\right) \\subset U\n\n$$\n\nTherefore the open subsets $U_{i, n} \\cap X-X_j$ and $V_{1, n_1} \\cap V_{2, n_2}$ form a countable base for $X$.\n\n\\end{proof}"}
{"id": "Munkres|exercise_43_2", "formal_statement": "theorem exercise_43_2 {X : Type*} [metric_space X]\n  {Y : Type*} [metric_space Y] [complete_space Y] (A : set X)\n  (f : X \u2192 Y) (hf : uniform_continuous_on f A) :\n  \u2203! (g : X \u2192 Y), continuous_on g (closure A) \u2227\n  uniform_continuous_on g (closure A) \u2227 \u2200 (x : A), g x = f x :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Let $(X, d_X)$ and $(Y, d_Y)$ be metric spaces; let $Y$ be complete. Let $A \\subset X$. Show that if $f \\colon A \\rightarrow Y$ is uniformly continuous, then $f$ can be uniquely extended to a continuous function $g \\colon \\bar{A} \\rightarrow Y$, and $g$ is uniformly continuous.\n", "nl_proof": "\\begin{proof}\n\n    Let $\\left(X, d_X\\right)$ and $\\left(Y, d_Y\\right)$ be metric spaces; let $Y$ be complete. Let $A \\subset X$. It is also given that $f: A \\longrightarrow X$ is a uniformly continuous function. Then we have to show that $f$ can be uniquely extended to a continuous function $g: \\bar{A} \\longrightarrow Y$, and $g$ is uniformly coninuous.\n\nWe define the function $g$ as follows:\n\n$$\n\ng(x)= \\begin{cases}f(x) & \\text { if } x \\in A \\\\ \\lim _{n \\rightarrow \\infty} f\\left(x_n\\right) & \\text { if } x \\in \\bar{A},\\end{cases}\n\n$$\n\nwhere $\\left\\{x_n\\right\\} \\subset A$ is some sequence in $A$ such that $\\lim _{n \\rightarrow \\infty} x_n=x$. Now we have to check that the above definition is well defined. We first note that since $\\left\\{x_n\\right\\} \\subset A$ is convergent in $X,\\left\\{x_n\\right\\}$ is Cauchy in $A$ and since $f$ is uniformly continuous on $A,\\left\\{f\\left(x_n\\right)\\right\\}$ is Cauchy in $Y$. Further, since $Y$ is complete $\\left\\{f\\left(x_n\\right)\\right\\}$ is convergent. So let us now consider two sequences $\\left\\{x_n\\right\\}$ and $\\left\\{y_n\\right\\}$ such that $\\lim _{n \\rightarrow \\infty} x_n=\\lim _{n \\rightarrow \\infty} y_n=x$. Then we need to prove that $\\lim _{n \\rightarrow \\infty} f\\left(x_n\\right)=\\lim _{n \\rightarrow \\infty} f\\left(y_n\\right)=f(x)$. Let\n\n$$\n\n\\lim _{n \\rightarrow \\infty} f\\left(x_n\\right)=a \\text { and } b=\\lim _{n \\rightarrow \\infty} f\\left(y_n\\right)\n\n$$\n\nNow since $f$ is uniformly continuous for any given $\\epsilon>0$ there exists $\\delta>0$ such that\n\n$$\n\nd_Y(f(x), f(y))<\\epsilon \\text { whenever } d_X(x, y)<\\delta \\text { and } x, y \\in A\n\n$$\n\nSo for this $\\delta>0$ there exists $N \\in \\mathbb{N}$ such that\n\n$$\n\nd_X\\left(x_n, x\\right)<\\frac{\\delta}{2} \\text { and } d_X\\left(y_n, x\\right)<\\frac{\\delta}{2}, \\text { foe all } n \\geq N .\n\n$$\n\nTherefore, we have that for all $n \\geq N$,\n\n$$\n\nd_X\\left(x_n, y_n\\right)<\\delta\n\n$$\n\nThus the equation (1) yields us that\n\n$$\n\nd_Y\\left(f\\left(x_n\\right), f\\left(y_n\\right)\\right)<\\epsilon \\text { for all } n \\geq N .\n\n$$\n\nNow since $\\lim _{n \\rightarrow \\infty} f\\left(x_n\\right)=a$ and $b=\\lim _{n \\rightarrow \\infty} f\\left(y_n\\right)$, so for the above $\\epsilon>0$ we have a natural number $K \\geq N$ such that\n\n$$\n\n\\begin{gathered}\n\nd_Y\\left(f\\left(x_n\\right), a\\right)<\\epsilon \\text { for all } n \\geq K \\text { and } \\\\\n\nd_Y\\left(f\\left(y_n\\right), b\\right)<\\epsilon \\text { for all } n \\geq K .\n\n\\end{gathered}\n\n$$\n\nMoreover, since $K \\geq N$, from $(2)$ we get\n\n$$\n\nd_Y\\left(f\\left(x_n\\right), f\\left(y_n\\right)\\right)<\\epsilon \\text { for all } n \\geq K .\n\n$$\n\nNow we calculate the following, for $n \\geq K$,\n\n$$\n\n\\begin{array}{rlr}\n\nd_Y(a, b) & \\leq & d_Y\\left(a, f\\left(x_n\\right)\\right)+d_Y\\left(f\\left(x_n\\right), f\\left(y_n\\right)\\right)+d\\left(f\\left(y_n\\right), b\\right) \\\\\n\n& < & \\epsilon+\\epsilon+\\epsilon \\text { by }(3),(4) \\text { and }(5) \\\\\n\n& = & 3 \\epsilon\n\n\\end{array}\n\n$$\n\nwhere the first inequality holds because of triangular inequality. Since $\\epsilon>0$ is arbitrary the above calculation shows that $d_Y(a, b)=0$. Thus, the above definition is independent of the choice of the sequence $\\left\\{x_n\\right\\}$ and hence the map $g$ is well defined. Moreover, from the construction it follows that $g$ is continuous on $\\bar{A}$.\n\nMoreover, we observe that $g$ is unique extension of $f$ by the construction.\n\nSo it remains to show that $g$ is uniformly continuous. In order to that we take a Cauchy sequence $\\left\\{a_n\\right\\} \\subset \\bar{A}$. Then since $\\bar{A}$ is a closed set so the sequence $\\left\\{a_n\\right\\}$ is convergent and hence $\\left\\{g\\left(a_n\\right)\\right\\}$ is also a convergent sequence as $g$ is continuous on $\\bar{A}$. So $\\left\\{g\\left(a_n\\right)\\right\\}$ is a Cauchy sequence in $Y$. Since a function is uniformly continuous if and only if it sends Cauchy sequences to Cauchy sequences, we conclude that $g$ is uniformly continuous.\n\n\\end{proof}"}
{"id": "Axler|exercise_1_3", "formal_statement": "theorem exercise_1_3 {F V : Type*} [add_comm_group V] [field F]\n  [module F V] {v : V} : -(-v) = v :=", "src_header": "import .common \n\nopen set fintype complex polynomial submodule linear_map finite_dimensional\nopen module module.End inner_product_space\n\nopen_locale big_operators\n\n", "nl_statement": "Prove that $-(-v) = v$ for every $v \\in V$.\n", "nl_proof": "\\begin{proof}\n\n    By definition, we have\n\n$$\n\n(-v)+(-(-v))=0 \\quad \\text { and } \\quad v+(-v)=0 .\n\n$$\n\nThis implies both $v$ and $-(-v)$ are additive inverses of $-v$, by the uniqueness of additive inverse, it follows that $-(-v)=v$.\n\n\\end{proof}"}
{"id": "Axler|exercise_1_6", "formal_statement": "theorem exercise_1_6 : \u2203 U : set (\u211d \u00d7 \u211d),\n  (U \u2260 \u2205) \u2227\n  (\u2200 (u v : \u211d \u00d7 \u211d), u \u2208 U \u2227 v \u2208 U \u2192 u + v \u2208 U) \u2227\n  (\u2200 (u : \u211d \u00d7 \u211d), u \u2208 U \u2192 -u \u2208 U) \u2227\n  (\u2200 U' : submodule \u211d (\u211d \u00d7 \u211d), U \u2260 \u2191U') :=", "src_header": "import .common \n\nopen set fintype complex polynomial submodule linear_map finite_dimensional\nopen module module.End inner_product_space\n\nopen_locale big_operators\n\n", "nl_statement": "Give an example of a nonempty subset $U$ of $\\mathbf{R}^2$ such that $U$ is closed under addition and under taking additive inverses (meaning $-u \\in U$ whenever $u \\in U$), but $U$ is not a subspace of $\\mathbf{R}^2$.\n", "nl_proof": "\\begin{proof}\n\n    \\[U=\\mathbb{Z}^2=\\left\\{(x, y) \\in \\mathbf{R}^2: x, y \\text { are integers }\\right\\}\\]\n\n$U=\\mathbb{Z}^2$ satisfies the desired properties. To come up with this, note by assumption, $U$ must be closed under addition and subtraction, so in particular, it must contain 0 . We need to find a set which fails scalar multiplication. A discrete set like $\\mathbb{Z}^2$ does this.\n\n\\end{proof}"}
{"id": "Axler|exercise_1_8", "formal_statement": "theorem exercise_1_8 {F V : Type*} [add_comm_group V] [field F]\n  [module F V] {\u03b9 : Type*} (u : \u03b9 \u2192 submodule F V) :\n  \u2203 U : submodule F V, (\u22c2 (i : \u03b9), (u i).carrier) = \u2191U :=", "src_header": "import .common \n\nopen set fintype complex polynomial submodule linear_map finite_dimensional\nopen module module.End inner_product_space\n\nopen_locale big_operators\n\n", "nl_statement": "Prove that the intersection of any collection of subspaces of $V$ is a subspace of $V$.\n", "nl_proof": "\\begin{proof}\n\nLet $V_1, V_2, \\ldots, V_n$ be subspaces of the vector space $V$ over the field $F$. We must show that their intersection $V_1 \\cap V_2 \\cap \\ldots \\cap V_n$ is also a subspace of $V$.\n\n\n\nTo begin, we observe that the additive identity $0$ of $V$ is in $V_1 \\cap V_2 \\cap \\ldots \\cap V_n$. This is because $0$ is in each subspace $V_i$, as they are subspaces and hence contain the additive identity.\n\n\n\nNext, we show that the intersection of subspaces is closed under addition. Let $u$ and $v$ be vectors in $V_1 \\cap V_2 \\cap \\ldots \\cap V_n$. By definition, $u$ and $v$ belong to each of the subspaces $V_i$. Since each $V_i$ is a subspace and therefore closed under addition, it follows that $u+v$ belongs to each $V_i$. Thus, $u+v$ belongs to the intersection $V_1 \\cap V_2 \\cap \\ldots \\cap V_n$.\n\n\n\nFinally, we show that the intersection of subspaces is closed under scalar multiplication. Let $a$ be a scalar in $F$ and let $v$ be a vector in $V_1 \\cap V_2 \\cap \\ldots \\cap V_n$. Since $v$ belongs to each $V_i$, we have $av$ belongs to each $V_i$ as well, as $V_i$ are subspaces and hence closed under scalar multiplication. Therefore, $av$ belongs to the intersection $V_1 \\cap V_2 \\cap \\ldots \\cap V_n$.\n\n\n\nThus, we have shown that $V_1 \\cap V_2 \\cap \\ldots \\cap V_n$ is a subspace of $V$.\n\n\\end{proof}"}
{"id": "Axler|exercise_3_1", "formal_statement": "theorem exercise_3_1 {F V : Type*}  \n  [add_comm_group V] [field F] [module F V] [finite_dimensional F V]\n  (T : V \u2192\u2097[F] V) (hT : finrank F V = 1) :\n  \u2203 c : F, \u2200 v : V, T v = c \u2022 v:=", "src_header": "import .common \n\nopen set fintype complex polynomial submodule linear_map finite_dimensional\nopen module module.End inner_product_space\n\nopen_locale big_operators\n\n", "nl_statement": "Show that every linear map from a one-dimensional vector space to itself is multiplication by some scalar. More precisely, prove that if $\\operatorname{dim} V=1$ and $T \\in \\mathcal{L}(V, V)$, then there exists $a \\in \\mathbf{F}$ such that $T v=a v$ for all $v \\in V$.\n", "nl_proof": "\\begin{proof}\n\n    If $\\operatorname{dim} V=1$, then in fact, $V=\\mathbf{F}$ and it is spanned by $1 \\in \\mathbf{F}$.\n\nLet $T$ be a linear map from $V$ to itself. Let $T(1)=\\lambda \\in V(=\\mathbf{F})$.\n\nStep 2\n\n2 of 3\n\nEvery $v \\in V$ is a scalar. Therefore,\n\n$$\n\n\\begin{aligned}\n\nT(v) & =T(v \\cdot 1) \\\\\n\n& =v T(1) \\ldots .(\\text { By the linearity of } T) \\\\\n\n& =v \\lambda\n\n\\end{aligned}\n\n$$\n\nHence, $T v=\\lambda v$ for every $v \\in V$.\n\n\\end{proof}"}
{"id": "Axler|exercise_4_4", "formal_statement": "theorem exercise_4_4 (p : polynomial \u2102) :\n  p.degree = @card (root_set p \u2102) (polynomial.root_set_fintype p \u2102) \u2194\n  disjoint\n  (@card (root_set p.derivative \u2102) (polynomial.root_set_fintype p.derivative \u2102))\n  (@card (root_set p \u2102) (polynomial.root_set_fintype p \u2102)) :=", "src_header": "import .common \n\nopen set fintype complex polynomial submodule linear_map finite_dimensional\nopen module module.End inner_product_space\n\nopen_locale big_operators\n\n", "nl_statement": "Suppose $p \\in \\mathcal{P}(\\mathbf{C})$ has degree $m$. Prove that $p$ has $m$ distinct roots if and only if $p$ and its derivative $p^{\\prime}$ have no roots in common.\n", "nl_proof": "\\begin{proof}\n\n    First, let $p$ have $m$ distinct roots. Since $p$ has the degree of $m$, then this could imply that $p$ can be actually written in the form of $p(z)=c\\left(z-\\lambda_1\\right) \\ldots\\left(z-\\lambda_m\\right)$, which you have $\\lambda_1, \\ldots, \\lambda_m$ being distinct.\n\nTo prove that both $p$ and $p^{\\prime}$ have no roots in commons, we must now show that $p^{\\prime}\\left(\\lambda_j\\right) \\neq 0$ for every $j$. So, to do so, just fix $j$. The previous expression for $p$ shows that we can now write $p$ in the form of $p(z)=\\left(z-\\lambda_j\\right) q(z)$, which $q$ is a polynomial such that $q\\left(\\lambda_j\\right) \\neq 0$.\n\n\n\nWhen you differentiate both sides of the previous equation, then you would then have $p^{\\prime}(z)=(z-$ $\\left.\\lambda_j\\right) q^{\\prime}(z)+q(z)$\n\n\n\nTherefore: $\\left.=p^{\\prime}\\left(\\lambda_j\\right)=q \\lambda_j\\right)$\n\nEquals: $p^{\\prime}\\left(\\lambda_j\\right) \\neq 0$\n\n\n\nNow, to prove the other direction, we would now prove the contrapositive, which means that we will be proving that if $p$ has actually less than $m$ distinct roots, then both $p$ and $p^{\\prime}$ have at least one root in common.\n\n\n\nNow, for some root of $\\lambda$ of $p$, we can write $p$ is in the form of $\\left.p(z)=(z-\\lambda)^n q(z)\\right)$, which is where both $n \\geq 2$ and $q$ is a polynomial. When differentiating both sides of the previous equations, we would then have $p^{\\prime}(z)=(z-\\lambda)^n q^{\\prime}(z)+n(z-\\lambda)^{n-1} q(z)$.\n\nTherefore, $p^{\\prime}(\\lambda)=0$, which would make $\\lambda$ is a common root of both $p$ and $p^{\\prime}$.\n\n\\end{proof}"}
{"id": "Axler|exercise_5_4", "formal_statement": "theorem exercise_5_4 {F V : Type*} [add_comm_group V] [field F]\n  [module F V] (S T : V \u2192\u2097[F] V) (hST : S \u2218 T = T \u2218 S) (c : F):\n  map S (T - c \u2022 id).ker = (T - c \u2022 id).ker :=", "src_header": "import .common \n\nopen set fintype complex polynomial submodule linear_map finite_dimensional\nopen module module.End inner_product_space\n\nopen_locale big_operators\n\n", "nl_statement": "Suppose that $S, T \\in \\mathcal{L}(V)$ are such that $S T=T S$. Prove that $\\operatorname{null} (T-\\lambda I)$ is invariant under $S$ for every $\\lambda \\in \\mathbf{F}$.\n", "nl_proof": "\\begin{proof}\n\n    First off, fix $\\lambda \\in F$. Secondly, let $v \\in \\operatorname{null}(T-\\lambda I)$. If so, then $(T-\\lambda I)(S v)=T S v-\\lambda S v=$ $S T v-\\lambda S v=S(T v-\\lambda v)=0$. Therefore, $S v \\in \\operatorname{null}(T-\\lambda I)$ since $n u l l(T-\\lambda I)$ is actually invariant under $S$.\n\n\\end{proof}"}
{"id": "Axler|exercise_5_12", "formal_statement": "theorem exercise_5_12 {F V : Type*} [add_comm_group V] [field F]\n  [module F V] {S : End F V}\n  (hS : \u2200 v : V, \u2203 c : F, v \u2208 eigenspace S c) :\n  \u2203 c : F, S = c \u2022 id :=", "src_header": "import .common \n\nopen set fintype complex polynomial submodule linear_map finite_dimensional\nopen module module.End inner_product_space\n\nopen_locale big_operators\n\n", "nl_statement": "Suppose $T \\in \\mathcal{L}(V)$ is such that every vector in $V$ is an eigenvector of $T$. Prove that $T$ is a scalar multiple of the identity operator.\n", "nl_proof": "\\begin{proof}\n\n    For every single $v \\in V$, there does exist $a_v \\in F$ such that $T v=a_v v$. Since $T 0=0$, then we have to make $a_0$ be the any number in F. However, for every single $v \\in V\\{0\\}$, then the value of $a_V$ is uniquely determined by the previous equation of $T v=a_v v$.\n\n\n\nNow, to show that $T$ is a scalar multiple of the identity, then me must show that $a_v$ is independent of $v$ for $v \\in V\\{0\\}$. We would now want to show that $a_v=a_w$.\n\n\n\nFirst, just make the case of where $(v, w)$ is linearly dependent. Then, there does exist $b \\in F$ such that $w=b v$. Now, you would have the following: $a_W w=T w=T(b v)=b T v=b\\left(a_v v\\right)=a_v w$. This is showing that $a_v=a_w$.\n\nFinally, make the consideration to make $(v, w)$ be linearly independent. Now, we would have the following: $\\left.a_{(} v+w\\right)(v+w)=T(v+w)=T v+T w=a_v v+a_w w$.\n\n\n\nThat previous equation implies the following: $\\left.\\left.\\left(a_{(} v+w\\right)-a_v\\right) v+\\left(a_{(} v+w\\right)-a_w\\right) w=0$. Since $(v, w)$ is linearly independent, this would imply that both $\\left.a_{(} v+w\\right)=a_v$ and $\\left.a_{(} v+w\\right)=a_w$. Therefore, $a_v=a_w$.\n\n\\end{proof}"}
{"id": "Axler|exercise_5_20", "formal_statement": "theorem exercise_5_20 {F V : Type*} [add_comm_group V] [field F]\n  [module F V] [finite_dimensional F V] {S T : End F V}\n  (h1 : @card T.eigenvalues (eigenvalues.fintype T) = finrank F V)\n  (h2 : \u2200 v : V, \u2203 c : F, v \u2208 eigenspace S c \u2194 \u2203 c : F, v \u2208 eigenspace T c) :\n  S * T = T * S :=", "src_header": "import .common \n\nopen set fintype complex polynomial submodule linear_map finite_dimensional\nopen module module.End inner_product_space\n\nopen_locale big_operators\n\n", "nl_statement": "Suppose that $T \\in \\mathcal{L}(V)$ has $\\operatorname{dim} V$ distinct eigenvalues and that $S \\in \\mathcal{L}(V)$ has the same eigenvectors as $T$ (not necessarily with the same eigenvalues). Prove that $S T=T S$.\n", "nl_proof": "\\begin{proof}\n\n    First off, let $n=\\operatorname{dim} V$. so, there is a basis of $\\left(v_1, \\ldots, v_j\\right)$ of $V$ that consist of eigenvectors of $T$. Now, let $\\lambda_1, \\ldots, \\lambda_n$ be the corresponding eigenvalues, then we would have $T v_j=\\lambda_1 v_j$ for every single $j$.\n\n\n\nNow, for every $v_j$ is also an eigenvector of S, so $S v_j=a_j v_j$ for some $a_j \\in F$. For each $j$, we would then have $(S T) v_j=S\\left(T v_j\\right)=\\lambda_j S v_j=a_j \\lambda_j v_j$ and $(T S) v_j=T\\left(S v_j\\right)=a_j T v_j=a_j \\lambda_j v_j$. Since both operators, which are $S T$ and $T S$, agree on a basis, then both are equal.\n\n\\end{proof}"}
{"id": "Axler|exercise_6_2", "formal_statement": "theorem exercise_6_2 {V : Type*} [add_comm_group V] [module \u2102 V]\n  [inner_product_space \u2102 V] (u v : V) :\n  \u27eau, v\u27eb_\u2102 = 0 \u2194 \u2200 (a : \u2102), \u2225u\u2225 \u2264 \u2225u + a \u2022 v\u2225 :=", "src_header": "import .common \n\nopen set fintype complex polynomial submodule linear_map finite_dimensional\nopen module module.End inner_product_space\n\nopen_locale big_operators\n\n", "nl_statement": "Suppose $u, v \\in V$. Prove that $\\langle u, v\\rangle=0$ if and only if $\\|u\\| \\leq\\|u+a v\\|$ for all $a \\in \\mathbf{F}$.\n", "nl_proof": "\\begin{proof}\n\n    First off, let us suppose that $(u, v)=0$.\n\nNow, let $a \\in \\mathbb{F}$. Next, $u, a v$ are orthogonal.\n\nThe Pythagorean theorem thus implies that\n\n$$\n\n\\begin{aligned}\n\n\\|u+a v\\|^2 & =\\|u\\|^2+\\|a v\\|^2 \\\\\n\n& \\geq\\|u\\|^2\n\n\\end{aligned}\n\n$$\n\nSo, by taking the square roots, this will now give us $\\|u\\| \\leq\\|u+a v\\|$.\n\nNow, to prove the implication in the other direction, we must now let $\\|u\\| \\leq$ $\\|u+a v\\|$ for all $a \\in \\mathbb{F}$. Squaring this inequality, we get both:\n\n$$\n\n\\begin{gathered}\n\n\\|u\\|^2 a n d \\leq\\|u+a v\\|^2 \\\\\n\n=(u+a v, u+a v) \\\\\n\n=(u, u)+(u, a v)+(a v, u)+(a v, a v) \\\\\n\n=\\|u\\|^2+\\bar{a}(u, v)+a \\overline{(u, v)}+|a|^2\\|v\\|^2 \\\\\n\n\\|u\\|^2+2 \\Re \\bar{a}(u, v)+|a|^2\\|v\\|^2\n\n\\end{gathered}\n\n$$\n\nfor all $a \\in \\mathbb{F}$.\n\nTherefore,\n\n$$\n\n-2 \\Re \\bar{a}(u, v) \\leq|a|^2\\|v\\|^2\n\n$$\n\nfor all $a \\in \\mathbb{F}$. In particular, we can let $a$ equal $-t(u, v)$ for $t>0$. Substituting this value for $a$ into the inequality above gives\n\n$$\n\n2 t|(u, v)|^2 \\leq t^2|(u, v)|^2\\|v\\|^2\n\n$$\n\nfor all $t>0$.\n\nStep 4\n\n4 of 4\n\nDivide both sides of the inequality above by $t$, getting\n\n$$\n\n2|(u, v)|^2 \\leq t \\mid(u, v)^2\\|v\\|^2\n\n$$\n\nfor all $t>0$. If $v=0$, then $(u, v)=0$, as desired. If $v \\neq 0$, set $t$ equal to $1 /\\|v\\|^2$ in the inequality above, getting\n\n$$\n\n2|(u, v)|^2 \\leq|(u, v)|^2,\n\n$$\n\nwhich implies that $(u, v)=0$.\n\n\\end{proof}"}
{"id": "Axler|exercise_6_7", "formal_statement": "theorem exercise_6_7 {V : Type*} [inner_product_space \u2102 V] (u v : V) :\n  \u27eau, v\u27eb_\u2102 = (\u2225u + v\u2225^2 - \u2225u - v\u2225^2 + I*\u2225u + I\u2022v\u2225^2 - I*\u2225u-I\u2022v\u2225^2) / 4 :=", "src_header": "import .common \n\nopen set fintype complex polynomial submodule linear_map finite_dimensional\nopen module module.End inner_product_space\n\nopen_locale big_operators\n\n", "nl_statement": "Prove that if $V$ is a complex inner-product space, then $\\langle u, v\\rangle=\\frac{\\|u+v\\|^{2}-\\|u-v\\|^{2}+\\|u+i v\\|^{2} i-\\|u-i v\\|^{2} i}{4}$ for all $u, v \\in V$.\n", "nl_proof": "\\begin{proof}\n\nLet $V$ be an inner-product space and $u, v\\in V$. Then \n\n$$\n\n\\begin{aligned}\n\n\\|u+v\\|^2 & =\\langle u+v, v+v\\rangle \\\\\n\n& =\\|u\\|^2+\\langle u, v\\rangle+\\langle v, u\\rangle+\\|v\\|^2 \\\\\n\n-\\|u-v\\|^2 & =-\\langle u-v, u-v\\rangle \\\\\n\n& =-\\|u\\|^2+\\langle u, v\\rangle+\\langle v, u\\rangle-\\|v\\|^2 \\\\\n\ni\\|u+i v\\|^2 & =i\\langle u+i v, u+i v\\rangle \\\\\n\n& =i\\|u\\|^2+\\langle u, v\\rangle-\\langle v, u\\rangle+i\\|v\\|^2 \\\\\n\n-i\\|u-i v\\|^2 & =-i\\langle u-i v, u-i v\\rangle \\\\\n\n& =-i\\|u\\|^2+\\langle u, v\\rangle-\\langle v, u\\rangle-i\\|v\\|^2 .\n\n\\end{aligned}\n\n$$\n\nThus $\\left(\\|u+v\\|^2\\right)-\\|u-v\\|^2+\\left(i\\|u+i v\\|^2\\right)-i\\|u-i v\\|^2=4\\langle u, v\\rangle.$\n\n\\end{proof}"}
{"id": "Axler|exercise_6_16", "formal_statement": "theorem exercise_6_16 {K V : Type*} [is_R_or_C K] [inner_product_space K V]\n  {U : submodule K V} : \n  U.orthogonal = \u22a5  \u2194 U = \u22a4 :=", "src_header": "import .common \n\nopen set fintype complex polynomial submodule linear_map finite_dimensional\nopen module module.End inner_product_space\n\nopen_locale big_operators\n\n", "nl_statement": "Suppose $U$ is a subspace of $V$. Prove that $U^{\\perp}=\\{0\\}$ if and only if $U=V$\n", "nl_proof": "\\begin{proof}\n\n    $V=U \\bigoplus U^{\\perp}$, therefore $U^\\perp = \\{0\\}$ iff $U=V$. \n\n\\end{proof}"}
{"id": "Axler|exercise_7_6", "formal_statement": "theorem exercise_7_6 {V : Type*} [inner_product_space \u2102 V]\n  [finite_dimensional \u2102 V] (T : End \u2102 V)\n  (hT : T * T.adjoint = T.adjoint * T) :\n  T.range = T.adjoint.range :=", "src_header": "import .common \n\nopen set fintype complex polynomial submodule linear_map finite_dimensional\nopen module module.End inner_product_space\n\nopen_locale big_operators\n\n", "nl_statement": "Prove that if $T \\in \\mathcal{L}(V)$ is normal, then $\\operatorname{range} T=\\operatorname{range} T^{*}.$\n", "nl_proof": "\\begin{proof}\n\n    Let $T \\in \\mathcal{L}(V)$ to be a normal operator.\n\nSuppose $u \\in \\operatorname{null} T$. Then, by $7.20$,\n\n$$\n\n0=\\|T u\\|=\\left\\|T^* u\\right\\|,\n\n$$\n\nwhich implies that $u \\in \\operatorname{null} T^*$.\n\nHence\n\n$$\n\n\\operatorname{null} T=\\operatorname{null} T^*\n\n$$\n\nbecause $\\left(T^*\\right)^*=T$ and the same argument can be repeated.\n\nNow we have\n\n$$\n\n\\begin{aligned}\n\n\\text { range } T & =\\left(\\text { null } T^*\\right)^{\\perp} \\\\\n\n& =(\\text { null } T)^{\\perp} \\\\\n\n& =\\operatorname{range} T^*,\n\n\\end{aligned}\n\n$$\n\nwhere the first and last equality follow from items (d) and (b) of 7.7.\n\nHence, range $T=$ range $T^*$.\n\n\\end{proof}"}
{"id": "Axler|exercise_7_10", "formal_statement": "theorem exercise_7_10 {V : Type*} [inner_product_space \u2102 V]\n  [finite_dimensional \u2102 V] (T : End \u2102 V)\n  (hT : T * T.adjoint = T.adjoint * T) (hT1 : T^9 = T^8) :\n  is_self_adjoint T \u2227 T^2 = T :=", "src_header": "import .common \n\nopen set fintype complex polynomial submodule linear_map finite_dimensional\nopen module module.End inner_product_space\n\nopen_locale big_operators\n\n", "nl_statement": "Suppose $V$ is a complex inner-product space and $T \\in \\mathcal{L}(V)$ is a normal operator such that $T^{9}=T^{8}$. Prove that $T$ is self-adjoint and $T^{2}=T$.\n", "nl_proof": "\\begin{proof}\n\n    Based on the complex spectral theorem, there is an orthonormal basis of $\\left(e_1, \\ldots, e_n\\right)$ of $V$ consisting of eigenvectors of $T$. Now, let $\\lambda_1, \\ldots, \\lambda_n$ be the corresponding eigenvalues. Therefore,\n\n$$\n\nT e_1=\\lambda_j e_j\n\n$$\n\nfor $j=1 \\ldots n$.\n\n\n\nNext, by applying $T$ repeatedly to both sides of the equation above, we get $T^9 e_j=\\left(\\lambda_j\\right)^9 e_j$ and rei =8ej. Thus $T^8 e_j=\\left(\\lambda_j\\right)^8 e_j$, which implies that $\\lambda_j$ equals 0 or 1 . In particular, all the eigenvalues of $T$ are real. This would then imply that $T$ is self-adjoint.\n\n\n\nNow, by applying $T$ to both sides of the equation above, we get\n\n$$\n\n\\begin{aligned}\n\nT^2 e_j & =\\left(\\lambda_j\\right)^2 e_j \\\\\n\n& =\\lambda_j e_j \\\\\n\n& =T e_j\n\n\\end{aligned}\n\n$$\n\nwhich is where the second equality holds because $\\lambda_j$ equals 0 or 1 . Because $T^2$ and $T$ agree on a basis, they must be equal.\n\n\\end{proof}"}
{"id": "Axler|exercise_7_14", "formal_statement": "theorem exercise_7_14 {\ud835\udd5c V : Type*} [is_R_or_C \ud835\udd5c]\n  [inner_product_space \ud835\udd5c V] [finite_dimensional \ud835\udd5c V]\n  {T : End \ud835\udd5c V} (hT : is_self_adjoint T)\n  {l : \ud835\udd5c} {\u03b5 : \u211d} (he : \u03b5 > 0) : \u2203 v : V, \u2016v\u2016= 1 \u2227 (\u2016T v - l \u2022 v\u2016 < \u03b5 \u2192\n  (\u2203 l' : T.eigenvalues, \u2016l - l'\u2016 < \u03b5)) :=", "src_header": "import .common \n\nopen set fintype complex polynomial submodule linear_map finite_dimensional\nopen module module.End inner_product_space\n\nopen_locale big_operators\n\n", "nl_statement": "Suppose $T \\in \\mathcal{L}(V)$ is self-adjoint, $\\lambda \\in \\mathbf{F}$, and $\\epsilon>0$. Prove that if there exists $v \\in V$ such that $\\|v\\|=1$ and $\\|T v-\\lambda v\\|<\\epsilon,$ then $T$ has an eigenvalue $\\lambda^{\\prime}$ such that $\\left|\\lambda-\\lambda^{\\prime}\\right|<\\epsilon$.\n", "nl_proof": "\\begin{proof}\n\n    Let $T \\in \\mathcal{L}(V)$ be a self-adjoint, and let $\\lambda \\in \\mathbf{F}$ and $\\epsilon>0$.\n\nBy the Spectral Theorem, there is $e_1, \\ldots, e_n$ an orthonormal basis of $V$ consisting of eigenvectors of $T$ and let $\\lambda_1, \\ldots, \\lambda_n$ denote their corresponding eigenvalues.\n\nChoose an eigenvalue $\\lambda^{\\prime}$ of $T$ such that $\\left|\\lambda^{\\prime}-\\lambda\\right|^2$ is minimized.\n\nThere are $a_1, \\ldots, a_n \\in \\mathbb{F}$ such that\n\n$$\n\nv=a_1 e_1+\\cdots+a_n e_n .\n\n$$\n\nThus, we have\n\n$$\n\n\\begin{aligned}\n\n\\epsilon^2 & >|| T v-\\left.\\lambda v\\right|^2 \\\\\n\n& =\\left|\\left\\langle T v-\\lambda v, e_1\\right\\rangle\\right|^2+\\cdots+\\left|\\left\\langle T v-\\lambda v, e_n\\right\\rangle\\right|^2 \\\\\n\n& =\\left|\\lambda_1 a_1-\\lambda a_1\\right|^2+\\cdots+\\left|\\lambda_n a_n-\\lambda a_n\\right|^2 \\\\\n\n& =\\left|a_1\\right|^2\\left|\\lambda_1-\\lambda\\right|^2+\\cdots+\\left|a_n\\right|^2\\left|\\lambda_n-\\lambda\\right|^2 \\\\\n\n& \\geq\\left|a_1\\right|^2\\left|\\lambda^{\\prime}-\\lambda\\right|^2+\\cdots+\\left|a_n\\right|^2\\left|\\lambda^{\\prime}-\\lambda\\right|^2 \\\\\n\n& =\\left|\\lambda^{\\prime}-\\lambda\\right|^2\n\n\\end{aligned}\n\n$$\n\nwhere the second and fifth lines follow from $6.30$ (the fifth because $\\|v\\|=1$ ). Now, we taking the square root.\n\nHence, $T$ has an eigenvalue $\\lambda^{\\prime}$ such that $\\left|\\lambda^{\\prime}-\\lambda\\right|<\\epsilon$\n\n\\end{proof}"}
{"id": "Ireland-Rosen|exercise_1_30", "formal_statement": "theorem exercise_1_30 {n : \u2115} : \n  \u00ac \u2203 a : \u2124, \u2211 (i : fin n), (1 : \u211a) / (n+2) = a :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen zsqrtd gaussian_int char_p nat.arithmetic_function \n\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that $\\frac{1}{2}+\\frac{1}{3}+\\cdots+\\frac{1}{n}$ is not an integer.\n", "nl_proof": "\\begin{proof}\n\nLet $2^s$ be the largest power of 2 occuring as a denominator in $H_n$, say $2^s=k \\leqslant n$. Write $H_n=$ $\\frac{1}{2^s}+\\left(1+1 / 2+\\ldots+1 /(k-1)+1 /(k+1)+\\ldots+1 / n\\right.$. The sum in parentheses can be written as $1 / 2^{s-1}$ times sum of fractions with odd denominators, so the denominator of the sum in parentheses will not be divisible by $2^s$, but it must equal $2^s$ by Ex $1.29$.\n\n\\end{proof}"}
{"id": "Ireland-Rosen|exercise_2_4", "formal_statement": "theorem exercise_2_4 {a : \u2124} (ha : a \u2260 0) \n  (f_a :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen zsqrtd gaussian_int char_p nat.arithmetic_function \n\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "If $a$ is a nonzero integer, then for $n>m$ show that $\\left(a^{2^{n}}+1, a^{2^{m}}+1\\right)=1$ or 2 depending on whether $a$ is odd or even.\n", "nl_proof": "\\begin{proof}    \n\n\\begin{align*}\n\n\\operatorname{ord}_p\\, n!  &= \\sum_{k\\geq 1} \\left \\lfloor \\frac{n}{p^{k}}\\right \\rfloor \\leq  \\sum_{k\\geq 1}  \\frac{n}{p^{k}} = \\frac{n}{p} \\frac{1}{1 - \\frac{1}{p}} = \\frac{n}{p-1}\n\n\\end{align*}\n\n\n\nThe decomposition of $n!$ in prime factors is\n\n\n\n$n! = p_1^{\\alpha_1}p_2^{\\alpha_2}\\cdots p_k^{\\alpha_k}$ \n\nwhere $\\alpha_i = \\operatorname{ord}_{p_i}\\, n! \\leq \\frac{n}{p_i-1}$, and $p_i \\leq n, \\ i=1,2,\\cdots,k$.\n\n\n\nThen\n\n\\begin{align*}\n\nn! &\\leq p_1^{\\frac{n}{p_1-1}}p_2^{\\frac{n}{p_2-1}}\\cdots p_k^{\\frac{n}{p_n-1}}\\\\\n\n\\sqrt[n]{n!} &\\leq p_1^{\\frac{1}{p_1-1}}p_2^{\\frac{1}{p_2-1}}\\cdots p_k^{\\frac{1}{p_n-1}}\\\\\n\n&\\leq \\prod_{p\\leq n} p^{\\frac{1}{p-1}}\n\n\\end{align*}\n\n(the values of $p$ in this product describe all prime numbers $p\\leq n$.)\n\n\\end{proof}"}
{"id": "Ireland-Rosen|exercise_2_27a", "formal_statement": "theorem exercise_2_27a : \n  \u00ac summable (\u03bb i : {p : \u2124 // squarefree p}, (1 : \u211a) / i) :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen zsqrtd gaussian_int char_p nat.arithmetic_function \n\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Show that $\\sum^{\\prime} 1 / n$, the sum being over square free integers, diverges.\n", "nl_proof": "\\begin{proof}\n\n    \n\nLet $S \\subset \\mathbb{N}^*$ the set of square free integers.\n\n\n\nLet $N \\in \\mathbb{N}^*$. Every integer $n, \\, 1\\leq n \\leq N$ can be written as $n = a b^2$, where $a,b$ are integers and $a$ is square free. Then $1\\leq a \\leq N$, and $1\\leq b \\leq \\sqrt{N}$, so\n\n$$\\sum_{n\\leq N} \\frac{1}{n} \\leq \\sum_{a \\in S, a\\leq N}\\  \\sum_{1\\leq b \\leq \\sqrt{N}} \\frac{1}{ab^2} \\leq  \\sum_{a \\in S, a\\leq N}\\ \\frac{1}{a} \\, \\sum_{b=1}^\\infty  \\frac{1}{b^2} = \\frac{\\pi^2}{6} \\sum_{a \\in S, a\\leq N}\\ \\frac{1}{a}.$$\n\nSo $$\\sum_{a \\in S, a\\leq N} \\frac{1}{a}  \\geq \\frac{6}{\\pi^2} \\sum_{n\\leq N} \\frac{1}{n}.$$\n\nAs $\\sum_{n=1}^\\infty \\frac{1}{n}$ diverges, $\\lim\\limits_{N \\to \\infty} \\sum\\limits_{a \\in S, a\\leq N} \\frac{1}{a} = +\\infty$, so the family $\\left(\\frac{1}{a}\\right)_{a\\in S}$ of the inverse of square free integers is not summable.\n\n\n\nLet $S_N = \\prod_{p<N}(1+1/p)$ , and $p_1,p_2,\\ldots, p_l\\ (l = l(N))$ all prime integers less than $N$. Then\n\n\\begin{align*}\n\nS_N &= \\left(1+\\frac{1}{p_1}\\right) \\cdots \\left(1+\\frac{1}{p_l}\\right)\\\\\n\n&=\\sum_{(\\varepsilon_1,\\cdots,\\varepsilon_l) \\in \\{0,1\\}^l } \\frac{1}{p_1^{\\varepsilon_1} \\cdots p_l^{\\varepsilon_l}}\n\n\\end{align*}\n\nWe prove this last formula  by induction. This is true for $l=1$ : $\\sum_{\\varepsilon \\in \\{0,1\\}} 1/p_1^\\varepsilon = 1 + 1/p_1$.\n\n\n\nIf it is true for the integer $l$, then \n\n\\begin{align*}\n\n\\left(1+\\frac{1}{p_1}\\right) \\cdots \\left(1+\\frac{1}{p_l}\\right)\\left(1+\\frac{1}{p_{l+1}}\\right) &= \\sum_{(\\varepsilon_1,\\ldots,\\varepsilon_l) \\in \\{0,1\\}^l } \\frac{1}{p_1^{\\varepsilon_1} \\cdots p_l^{\\varepsilon_l}} \\left(1+\\frac{1}{p_{l+1}}\\right)\\\\\n\n&=\\sum_{(\\varepsilon_1,\\ldots,\\varepsilon_l) \\in \\{0,1\\}^l } \\frac{1}{p_1^{\\varepsilon_1} \\cdots p_l^{\\varepsilon_l}} + \\sum_{(\\varepsilon_1,\\ldots,\\varepsilon_l) \\in \\{0,1\\}^l } \\frac{1}{p_1^{\\varepsilon_1} \\cdots p_l^{\\varepsilon_l}p_{l+1}}\\\\\n\n&=\\sum_{(\\varepsilon_1,\\ldots,\\varepsilon_l,\\varepsilon_{l+1}) \\in \\{0,1\\}^{l+1} } \\frac{1}{p_1^{\\varepsilon_1} \\cdots p_l^{\\varepsilon_l}p_{l+1}^{\\varepsilon_{l+1}}} \n\n\\end{align*}\n\nSo it is true for all $l$. \n\n\n\nThus $S_N = \\sum_{n\\in \\Delta} \\frac{1}{n}$, where $\\Delta$ is the set of square free integers whose prime factors are less than $N$.\n\n\n\nAs $\\sum 1/n$, the sum being over square free integers, diverges, $\\lim\\limits_{N\\to \\infty} S_N = + \\infty$ :\n\n$$\\lim_{N \\to \\infty} \\prod_{p<N} \\left(1+\\frac{1}{p}\\right) = +\\infty.$$\n\n $e^x \\geq 1+x, x \\geq \\log (1+x)$ for $x>0$, so\n\n$$\\log S_N = \\sum_{k=1}^{l(N)} \\log\\left(1+\\frac{1}{p_k}\\right) \\leq \\sum_{k=1}^{l(N)} \\frac{1}{p_k}.$$\n\n$\\lim\\limits_{N\\to \\infty} \\log S_N = +\\infty$ and $\\lim\\limits_{N\\to \\infty} l(N) = +\\infty$, so\n\n$$\\lim_{N\\to \\infty} \\sum_{p<N} \\frac{1}{p} = +\\infty.$$\n\n\\end{proof}"}
{"id": "Ireland-Rosen|exercise_3_4", "formal_statement": "theorem exercise_3_4 : \u00ac \u2203 x y : \u2124, 3*x^2 + 2 = y^2 :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen zsqrtd gaussian_int char_p nat.arithmetic_function \n\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Show that the equation $3 x^{2}+2=y^{2}$ has no solution in integers.\n", "nl_proof": "\\begin{proof}    \n\nIf $3x^2+2 = y^2$, then  $\\overline{y}^2 = \\overline{2}$ in $\\mathbb{Z}/3\\mathbb{Z}$.\n\n\n\n\n\nAs $\\{-1,0,1\\}$ is a complete set of residues modulo $3$, the squares in $\\mathbb{Z}/3\\mathbb{Z}$ are $\\overline{0} = \\overline{0}^2$ and  $\\overline{1} = \\overline{1}^2 = (\\overline{-1})^2$, so $\\overline{2}$ is not a square in $\\mathbb{Z}/3\\mathbb{Z}$ : $\\overline{y}^2 = \\overline{2}$ is impossible in $\\mathbb{Z}/3\\mathbb{Z}$.\n\n\n\nThus $3x^2+2 = y^2$ has no solution in integers.\n\n\\end{proof}"}
{"id": "Ireland-Rosen|exercise_3_10", "formal_statement": "theorem exercise_3_10 {n : \u2115} (hn0 : \u00ac n.prime) (hn1 : n \u2260 4) : \n  factorial (n-1) \u2261 0 [MOD n] :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen zsqrtd gaussian_int char_p nat.arithmetic_function \n\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "If $n$ is not a prime, show that $(n-1) ! \\equiv 0(n)$, except when $n=4$.\n", "nl_proof": "\\begin{proof}    \n\nSuppose that $n >1$ is not a prime. Then $n = uv$, where $2 \\leq u \\leq v \\leq n-1$.\n\n\n\n$\\bullet$ If $u \\neq v$, then $n = uv \\mid (n-1)! = 1\\times 2 \\times\\cdots \\times u \\times\\cdots \\times v \\times \\cdots \\times (n-1)$ (even if $u\\wedge v \\neq 1$ !).\n\n\n\n$\\bullet$ If $u=v$, $n = u^2$ is a square.\n\n\n\nIf $u$ is not prime, $u =st,\\ 2\\leq s \\leq t \\leq u-1 \\leq n-1$, and $n = u' v'$, where $u' =s,v' =st^2$ verify  $2 \\leq u' < v' \\leq n-1$. As in the first case, $n = u'v' \\mid (n-1)!$.  \n\n\n\nIf $u = p$ is a prime, then $n =p^2$.\n\n\n\nIn the case $p = 2$, $n = 4$ and $n=4  \\nmid (n-1)! = 6$. In the other case $p >2$, and $(n-1)! = (p^2 - 1)!$ contains the factors $p < 2p < p^2$, so $p^2 \\mid (p^2-1)!, n \\mid (n-1)!$.\n\n\n\nConclusion : if $n$ is not a prime, $(n - 1)! \\equiv 0 \\pmod n$, except when $n=4$.\n\n\\end{proof}"}
{"id": "Ireland-Rosen|exercise_4_4", "formal_statement": "theorem exercise_4_4 {p t: \u2115} (hp0 : p.prime) (hp1 : p = 4*t + 1) \n  (a : zmod p) : \n  is_primitive_root a p \u2194 is_primitive_root (-a) p :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen zsqrtd gaussian_int char_p nat.arithmetic_function \n\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Consider a prime $p$ of the form $4 t+1$. Show that $a$ is a primitive root modulo $p$ iff $-a$ is a primitive root modulo $p$.\n", "nl_proof": "\\begin{proof}\n\n    Suppose that $a$ is a primitive root modulo $p$. As $p-1$ is even, $(-a)^{p-1}=a^{p-1} \\equiv 1$ $(\\bmod p)$\n\nIf $(-a)^n \\equiv 1(\\bmod p)$, with $n \\in \\mathbb{N}$, then $a^n \\equiv(-1)^n(\\bmod p)$.\n\nTherefore $a^{2 n} \\equiv 1(\\bmod p)$. As $a$ is a primitive root modulo $p, p-1|2 n, 2 t| n$, so $n$ is even.\n\n\n\nHence $a^n \\equiv 1(\\bmod p)$, and $p-1 \\mid n$. So the least $n \\in \\mathbb{N}^*$ such that $(-a)^n \\equiv 1$ $(\\bmod p)$ is $p-1:$ the order of $-a$ modulo $p$ is $p-1,-a$ is a primitive root modulo $p$. Conversely, if $-a$ is a primitive root modulo $p$, we apply the previous result at $-a$ to to obtain that $-(-a)=a$ is a primitive root.\n\n\\end{proof}"}
{"id": "Ireland-Rosen|exercise_4_6", "formal_statement": "theorem exercise_4_6 {p n : \u2115} (hp : p.prime) (hpn : p = 2^n + 1) : \n  is_primitive_root 3 p :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen zsqrtd gaussian_int char_p nat.arithmetic_function \n\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "If $p=2^{n}+1$ is a Fermat prime, show that 3 is a primitive root modulo $p$.\n", "nl_proof": "\\begin{proof}\n\n \\newcommand{\\legendre}[2]{\\genfrac{(}{)}{}{}{#1}{#2}}\n\nWrite $p = 2^k + 1$, with $k = 2^n$.\n\n\n\nWe suppose that $n>0$, so $k\\geq 2, p \\geq 5$. As $p$ is prime, $3^{p-1} \\equiv 1 \\pmod p$. \n\n\n\nIn other words, $3^{2^k} \\equiv 1 \\pmod p$ : the order of $3$ is a divisor of $2^k$, a power of $2$.\n\n\n\n$3$ has order $2^k$ modulo $p$ iff $3^{2^{k-1}} \\not \\equiv 1 \\pmod p$. As $\\left (3^{2^{k-1}} \\right)^2 \\equiv 1 \\pmod p$, where $p$ is prime, this is equivalent to $3^{2^{k-1}}  \\equiv -1 \\pmod p$, which remains to prove.\n\n\n\n$3^{2^{k-1}} = 3^{(p-1)/2} \\equiv \\legendre{3}{p} \\pmod p$.\n\n\n\nAs the result is true for $p=5$, we can suppose $n\\geq 2$.\n\nFrom the law of quadratic reciprocity :\n\n$$\\legendre{3}{p} \\legendre{p}{3} = (-1)^{(p-1)/2} = (-1)^{2^{k-1}} = 1.$$\n\nSo $\\legendre{3}{p} = \\legendre{p}{3}$\n\n \n\n\\begin{align*}\n\np = 2^{2^n}+1 &\\equiv (-1)^{2^n} + 1 \\pmod 3\\\\\n\n&\\equiv 2 \\equiv -1 \\pmod 3,\n\n\\end{align*}\n\nso $\\legendre{3}{p} = \\legendre {p}{3} = -1$, that is to say\n\n$$3^{2^{k-1}}  \\equiv -1 \\pmod p.$$\n\nThe order of $3$ modulo $p = 2^{2^n} + 1$ is $p-1 = 2^{2^n}$ : $3$ is a primitive root modulo $p$.\n\n\n\n(On the other hand, if $3$ is of order $p-1$ modulo $p$, then $p$ is prime, so\n\n$$ F_n = 2^{2^n} + 1 \\ \\mathrm{is}\\ \\mathrm{prime}\\ \\iff 3^{(F_n-1)/2} = 3^{2^{2^n - 1}} \\equiv -1 \\pmod {F_n}.)$$\n\n\\end{proof}"}
{"id": "Ireland-Rosen|exercise_4_11", "formal_statement": "theorem exercise_4_11 {p : \u2115} (hp : p.prime) (k s: \u2115) \n  (s :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen zsqrtd gaussian_int char_p nat.arithmetic_function \n\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that $1^{k}+2^{k}+\\cdots+(p-1)^{k} \\equiv 0(p)$ if $p-1 \\nmid k$ and $-1(p)$ if $p-1 \\mid k$.\n", "nl_proof": "\\begin{proof}    \n\nLet $S_k = 1^k+2^k+\\cdots+(p-1)^k$.\n\n\n\nLet $g$ a primitive root modulo $p$ : $\\overline{g}$ a generator of $\\mathbb{F}_p^*$.\n\n\n\nAs $(\\overline{1},\\overline{g},\\overline{g}^{2}, \\ldots, \\overline{g}^{p-2}) $ is a permutation of $ (\\overline{1},\\overline{2}, \\ldots,\\overline{p-1})$,\n\n\\begin{align*}\n\n\\overline{S_k} &= \\overline{1}^k + \\overline{2}^k+\\cdots+ \\overline{p-1}^k\\\\\n\n&= \\sum_{i=0}^{p-2} \\overline{g}^{ki} =\n\n\\left\\{\n\n\\begin{array}{ccc}\n\n\\overline{ p-1} = -\\overline{1} &  \\mathrm{if} &  p-1 \\mid k  \\\\\n\n \\frac{ \\overline{g}^{(p-1)k} -1}{ \\overline{g}^k -1} = \\overline{0}&  \\mathrm{if}  &   p-1 \\nmid k\n\n\\end{array}\n\n\\right.\n\n\\end{align*}\n\nsince $p-1 \\mid k \\iff \\overline{g}^k = \\overline{1}$.\n\n\n\nConclusion :\n\n\\begin{align*}\n\n1^k+2^k+\\cdots+(p-1)^k&\\equiv 0 \\pmod p\\ \\mathrm{if} \\ p-1 \\nmid k\\\\\n\n1^k+2^k+\\cdots+(p-1)^k&\\equiv -1 \\pmod p\\ \\mathrm{if} \\ p-1 \\mid k\\\\\n\n\\end{align*}\n\n\\end{proof}"}
{"id": "Ireland-Rosen|exercise_5_28", "formal_statement": "theorem exercise_5_28 {p : \u2115} (hp : p.prime) (hp1 : p \u2261 1 [MOD 4]): \n  \u2203 x, x^4 \u2261 2 [MOD p] \u2194 \u2203 A B, p = A^2 + 64*B^2 :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen zsqrtd gaussian_int char_p nat.arithmetic_function \n\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Show that $x^{4} \\equiv 2(p)$ has a solution for $p \\equiv 1(4)$ iff $p$ is of the form $A^{2}+64 B^{2}$.\n", "nl_proof": "\\begin{proof}    \n\nIf  $p\\equiv 1\\ [4]$ and if there exists $x \\in \\mathbb{Z}$ such that $x^4 \\equiv 2\\ [p]$, then\n\n$$2^{\\frac{p-1}{4} }\\equiv  x^{p-1} \\equiv 1 \\ [p].$$ \n\n\n\nFrom Ex. 5.27, where $p = a^2 +b^2, a$ odd,  we know that $$f^{\\frac{ab}{2}} \\equiv 2^{\\frac{p-1}{4} } \\equiv 1 \\ [p].$$\n\n\n\nSince $f^2 \\equiv -1\\ [p]$, the order of $f$ modulo $p$ is 4, thus $4 \\mid \\frac{ab}{2}$, so $8\\mid ab$.\n\n\n\nAs $a$ is odd, $8 | b$, then $p = A^2 + 64 B^2$ (with $A = a, B = b/8$).\n\n\n\n\\bigskip\n\n\n\nConversely, if $p=A^2+64 B^2$, then $p\\equiv A^2 \\equiv 1 \\ [4]$.\n\n\n\nLet $a=A,b=8B$. Then $$2^{\\frac{p-1}{4} } \\equiv f^{\\frac{ab}{2}} \\equiv f^{4AB} \\equiv (-1)^{2AB} \\equiv 1 \\ [p].$$\n\n\n\nAs $2^{\\frac{p-1}{4} } \\equiv 1 \\ [p]$, $x^4 \\equiv 2 \\ [p]$ has a solution in $\\mathbb{Z}$ (Prop. 4.2.1) : $2$ is a biquadratic residue modulo $p$.\n\n\n\nConclusion : \n\n\n\n$$\\exists A \\in \\mathbb{Z}, \\exists B \\in \\mathbb{Z}\\,, p = A^2+64 B^2 \\iff( p\\equiv 1 \\ [4] \\ \\mathrm{and} \\ \\exists x \\in \\mathbb{Z}, \\, x^4 \\equiv 2 \\ [p]).$$\n\n\\end{proof}"}
{"id": "Ireland-Rosen|exercise_12_12", "formal_statement": "theorem exercise_12_12 : is_algebraic \u211a (sin (real.pi/12)) :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen zsqrtd gaussian_int char_p nat.arithmetic_function \n\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Show that $\\sin (\\pi / 12)$ is an algebraic number.\n", "nl_proof": "\\begin{proof}\n\n$$\n\n\\begin{aligned}\n\n    \\sin \\pi/12=\\sin \\left(\\pi/4-\\pi/6\\right) & =\\sin \\pi/4 \\cos \\pi/6-\\cos \\pi/4 \\sin \\pi/6 \\\\\n\n& =\\frac{\\sqrt{3}}{2 \\sqrt{2}}-\\frac{1}{2 \\sqrt{2}} \\\\\n\n& =\\frac{\\sqrt{3}-1}{2 \\sqrt{2}}\n\n\\end{aligned}\n\n$$\n\n\\end{proof}"}
{"id": "Shakarchi|exercise_1_13a", "formal_statement": "theorem exercise_1_13a {f : \u2102 \u2192 \u2102} (\u03a9 : set \u2102) (a b : \u03a9) (h : is_open \u03a9)\n  (hf : differentiable_on \u2102 f \u03a9) (hc : \u2203 (c : \u211d), \u2200 z \u2208 \u03a9, (f z).re = c) :\n  f a = f b :=", "src_header": "import .common \n\nopen complex filter function interval_integral metric\n\nopen_locale big_operators\nopen_locale filter\nopen_locale topology\n\n", "nl_statement": "Suppose that $f$ is holomorphic in an open set $\\Omega$. Prove that if $\\text{Re}(f)$ is constant, then $f$ is constant.\n", "nl_proof": "\\begin{proof}\n\nLet $f(z)=f(x, y)=u(x, y)+i v(x, y)$, where $z=x+i y$.\n\nSince $\\operatorname{Re}(f)=$ constant,\n\n$$\n\n\\frac{\\partial u}{\\partial x}=0, \\frac{\\partial u}{\\partial y}=0 .\n\n$$\n\nBy the Cauchy-Riemann equations,\n\n$$\n\n\\frac{\\partial v}{\\partial x}=-\\frac{\\partial u}{\\partial y}=0 .\n\n$$\n\nThus, in $\\Omega$,\n\n$$\n\nf^{\\prime}(z)=\\frac{\\partial f}{\\partial x}=\\frac{\\partial u}{\\partial x}+i \\frac{\\partial v}{\\partial x}=0+0=0 .\n\n$$\n\n3\n\nThus $f(z)$ is constant.\n\n\\end{proof}"}
{"id": "Shakarchi|exercise_1_13c", "formal_statement": "theorem exercise_1_13c {f : \u2102 \u2192 \u2102} (\u03a9 : set \u2102) (a b : \u03a9) (h : is_open \u03a9)\n  (hf : differentiable_on \u2102 f \u03a9) (hc : \u2203 (c : \u211d), \u2200 z \u2208 \u03a9, abs (f z) = c) :\n  f a = f b :=", "src_header": "import .common \n\nopen complex filter function interval_integral metric\n\nopen_locale big_operators\nopen_locale filter\nopen_locale topology\n\n", "nl_statement": "Suppose that $f$ is holomorphic in an open set $\\Omega$. Prove that if $|f|$ is constant, then $f$ is constant.\n", "nl_proof": "\\begin{proof}\n\nLet $f(z)=f(x, y)=u(x, y)+i v(x, y)$, where $z=x+i y$.\n\nWe first give a mostly correct argument; the reader should pay attention to find the difficulty. Since $|f|=\\sqrt{u^2+v^2}$ is constant,\n\n$$\n\n\\left\\{\\begin{array}{l}\n\n0=\\frac{\\partial\\left(u^2+v^2\\right)}{\\partial x}=2 u \\frac{\\partial u}{\\partial x}+2 v \\frac{\\partial v}{\\partial x} . \\\\\n\n0=\\frac{\\partial\\left(u^2+v^2\\right)}{\\partial y}=2 u \\frac{\\partial u}{\\partial y}+2 v \\frac{\\partial v}{\\partial y} .\n\n\\end{array}\\right.\n\n$$\n\nPlug in the Cauchy-Riemann equations and we get\n\n$$\n\n\\begin{gathered}\n\nu \\frac{\\partial v}{\\partial y}+v \\frac{\\partial v}{\\partial x}=0 \\\\\n\n-u \\frac{\\partial v}{\\partial x}+v \\frac{\\partial v}{\\partial y}=0 \\\\\n\n(1.14) \\Rightarrow \\frac{\\partial v}{\\partial x}=\\frac{v}{u} \\frac{\\partial v}{\\partial y}\n\n\\end{gathered}\n\n$$\n\nPlug (1.15) into (1.13) and we get\n\n$$\n\n\\frac{u^2+v^2}{u} \\frac{\\partial v}{\\partial y}=0 .\n\n$$\n\nSo $u^2+v^2=0$ or $\\frac{\\partial v}{\\partial y}=0$.\n\n\n\nIf $u^2+v^2=0$, then, since $u, v$ are real, $u=v=0$, and thus $f=0$ which is constant.\n\n\n\nThus we may assume $u^2+v^2$ equals a non-zero constant, and we may divide by it. We multiply both sides by $u$ and find $\\frac{\\partial v}{\\partial y}=0$, then by (1.15), $\\frac{\\partial v}{\\partial x}=0$, and by Cauchy-Riemann, $\\frac{\\partial u}{\\partial x}=0$.\n\n$$\n\nf^{\\prime}=\\frac{\\partial f}{\\partial x}=\\frac{\\partial u}{\\partial x}+i \\frac{\\partial v}{\\partial x}=0 .\n\n$$\n\nThus $f$ is constant.\n\nWhy is the above only mostly a proof? The problem is we have a division by $u$, and need to make sure everything is well-defined. Specifically, we need to know that $u$ is never zero. We do have $f^{\\prime}=0$ except at points where $u=0$, but we would need to investigate that a bit more.\n\nLet's return to\n\n$$\n\n\\left\\{\\begin{array}{l}\n\n0=\\frac{\\partial\\left(u^2+v^2\\right)}{\\partial x}=2 u \\frac{\\partial u}{\\partial x}+2 v \\frac{\\partial v}{\\partial x} . \\\\\n\n0=\\frac{\\partial\\left(u^2+v^2\\right)}{\\partial y}=2 u \\frac{\\partial u}{\\partial y}+2 v \\frac{\\partial v}{\\partial y} .\n\n\\end{array}\\right.\n\n$$\n\nPlug in the Cauchy-Riemann equations and we get\n\n$$\n\n\\begin{array}{r}\n\nu \\frac{\\partial v}{\\partial y}+v \\frac{\\partial v}{\\partial x}=0 \\\\\n\n-u \\frac{\\partial v}{\\partial x}+v \\frac{\\partial v}{\\partial y}=0 .\n\n\\end{array}\n\n$$\n\nWe multiply the first equation $u$ and the second by $v$, and obtain\n\n$$\n\n\\begin{aligned}\n\nu^2 \\frac{\\partial v}{\\partial y}+u v \\frac{\\partial v}{\\partial x} & =0 \\\\\n\n-u v \\frac{\\partial v}{\\partial x}+v^2 \\frac{\\partial v}{\\partial y} & =0 .\n\n\\end{aligned}\n\n$$\n\nAdding the two yields\n\n$$\n\nu^2 \\frac{\\partial v}{\\partial y}+v^2 \\frac{\\partial v}{\\partial y}=0,\n\n$$\n\nor equivalently\n\n$$\n\n\\left(u^2+v^2\\right) \\frac{\\partial v}{\\partial y}=0 .\n\n$$\n\nWe now argue in a similar manner as before, except now we don't have the annoying $u$ in the denominator. If $u^2+v^2=0$ then $u=v=0$, else we can divide by $u^2+v^2$ and find $\\partial v / \\partial y=0$. Arguing along these lines finishes the proof.\n\n\\end{proof}"}
{"id": "Shakarchi|exercise_1_19b", "formal_statement": "theorem exercise_1_19b (z : \u2102) (hz : abs z = 1) (s : \u2115 \u2192 \u2102)\n    (h : s = (\u03bb n, \u2211 i in (finset.range n), i * z / i ^ 2)) :\n    \u2203 y, tendsto s at_top (\ud835\udcdd y) :=", "src_header": "import .common \n\nopen complex filter function interval_integral metric\n\nopen_locale big_operators\nopen_locale filter\nopen_locale topology\n\n", "nl_statement": "Prove that the power series $\\sum zn/n^2$ converges at every point of the unit circle.\n", "nl_proof": "\\begin{proof}\n\n    Since $\\left|z^n / n^2\\right|=1 / n^2$ for all $|z|=1$, then $\\sum z^n / n^2$ converges at every point in the unit circle as $\\sum 1 / n^2$ does ( $p$-series $p=2$.)\n\n\\end{proof}"}
{"id": "Shakarchi|exercise_1_26", "formal_statement": "theorem exercise_1_26\n  (f F\u2081 F\u2082 : \u2102 \u2192 \u2102) (\u03a9 : set \u2102) (h1 : is_open \u03a9) (h2 : is_connected \u03a9)\n  (hF\u2081 : differentiable_on \u2102 F\u2081 \u03a9) (hF\u2082 : differentiable_on \u2102 F\u2082 \u03a9)\n  (hdF\u2081 : \u2200 x \u2208 \u03a9, deriv F\u2081 x = f x) (hdF\u2082 : \u2200 x \u2208 \u03a9, deriv F\u2082 x = f x)\n  : \u2203 c : \u2102, \u2200 x, F\u2081 x = F\u2082 x + c :=", "src_header": "import .common \n\nopen complex filter function interval_integral metric\n\nopen_locale big_operators\nopen_locale filter\nopen_locale topology\n\n", "nl_statement": "Suppose $f$ is continuous in a region $\\Omega$. Prove that any two primitives of $f$ (if they exist) differ by a constant.\n", "nl_proof": "\\begin{proof}\n\n    Suppose $F_1$ adn $F_2$ are primitives of $F$. Then $(F_1-F_2)^\\prime = f - f = 0$, therefore $F_1$ and $F_2$ differ by a constant. \n\n\\end{proof}"}
{"id": "Shakarchi|exercise_2_9", "formal_statement": "theorem exercise_2_9\n  {f : \u2102 \u2192 \u2102} (\u03a9 : set \u2102) (b : metric.bounded \u03a9) (h : is_open \u03a9)\n  (hf : differentiable_on \u2102 f \u03a9) (z \u2208 \u03a9) (hz : f z = z) (h'z : deriv f z = 1) :\n  \u2203 (f_lin : \u2102 \u2192L[\u2102] \u2102), \u2200 x \u2208 \u03a9, f x = f_lin x :=", "src_header": "import .common \n\nopen complex filter function interval_integral metric\n\nopen_locale big_operators\nopen_locale filter\nopen_locale topology\n\n", "nl_statement": "Let $\\Omega$ be a bounded open subset of $\\mathbb{C}$, and $\\varphi: \\Omega \\rightarrow \\Omega$ a holomorphic function. Prove that if there exists a point $z_{0} \\in \\Omega$ such that $\\varphi\\left(z_{0}\\right)=z_{0} \\quad \\text { and } \\quad \\varphi^{\\prime}\\left(z_{0}\\right)=1$ then $\\varphi$ is linear.\n", "nl_proof": "\\begin{proof}\n\n    When $\\Omega$ is connected, if $\\varphi$ is not linear, then there exists $n \\geq 2$ and $a_n \\neq 0$, such that\n\n$$\n\n\\varphi(z)=z+a_n\\left(z-z_0\\right)^n+O\\left(\\left(z-z_0\\right)^{n+1}\\right) .\n\n$$\n\nAs you have noticed, by induction, it follows that for every $k \\geq 1$,\n\n$$\n\n\\varphi^k(z)=z+k a_n\\left(z-z_0\\right)^n+O\\left(\\left(z-z_0\\right)^{n+1}\\right) .\n\n$$\n\nLet $r>0$ be such that when $\\left|z-z_0\\right| \\leq r$, then $z \\in \\Omega$. Then by (1),\n\n$$\n\nk a_n=\\frac{1}{2 \\pi i} \\int_{\\left|z-z_0\\right|=r} \\frac{\\varphi^k(z)}{\\left(z-z_0\\right)^{n+1}} d z .\n\n$$\n\nSince $\\varphi^k(\\Omega) \\subset \\Omega$ and since $\\Omega$ is bounded, there exists $M>0$, independent of $k$, such that $\\left|\\varphi^k\\right| \\leq M$ on $\\Omega$. Then by (2),\n\n$$\n\nk\\left|a_n\\right| \\leq M r^{-n} .\n\n$$\n\nSince $k$ is arbitrary, $a_n=0$, a contradiction.\n\n\\end{proof}"}
{"id": "Shakarchi|exercise_3_3", "formal_statement": "theorem exercise_3_3 (a : \u211d) (ha : 0 < a) :\n    tendsto (\u03bb y, \u222b x in -y..y, real.cos x / (x ^ 2 + a ^ 2))\n    at_top (\ud835\udcdd (real.pi * (real.exp (-a) / a))) :=", "src_header": "import .common \n\nopen complex filter function interval_integral metric\n\nopen_locale big_operators\nopen_locale filter\nopen_locale topology\n\n", "nl_statement": "Show that $ \\int_{-\\infty}^{\\infty} \\frac{\\cos x}{x^2 + a^2} dx = \\pi \\frac{e^{-a}}{a}$ for $a > 0$.\n", "nl_proof": "\\begin{proof}\n\n    $\\cos x=\\frac{e^{i x}+e^{-i x}}{2}$. changing $x \\rightarrow-x$ we see that we can just integrate $e^{i x} /\\left(x^2+a^2\\right)$ and we'll get the same answer. Again, we use the same semicircle and part of the real line. The only pole is $x=i a$, it has order 1 and the residue at it is $\\lim _{x \\rightarrow i a} \\frac{e^{i x}}{x^2+a^2}(x-i a)=\\frac{e^{-a}}{2 i a}$, which multiplied by $2 \\pi i$ gives the answer.\n\n\\end{proof}"}
{"id": "Shakarchi|exercise_3_9", "formal_statement": "theorem exercise_3_9 : \u222b x in 0..1, real.log (real.sin (real.pi * x)) = - real.log 2 :=", "src_header": "import .common \n\nopen complex filter function interval_integral metric\n\nopen_locale big_operators\nopen_locale filter\nopen_locale topology\n\n", "nl_statement": "Show that $\\int_0^1 \\log(\\sin \\pi x) dx = - \\log 2$.\n", "nl_proof": "\\begin{proof}\n\nConsider\n\n$$\n\n\\begin{gathered}\n\nf(z)=\\log \\left(1-e^{2 \\pi z i}\\right)=\\log \\left(e^{\\pi z i}\\left(e^{-\\pi z i}-e^{\\pi z i}\\right)\\right)=\\log (-2 i)+\\pi z i+\\log \\\\\n\n(\\sin (\\pi z))\n\n\\end{gathered}\n\n$$\n\nThen we have\n\n$$\n\n\\begin{aligned}\n\n\\int_0^1 f(z) d z & =\\log (-2 i)+\\frac{i \\pi}{2}+\\int_0^1 \\log (\\sin (\\pi z)) d z \\\\\n\n& =\\int_0^1 \\log (\\sin (\\pi z)) d z+\\log (-2 i)+\\log (i) \\\\\n\n& =\\log (2)+\\int_0^1 \\log (\\sin (\\pi z)) d z\n\n\\end{aligned}\n\n$$\n\nNow it suffices to show that $\\int_0^1 f(z) d z=0$. Consider the contour $C(\\epsilon, R)$ (which is the contour given in your question) given by the following.\n\n1. $C_1(\\epsilon, R)$ : The vertical line along the imaginary axis from $i R$ to $i \\epsilon$.\n\n2. $C_2(\\epsilon)$ : The quarter turn of radius $\\epsilon$ about 0 .\n\n3. $C_3(\\epsilon)$ : Along the real axis from $(\\epsilon, 1-\\epsilon)$.\n\n4. $C_4(\\epsilon)$ : The quarter turn of radius $\\epsilon$ about 1 .\n\n5. $C_5(\\epsilon, R)$ : The vertical line from $1+i \\epsilon$ to $1+i R$.\n\n6. $C_6(R)$ : The horizontal line from $1+i R$ to $i R$.\n\n$f(z)$ is analytic inside the contour $C$ and hence $\\oint_C f(z)=0$. This gives us\n\n$$\n\n\\begin{aligned}\n\n\\int_{C_1(\\epsilon, R)} f d z+\\int_{C_2(\\epsilon)} f d z+\\int_{C_3(\\epsilon)} f d z+\\int_{C_4(\\epsilon)} f d z+\\int_{C_5(\\epsilon, R)} f d z+\\int_{C_6(R)} f d z \\\\\n\n=0\n\n\\end{aligned}\n\n$$\n\nNow the integral along 1 cancels with the integral along 5 due to symmetry. Integrals along 2 and 4 scale as $\\epsilon \\log (\\epsilon)$. Integral along 6 goes to 0 as $R \\rightarrow \\infty$. This gives us\n\n$$\n\n\\lim _{\\epsilon \\rightarrow 0} \\int_{C_3(\\epsilon)} f d z=0\n\n$$\n\nwhich is what we need.\n\n\\end{proof}"}
{"id": "Shakarchi|exercise_3_22", "formal_statement": "theorem exercise_3_22 (D : set \u2102) (hD : D = ball 0 1) (f : \u2102 \u2192 \u2102)\n    (hf : differentiable_on \u2102 f D) (hfc : continuous_on f (closure D)) :\n    \u00ac \u2200 z \u2208 (sphere (0 : \u2102) 1), f z = 1 / z :=", "src_header": "import .common \n\nopen complex filter function interval_integral metric\n\nopen_locale big_operators\nopen_locale filter\nopen_locale topology\n\n", "nl_statement": "Show that there is no holomorphic function $f$ in the unit disc $D$ that extends continuously to $\\partial D$ such that $f(z) = 1/z$ for $z \\in \\partial D$.\n", "nl_proof": "\\begin{proof}\n\n    Consider $g(r)=\\int_{|z|=r} f(z) d z$. Cauchy theorem implies that $g(r)=0$ for all $r<1$. Now since $\\left.f\\right|_{\\partial D}=1 / z$ we have $\\lim _{r \\rightarrow 1} \\int_{|z|=r} f(z) d z=\\int_{|z|=1} \\frac{1}{z} d z=\\frac{2}{\\pi i} \\neq 0$. Contradiction.\n\n\\end{proof}"}
{"id": "Putnam|exercise_2020_b5", "formal_statement": "theorem exercise_2020_b5 (z : fin 4 \u2192 \u2102) (hz0 : \u2200 n, \u2016z n\u2016 < 1) \n  (hz1 : \u2200 n : fin 4, z n \u2260 1) : \n  3 - z 0 - z 1 - z 2 - z 3 + (z 0) * (z 1) * (z 2) * (z 3) \u2260 0 :=", "src_header": "import .common \n\nopen real topological_space filter polynomial\nopen_locale topology big_operators complex_conjugate filter ennreal \n\n", "nl_statement": "For $j \\in\\{1,2,3,4\\}$, let $z_{j}$ be a complex number with $\\left|z_{j}\\right|=1$ and $z_{j} \\neq 1$. Prove that $3-z_{1}-z_{2}-z_{3}-z_{4}+z_{1} z_{2} z_{3} z_{4} \\neq 0 .$\n", "nl_proof": "\\begin{proof}\n\n    It will suffice to show that for any $z_1, z_2, z_3, z_4 \\in \\mathbb{C}$ of modulus 1 such that $|3-z_1-z_2-z_3-z_4| = |z_1z_2z_3z_4|$, at least one of $z_1, z_2, z_3$ is equal to 1.\n\n\n\nTo this end, let $z_1=e^{\\alpha i}, z_2=e^{\\beta i}, z_3=e^{\\gamma i}$ and \n\n\\[\n\nf(\\alpha, \\beta, \\gamma)=|3-z_1-z_2-z_3|^2-|1-z_1z_2z_3|^2.\n\n\\]\n\n A routine calculation shows that \n\n\\begin{align*}\n\nf(\\alpha, \\beta, \\gamma)&=\n\n10 - 6\\cos(\\alpha) - 6\\cos(\\beta) - 6\\cos(\\gamma) \\\\\n\n&\\quad + 2\\cos(\\alpha + \\beta + \\gamma) + 2\\cos(\\alpha - \\beta) \\\\\n\n&\\quad + 2\\cos(\\beta - \\gamma) + 2\\cos(\\gamma - \\alpha).\n\n\\end{align*}\n\nSince the function $f$ is continuously differentiable, and periodic in each variable, $f$ has a maximum and a minimum and it attains these values only at points where $\\nabla f=(0,0,0)$.  A routine calculation now shows that \n\n\\begin{align*}\n\n\\frac{\\partial f}{\\partial \\alpha} + \\frac{\\partial f}{\\partial \\beta} + \\frac{\\partial f}{\\partial \\gamma} &=\n\n6(\\sin(\\alpha) +\\sin(\\beta)+\\sin(\\gamma)-  \\sin(\\alpha + \\beta + \\gamma)) \\\\\n\n&=\n\n24\\sin\\left(\\frac{\\alpha+\\beta}{2}\\right) \\sin\\left(\\frac{\\beta+\\gamma}{2}\\right)\n\n\\sin\\left(\\frac{\\gamma+\\alpha}{2}\\right).\n\n\\end{align*}\n\nHence every critical point of $f$ must satisfy one of $z_1z_2=1$, $z_2z_3=1$, or $z_3z_1=1$. By symmetry, let us assume that $z_1z_2=1$. Then \n\n\\[\n\nf = |3-2\\mathrm{Re}(z_1)-z_3|^2-|1-z_3|^2;\n\n\\]\n\nsince $3-2\\mathrm{Re}(z_1)\\ge 1$, $f$ is nonnegative and can be zero only if the real part of $z_1$, and hence also $z_1$ itself, is equal to $1$. \n\n\\end{proof}"}
{"id": "Putnam|exercise_2018_b2", "formal_statement": "theorem exercise_2018_b2 (n : \u2115) (hn : n > 0) (f : \u2115 \u2192 \u2102 \u2192 \u2102) \n  (hf : \u2200 n : \u2115, f n = \u03bb z, (\u2211 (i : fin n), (n-i)* z^(i : \u2115))) : \n  \u00ac (\u2203 z : \u2102, \u2016z\u2016 \u2264 1 \u2227 f n z = 0) :=", "src_header": "import .common \n\nopen real topological_space filter polynomial\nopen_locale topology big_operators complex_conjugate filter ennreal \n\n", "nl_statement": "Let $n$ be a positive integer, and let $f_{n}(z)=n+(n-1) z+$ $(n-2) z^{2}+\\cdots+z^{n-1}$. Prove that $f_{n}$ has no roots in the closed unit disk $\\{z \\in \\mathbb{C}:|z| \\leq 1\\}$.\n", "nl_proof": "\\begin{proof}\n\n    Note first that $f_n(1) > 0$, so $1$ is not a root of $f_n$.\n\nNext, note that\n\n\\[\n\n(z-1)f_n(z) = z^n + \\cdots + z - n;\n\n\\]\n\nhowever, for $\\left| z \\right| \\leq 1$, we have \n\n$\\left| z^n + \\cdots + z \\right| \\leq n$ by the triangle inequality;\n\nequality can only occur if $z,\\dots,z^n$ have norm 1 and the same argument, which only happens for $z=1$.\n\nThus there can be no root of $f_n$ with $|z| \\leq 1$.\n\n\\end{proof}"}
{"id": "Putnam|exercise_2017_b3", "formal_statement": "theorem exercise_2017_b3 (f : \u211d \u2192 \u211d) (c : \u2115 \u2192 \u211d)\n  (hf : f = \u03bb x, (\u2211' (i : \u2115), (c i) * x^i)) \n  (hc : \u2200 n, c n = 0 \u2228 c n = 1)\n  (hf1 : f (2/3) = 3/2) : \n  irrational (f (1/2)) :=", "src_header": "import .common \n\nopen real topological_space filter polynomial\nopen_locale topology big_operators complex_conjugate filter ennreal \n\n", "nl_statement": "Suppose that $f(x)=\\sum_{i=0}^{\\infty} c_{i} x^{i}$ is a power series for which each coefficient $c_{i}$ is 0 or 1 . Show that if $f(2 / 3)=3 / 2$, then $f(1 / 2)$ must be irrational.\n", "nl_proof": "\\begin{proof}\n\n    Suppose by way of contradiction that $f(1/2)$ is rational. Then $\\sum_{i=0}^{\\infty} c_i 2^{-i}$ is the binary expansion of a rational number, and hence must be eventually periodic; that is, there exist some integers $m,n$ such that\n\n$c_i = c_{m+i}$ for all $i \\geq n$. We may then write\n\n\\[\n\nf(x) = \\sum_{i=0}^{n-1} c_i x^i + \\frac{x^n}{1-x^m} \\sum_{i=0}^{m-1} c_{n+i} x^i.\n\n\\]\n\nEvaluating at $x = 2/3$, we may equate $f(2/3) = 3/2$ with \n\n\\[\n\n\\frac{1}{3^{n-1}} \\sum_{i=0}^{n-1} c_i 2^i 3^{n-i-1} + \\frac{2^n 3^m}{3^{n+m-1}(3^m-2^m)} \\sum_{i=0}^{m-1} c_{n+i} 2^i 3^{m-1-i};\n\n\\]\n\nsince all terms on the right-hand side have odd denominator, the same must be true of the sum, a contradiction.\n\n\\end{proof}"}
{"id": "Putnam|exercise_2010_a4", "formal_statement": "theorem exercise_2010_a4 (n : \u2115) : \n  \u00ac nat.prime (10^10^10^n + 10^10^n + 10^n - 1) :=", "src_header": "import .common \n\nopen real topological_space filter polynomial\nopen_locale topology big_operators complex_conjugate filter ennreal \n\n", "nl_statement": "Prove that for each positive integer $n$, the number $10^{10^{10^n}}+10^{10^n}+10^n-1$ is not prime.\n", "nl_proof": "\\begin{proof}\n\n    Put\n\n\\[\n\nN = 10^{10^{10^n}} + 10^{10^n} + 10^n - 1.\n\n\\]\n\nWrite $n = 2^m k$ with $m$ a nonnegative integer and $k$ a positive odd integer.\n\nFor any nonnegative integer $j$,\n\n\\[\n\n10^{2^m j} \\equiv (-1)^j \\pmod{10^{2^m} + 1}.\n\n\\]\n\nSince $10^n \\geq n \\geq 2^m \\geq m+1$, $10^n$ is divisible by $2^n$ and hence by $2^{m+1}$,\n\nand similarly $10^{10^n}$ is divisible by $2^{10^n}$ and hence by $2^{m+1}$. It follows that\n\n\\[\n\nN \\equiv 1 + 1 + (-1) + (-1) \\equiv 0 \\pmod{10^{2^m} + 1}.\n\n\\]\n\nSince $N \\geq 10^{10^n} > 10^n + 1 \\geq 10^{2^m} + 1$, it follows that $N$ is composite.\n\n\\end{proof}"}
{"id": "Putnam|exercise_2000_a2", "formal_statement": "theorem exercise_2000_a2 : \n  \u2200 N : \u2115, \u2203 n : \u2115, n > N \u2227 \u2203 i : fin 6 \u2192 \u2115, n = (i 0)^2 + (i 1)^2 \u2227 \n  n + 1 = (i 2)^2 + (i 3)^2 \u2227 n + 2 = (i 4)^2 + (i 5)^2 :=", "src_header": "import .common \n\nopen real topological_space filter polynomial\nopen_locale topology big_operators complex_conjugate filter ennreal \n\n", "nl_statement": "Prove that there exist infinitely many integers $n$ such that $n, n+1, n+2$ are each the sum of the squares of two integers. \n", "nl_proof": "\\begin{proof}\n\n    It is well-known that the equation $x^2-2y^2=1$ has infinitely\n\nmany solutions (the so-called ``Pell'' equation).  Thus setting\n\n$n=2y^2$ (so that $n=y^2+y^2$, $n+1=x^2+0^2$, $n+2=x^2+1^2$)\n\nyields infinitely many $n$ with the desired property.\n\n\\end{proof}"}
{"id": "Putnam|exercise_1998_a3", "formal_statement": "theorem exercise_1998_a3 (f : \u211d \u2192 \u211d) (hf : cont_diff \u211d 3 f) : \n  \u2203 a : \u211d, (f a) * (deriv f a) * (iterated_deriv 2 f a) * (iterated_deriv 3 f a) \u2265 0 :=", "src_header": "import .common \n\nopen real topological_space filter polynomial\nopen_locale topology big_operators complex_conjugate filter ennreal \n\n", "nl_statement": "Let $f$ be a real function on the real line with continuous third derivative. Prove that there exists a point $a$ such that\n", "nl_proof": ""}
{"id": "Pugh|exercise_2_12a", "formal_statement": "theorem exercise_2_12a (f : \u2115 \u2192 \u2115) (p : \u2115 \u2192 \u211d) (a : \u211d)\n  (hf : injective f) (hp : tendsto p at_top (\ud835\udcdd a)) :\n  tendsto (\u03bb n, p (f n)) at_top (\ud835\udcdd a) :=", "src_header": "import .common\n\nopen set real filter function ring_hom topological_space\n\nopen_locale big_operators\nopen_locale filter\nopen_locale topology \nnoncomputable theory \n\n", "nl_statement": "Let $(p_n)$ be a sequence and $f:\\mathbb{N}\\to\\mathbb{N}$. The sequence $(q_k)_{k\\in\\mathbb{N}}$ with $q_k=p_{f(k)}$ is called a rearrangement of $(p_n)$. Show that if $f$ is an injection, the limit of a sequence is unaffected by rearrangement.\n", "nl_proof": "\\begin{proof}\n\n    Let $\\varepsilon>0$. Since $p_n \\rightarrow L$, we have that, for all $n$ except $n \\leq N$, $d\\left(p_n, L\\right)<\\epsilon$. Let $S=\\{n \\mid f(n) \\leq N\\}$, let $n_0$ be the largest $n \\in S$, we know there is such a largest $n$ because $f(n)$ is injective. Now we have that $\\forall n>n_0 f(n)>N$ which implies that $p_{f(n)} \\rightarrow L$, as required.\n\n\\end{proof}"}
{"id": "Pugh|exercise_2_29", "formal_statement": "theorem exercise_2_29 (M : Type*) [metric_space M]\n  (O C : set (set M))\n  (hO : O = {s | is_open s})\n  (hC : C = {s | is_closed s}) :\n  \u2203 f : O \u2192 C, bijective f :=", "src_header": "import .common\n\nopen set real filter function ring_hom topological_space\n\nopen_locale big_operators\nopen_locale filter\nopen_locale topology \nnoncomputable theory \n\n", "nl_statement": "Let $\\mathcal{T}$ be the collection of open subsets of a metric space $\\mathrm{M}$, and $\\mathcal{K}$ the collection of closed subsets. Show that there is a bijection from $\\mathcal{T}$ onto $\\mathcal{K}$.\n", "nl_proof": "\\begin{proof}\n\n   The bijection given by $x\\mapsto X^C$ suffices.  \n\n\\end{proof}"}
{"id": "Pugh|exercise_2_41", "formal_statement": "theorem exercise_2_41 (m : \u2115) {X : Type*} [normed_space \u211d ((fin m) \u2192 \u211d)] :\n  is_compact (metric.closed_ball 0 1) :=", "src_header": "import .common\n\nopen set real filter function ring_hom topological_space\n\nopen_locale big_operators\nopen_locale filter\nopen_locale topology \nnoncomputable theory \n\n", "nl_statement": "Let $\\|\\cdot\\|$ be any norm on $\\mathbb{R}^{m}$ and let $B=\\left\\{x \\in \\mathbb{R}^{m}:\\|x\\| \\leq 1\\right\\}$. Prove that $B$ is compact.\n", "nl_proof": "\\begin{proof}\n\n    Let us call $\\|\\cdot\\|_E$ the Euclidean norm in $\\mathbb{R}^m$. We start by claiming that there exist constants $C_1, C_2>0$ such that\n\n$$\n\nC_1\\|x\\|_E \\leq\\|x\\| \\leq C_2\\|x\\|_E, \\forall x \\in \\mathbb{R}^m .\n\n$$\n\nAssuming (1) to be true, let us finish the problem. First let us show that $B$ is bounded w.r.t. $d_E$, which is how we call the Euclidean distance in $\\mathbb{R}^m$. Indeed, given $x \\in B,\\|x\\|_E \\leq \\frac{1}{C_1}\\|x\\| \\leq \\frac{1}{C_1}$. Hence $B \\subset\\left\\{x \\in \\mathbb{R}^m: d_E(x, 0)<\\frac{1}{C_1}+1\\right\\}$, which means $B$ is bounded w.r.t $d_E$.\n\nNow let us show that $B$ is closed w.r.t. $d_E$. Let $x_n \\rightarrow x$ w.r.t. $d_E$, where $x_n \\in B$. Notice that this implies that $x_n \\rightarrow x$ w.r.t. $d(x, y)=\\|x-y\\|$, the distance coming from $\\|\\cdot\\|$, since by (1) we have\n\n$$\n\nd\\left(x_n, x\\right)=\\left\\|x_n-x\\right\\| \\leq C_2\\left\\|x_n-x\\right\\|_E \\rightarrow 0 .\n\n$$\n\nAlso, notice that\n\n$$\n\n\\|x\\| \\leq\\left\\|x_n-x\\right\\|+\\left\\|x_n\\right\\| \\leq\\left\\|x_n-x\\right\\|+1,\n\n$$\n\nhence passing to the limit we obtain that $\\|x\\| \\leq 1$, therefore $x \\in B$ and so $B$ is closed w.r.t. $d_E$. Since $B$ is closed and bounded w.r.t. $d_E$, it must be compact. Now we claim that the identity function, $i d:\\left(\\mathbb{R}^m, d_E\\right) \\rightarrow\\left(\\mathbb{R}^m, d\\right)$ where $\\left(\\mathbb{R}^m, d_E\\right)$ means we are using the distance $d_E$ in $\\mathbb{R}^m$ and $\\left(\\mathbb{R}^m, d\\right)$ means we are using the distance $d$ in $\\mathbb{R}^m$, is a homeomorphism. This follows by (1), since $i d$ is always a bijection, and it is continuous and its inverse is continuous by (1) (if $x_n \\rightarrow x$ w.r.t. $d_E$, then $x_n \\rightarrow x$ w.r.t. $d$ and vice-versa, by (1)). By a result we saw in class, since $B$ is compact in $\\left(\\mathbb{R}^m, d_E\\right)$ and $i d$ is a homeomorphism, then $i d(B)=B$ is compact w.r.t. $d$.\n\n\n\nWe are left with proving (1). Notice that it suffices to prove that $C_1 \\leq\\|x\\| \\leq$ $C_2, \\forall x \\in \\mathbb{R}^m$ with $\\|x\\|_E=1$. Indeed, if this is true, given $x \\in \\mathbb{R}^m$, either $\\|x\\|_E=0$ (which implies $x=0$ and (1) holds in this case), or $x /\\|x\\|_E=y$ is such that $\\|y\\|_E=1$, so $C_1 \\leq\\|y\\| \\leq C_2$, which implies $C_1\\|x\\|_E \\leq\\|x\\| \\leq C_2\\|x\\|_E$.\n\nWe want to show now that $\\|\\cdot\\|$ is continuous w.r.t. $d_E$, that is, given $\\varepsilon>0$ and $x \\in \\mathbb{R}^m$, there exists $\\delta>0$ such that if $d_E(x, y)<\\delta$, then $\\|\\mid x\\|-\\|y\\| \\|<\\varepsilon$.\n\n\n\nBy the triangle inequality, $\\|x\\|-\\|y\\| \\leq\\|x-y\\|$, and $\\|y\\|-\\|x\\| \\leq\\|x-y\\|$, therefore\n\n$$\n\n|\\|x||-\\| y|\\|\\leq\\| x-y \\| .\n\n$$\n\nWriting now $x=\\sum_{i=1}^m a_i e_i, y=\\sum_{i=1}^m b_i e_i$, where $e_i=(0, \\ldots, 1,0, \\ldots, 0)$ (with 1 in the i-th component), we obtain by the triangle inequality,\n\n$$\n\n\\begin{aligned}\n\n\\|x-y\\| & =\\left\\|\\sum_{i=1}^m\\left(a_i-b_i\\right) e_i\\right\\| \\leq \\sum_{i=1}^m\\left|a_i-b_i\\left\\|\\left|\\left\\|e_i\\right\\| \\leq \\max _{i=1, \\ldots, m}\\left\\|e_i\\right\\| \\sum_{i=1}^m\\right| a_i-b_i \\mid\\right.\\right. \\\\\n\n& =\\max _{i=1, \\ldots, m}\\left\\|e_i\\right\\| d_{s u m}(x, y) \\leq \\max _{i=1, \\ldots, m}\\left\\|e_i\\right\\| m d_{\\max }(x, y) \\\\\n\n& \\leq \\max _{i=1, \\ldots, m}\\left\\|e_i\\right\\| m d_E(x, y) .\n\n\\end{aligned}\n\n$$\n\nLet $\\delta=\\frac{\\varepsilon}{m \\max _{i=1, \\ldots, m}\\left\\|e_i\\right\\|}$. Then if $d_E(x, y)<\\delta,\\|x\\|-\\|y\\|||<\\varepsilon$.\n\nSince $\\|\\cdot\\|$ is continuous w.r.t. $d_E$ and $K=\\left\\{x \\in \\mathbb{R}^m:\\|x\\|_E=1\\right\\}$ is compact w.r.t. $d_E$, then the function $\\|\\cdot\\|$ achieves a maximum and a minimum value on $K$. Call $C_1=\\min _{x \\in K}\\|x\\|, C_2=\\max _{x \\in K}\\|x\\|$. Then\n\n$$\n\nC_1 \\leq\\|x\\| \\leq C_2, \\forall x \\in \\mathbb{R}^m \\text { such that }\\|x\\|_E=1,\n\n$$\n\nwhich is what we needed.\n\n\\end{proof}"}
{"id": "Pugh|exercise_2_57", "formal_statement": "theorem exercise_2_57 {X : Type*} [topological_space X]\n  : \u2203 (S : set X), is_connected S \u2227 \u00ac is_connected (interior S) :=", "src_header": "import .common\n\nopen set real filter function ring_hom topological_space\n\nopen_locale big_operators\nopen_locale filter\nopen_locale topology \nnoncomputable theory \n\n", "nl_statement": "Show that if $S$ is connected, it is not true in general that its interior is connected.\n", "nl_proof": "\\begin{proof}\n\n    Consider $X=\\mathbb{R}^2$ and\n\n$$\n\nA=([-2,0] \\times[-2,0]) \\cup([0,2] \\times[0,2])\n\n$$\n\nwhich is connected, while $\\operatorname{int}(A)$ is not connected.\n\nTo see this consider the continuous function $f: \\mathbb{R}^2 \\rightarrow \\mathbb{R}$ is defined by $f(x, y)=x+y$. Let $U=f^{-1}(0,+\\infty)$ which is open in $\\mathbb{R}^2$ and so $U \\cap \\operatorname{int}(A)$ is open in $\\operatorname{int}(A)$. Also, since $(0,0) \\notin \\operatorname{int}(A)$, so for all $(x, y) \\in \\operatorname{int}(A), f(x, y) \\neq 0$ and $U \\cap \\operatorname{int}(A)=f^{-1}[0,+\\infty) \\cap \\operatorname{int}(A)$ is closed in $\\operatorname{int}(A)$. Furthermore, $(1,1)=f^{-1}(2) \\in U \\cap \\operatorname{int}(A)$ shows that $U \\cap \\operatorname{int}(A) \\neq \\emptyset$ while $(-1,-1) \\in \\operatorname{int}(A)$ and $(-1,-1) \\notin U$ shows that $U \\cap \\operatorname{int}(A) \\neq \\operatorname{int}(A)$.\n\n\\end{proof}"}
{"id": "Pugh|exercise_2_126", "formal_statement": "theorem exercise_2_126 {E : set \u211d}\n  (hE : \u00ac set.countable E) : \u2203 (p : \u211d), cluster_pt p (\ud835\udcdf E) :=", "src_header": "import .common\n\nopen set real filter function ring_hom topological_space\n\nopen_locale big_operators\nopen_locale filter\nopen_locale topology \nnoncomputable theory \n\n", "nl_statement": "Suppose that $E$ is an uncountable subset of $\\mathbb{R}$. Prove that there exists a point $p \\in \\mathbb{R}$ at which $E$ condenses.\n", "nl_proof": "\\begin{proof}\n\n    I think this is the proof by contrapositive that you were getting at.\n\nSuppose that $E$ has no limit points at all. Pick an arbitrary point $x \\in E$. Then $x$ cannot be a limit point, so there must be some $\\delta>0$ such that the ball of radius $\\delta$ around $x$ contains no other points of $E$ :\n\n$$\n\nB_\\delta(x) \\cap E=\\{x\\}\n\n$$\n\nCall this \"point 1 \". For the next point, take the closest element to $x$ and on its left; that is, choose the point\n\n$$\n\n\\max [E \\cap(-\\infty, x)]\n\n$$\n\nif it exists (that is important - if not, skip to the next step). Note that by the argument above, this supremum, should it exist, cannot equal $x$ and is therefore a new point in $E$.\n\n\n\nCall this \"point 2 \". Now take the first point to the right of $x$ for \"point 3 \". Take the first point to the left of point 2 for \"point 4 \". And so on, ad infinitum.\n\n\n\nThis gives a countable list of unique points; we must show that it exhausts the entire set $E$. Suppose not. Suppose there is some element $a<x$ which is never included in the list (picking $a$ on the negative side of $x$ is arbitrary, and the same argument would work for the second case). Then the element closest and to the right of $a$ in $E$ (which exists, by the no-limit-points argument at the beginning) is also not in the list; if it was, $a$ would have been in one of the next two spots. And same with that point (call it $a_1$ ); there is a closest $a_2>a_1 \\in E$ such that $a_2$ is not in the list. Repeating, we generate an infinite monotone-increasing sequence $\\left\\{a_i\\right\\}$ of elements in $E$ and not in the list, which is clearly bounded above by $x$. By the Monotone\n\nConvergence Theorem this sequence has a limit. But that means the sequence $\\left\\{a_i\\right\\} \\subset E$ converges to a limit, and hence $E$ has a limit point, contradicting the assumption. Therefore our list exhausts $E$, and we have enumerated all its elements.\n\n\\end{proof}"}
{"id": "Pugh|exercise_3_4", "formal_statement": "theorem exercise_3_4 (n : \u2115) :\n  tendsto (\u03bb n, (sqrt (n + 1) - sqrt n)) at_top (\ud835\udcdd 0) :=", "src_header": "import .common\n\nopen set real filter function ring_hom topological_space\n\nopen_locale big_operators\nopen_locale filter\nopen_locale topology \nnoncomputable theory \n\n", "nl_statement": "Prove that $\\sqrt{n+1}-\\sqrt{n} \\rightarrow 0$ as $n \\rightarrow \\infty$.\n", "nl_proof": "\\begin{proof}\n\n    $$\n\n\\sqrt{n+1}-\\sqrt{n}=\\frac{(\\sqrt{n+1}-\\sqrt{n})(\\sqrt{n+1}+\\sqrt{n})}{\\sqrt{n+1}+\\sqrt{n}}=\\frac{1}{\\sqrt{n+1}+\\sqrt{n}}<\\frac{1}{2 \\sqrt{n}}\n\n$$\n\n\\end{proof}"}
{"id": "Pugh|exercise_3_63b", "formal_statement": "theorem exercise_3_63b (p : \u211d) (f : \u2115 \u2192 \u211d) (hp : p \u2264 1)\n  (h : f = \u03bb k, (1 : \u211d) / (k * (log k) ^ p)) :\n  \u00ac \u2203 l, tendsto f at_top (\ud835\udcdd l) :=", "src_header": "import .common\n\nopen set real filter function ring_hom topological_space\n\nopen_locale big_operators\nopen_locale filter\nopen_locale topology \nnoncomputable theory \n\n", "nl_statement": "Prove that $\\sum 1/k(\\log(k))^p$ diverges when $p \\leq 1$.\n", "nl_proof": "\\begin{proof} \n\n    Using the integral test, for a set $a$, we see\n\n$$\n\n\\lim _{b \\rightarrow \\infty} \\int_a^b \\frac{1}{x \\log (x)^c} d x=\\lim _{b \\rightarrow \\infty}\\left(\\frac{\\log (b)^{1-c}}{1-c}-\\frac{\\log (a)^{1-c}}{1-c}\\right)\n\n$$\n\nwhich goes to infinity if $c \\leq 1$ and converges if $c>1$. Thus,\n\n$$\n\n\\sum_{n=2}^{\\infty} \\frac{1}{n \\log (n)^c}\n\n$$\n\nconverges if and only if $c>1$. \n\n\\end{proof}"}
{"id": "Herstein|exercise_2_1_18", "formal_statement": "theorem exercise_2_1_18 {G : Type*} [group G] \n  [fintype G] (hG2 : even (fintype.card G)) :\n  \u2203 (a : G), a \u2260 1 \u2227 a = a\u207b\u00b9 :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "If $G$ is a finite group of even order, show that there must be an element $a \\neq e$ such that $a=a^{-1}$.\n", "nl_proof": "\\begin{proof}\n\n    First note that $a=a^{-1}$ is the same as saying $a^2=e$, where $e$ is the identity. I.e. the statement is that there exists at least one element of order 2 in $G$.\n\nEvery element $a$ of $G$ of order at least 3 has an inverse $a^{-1}$ that is not itself -- that is, $a \\neq a^{-1}$. So the subset of all such elements has an even cardinality (/size). There's exactly one element with order 1 : the identity $e^1=e$. So $G$ contains an even number of elements -call it $2 k$-- of which an even number are elements of order 3 or above -- call that $2 n$ where $n<k$-- and exactly one element of order 1 . Hence the number of elements of order 2 is\n\n$$\n\n2 k-2 n-1=2(k-n)-1\n\n$$\n\nThis cannot equal 0 as $2(k-n)$ is even and 1 is odd. Hence there's at least one element of order 2 in $G$, which concludes the proof.\n\n\\end{proof}"}
{"id": "Herstein|exercise_2_1_26", "formal_statement": "theorem exercise_2_1_26 {G : Type*} [group G] \n  [fintype G] (a : G) : \u2203 (n : \u2115), a ^ n = 1 :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "If $G$ is a finite group, prove that, given $a \\in G$, there is a positive integer $n$, depending on $a$, such that $a^n = e$.\n", "nl_proof": "\\begin{proof}\n\n    Because there are only a finite number of elements of $G$, it's clear that the set $\\left\\{a, a^2, a^3, \\ldots\\right\\}$ must be a finite set and in particular, there should exist some $i$ and $j$ such that $i \\neq j$ and $a^i=a^j$. WLOG suppose further that $i>j$ (just reverse the roles of $i$ and $j$ otherwise). Then multiply both sides by $\\left(a^j\\right)^{-1}=a^{-j}$ to get\n\n$$\n\na^i * a^{-j}=a^{i-j}=e\n\n$$\n\nThus the $n=i-j$ is a positive integer such that $a^n=e$.\n\n\\end{proof}"}
{"id": "Herstein|exercise_2_2_3", "formal_statement": "theorem exercise_2_2_3 {G : Type*} [group G]\n  {P : \u2115 \u2192 Prop} {hP : P = \u03bb i, \u2200 a b : G, (a*b)^i = a^i * b^i}\n  (hP1 : \u2203 n : \u2115, P n \u2227 P (n+1) \u2227 P (n+2)) : comm_group G :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "If $G$ is a group in which $(a b)^{i}=a^{i} b^{i}$ for three consecutive integers $i$, prove that $G$ is abelian.\n", "nl_proof": "\\begin{proof}\n\n    Let $G$ be a group, $a, b \\in G$ and $i$ be any integer. Then from given condition,\n\n$$\n\n\\begin{aligned}\n\n(a b)^i & =a^i b^i \\\\\n\n(a b)^{i+1} & =a^{i+1} b^{i+1} \\\\\n\n(a b)^{i+2} & =a^{i+2} b^{i+2}\n\n\\end{aligned}\n\n$$\n\nFrom first and second, we get\n\n$$\n\na^{i+1} b^{i+1}=(a b)^i(a b)=a^i b^i a b \\Longrightarrow b^i a=a b^i\n\n$$\n\nFrom first and third, we get\n\n$$\n\na^{i+2} b^{i+2}=(a b)^i(a b)^2=a^i b^i a b a b \\Longrightarrow a^2 b^{i+1}=b^i a b a\n\n$$\n\nThis gives\n\n$$\n\na^2 b^{i+1}=a\\left(a b^i\\right) b=a b^i a b=b^i a^2 b\n\n$$\n\nFinally, we get\n\n$$\n\nb^i a b a=b^i a^2 b \\Longrightarrow b a=a b\n\n$$\n\nThis shows that $G$ is Abelian.\n\n\\end{proof}"}
{"id": "Herstein|exercise_2_2_6c", "formal_statement": "theorem exercise_2_2_6c {G : Type*} [group G] {n : \u2115} (hn : n > 1) \n  (h : \u2200 (a b : G), (a * b) ^ n = a ^ n * b ^ n) :\n  \u2200 (a b : G), (a * b * a\u207b\u00b9 * b\u207b\u00b9) ^ (n * (n - 1)) = 1 :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $G$ be a group in which $(a b)^{n}=a^{n} b^{n}$ for some fixed integer $n>1$ for all $a, b \\in G$. For all $a, b \\in G$, prove that $\\left(a b a^{-1} b^{-1}\\right)^{n(n-1)}=e$. \n", "nl_proof": "\\begin{proof}\n\n    We start with the following two intermediate results.\n\n(1) $(a b)^{n-1}=b^{n-1} a^{n-1}$.\n\n(2) $a^n b^{n-1}=b^{n-1} a^n$.\n\nTo prove (1), notice by the given condition for all $a, b \\in G$\n\n$(b a)^n=b^n a^n$, for some fixed integers $n>1$.\n\nThen,\n\n$(b a)^n=b^n a^n \\Longrightarrow b .(a b)(a b) \\ldots .(a b) . a=b\\left(b^{n-1} a^{n-1}\\right) a$, where $(a b)$ occurs $n-1$ times $\\Longrightarrow(a b)^{n-1}=b^{n-1} a^{n-1}$, by cancellation law.\n\nHence, for all $a, b \\in G$\n\n$$\n\n(a b)^{n-1}=b^{n-1} a^{n-1} .\n\n$$\n\nTo prove (2), notice by the given condition for all $a, b \\in G$\n\n$(b a)^n=b^n a^n$, for some fixed integers $n>1$.\n\nThen we have\n\n$$\n\n\\begin{aligned}\n\n& (b a)^n=b^n a^n \\\\\n\n\\Longrightarrow & b \\cdot(a b)(a b) \\ldots(a b) \\cdot a=b\\left(b^{n-1} a^{n-1}\\right) a, \\text { where }(a b) \\text { occurs } n-1 \\text { times } \\\\\n\n\\Longrightarrow & (a b)^{n-1}=b^{n-1} a^{n-1}, \\text { by cancellation law } \\\\\n\n\\Longrightarrow & (a b)^{n-1}(a b)=\\left(b^{n-1} a^{n-1}\\right)(a b) \\\\\n\n\\Longrightarrow & (a b)^n=b^{n-1} a^n b \\\\\n\n\\Longrightarrow & a^n b^n=b^{n-1} a^n b, \\text { given condition } \\\\\n\n\\Longrightarrow & a^n b^{n-1}=b^{n-1} a^n, \\text { by cancellation law. }\n\n\\end{aligned}\n\n$$\n\nTherefore for all $a, b \\in G$ we have\n\n$$\n\na^n b^{n-1}=b^{n-1} a^n\n\n$$\n\nIn order to show that\n\n$$\n\n\\left(a b a^{-1} b^{-1}\\right)^{n(n-1)}=e, \\text { for all } a, b \\in G\n\n$$\n\nit is enough to show that\n\n$$\n\n(a b)^{n(n-1)}=(b a)^{n(n-1)}, \\quad \\forall x, y \\in G .\n\n$$\n\nStep 3\n\nThis is because of\n\n$$\n\n\\begin{aligned}\n\n(a b)^{n(n-1)}=(b a)^{n(n-1)} & \\left.\\Longrightarrow(b a)^{-1}\\right)^{n(n-1)}(a b)^{n(n-1)}=e \\\\\n\n& \\Longrightarrow\\left(a^{-1} b^{-1}\\right)^{n(n-1)}(a b)^{n(n-1)}=e \\\\\n\n& \\Longrightarrow\\left(\\left(a^{-1} b^{-1}\\right)^n\\right)^{n-1}\\left((a b)^n\\right)(n-1)=e \\\\\n\n& \\Longrightarrow\\left((a b)^n\\left(a^{-1} b^{-1}\\right)^n\\right)^{n-1}=e, \\text { by }(1) \\\\\n\n& \\Longrightarrow\\left(a b a^{-1} b^{-1}\\right)^{n(n-1)}=e, \\text { ( given condition) }\n\n\\end{aligned}\n\n$$\n\nNow, it suffices to show that\n\n$$\n\n(a b)^{n(n-1)}=(b a)^{n(n-1)}, \\quad \\forall x, y \\in G .\n\n$$\n\nNow, we have\n\n$$\n\n\\begin{aligned}\n\n(a b)^{n(n-1)} & =\\left(a^n b^n\\right)^{n-1}, \\text { by the given condition } \\\\\n\n& =\\left(a^n b^{n-1} b\\right)^{n-1} \\\\\n\n& =\\left(b^{n-1} a^n b\\right)^{n-1}, \\text { by }(2) \\\\\n\n& =\\left(a^n b\\right)^{n-1}\\left(b^{n-1}\\right)^{n-1}, \\text { by }(1) \\\\\n\n& =b^{n-1}\\left(a^n\\right)^{n-1}\\left(b^{n-1}\\right)^{n-1}, \\text { by }(1) \\\\\n\n& =\\left(b^{n-1}\\left(a^{n-1}\\right)^n\\right)\\left(b^{n-1}\\right)^{n-1} \\\\\n\n& =\\left(a^{n-1}\\right)^n b^{n-1}\\left(b^{n-1}\\right)^{n-1}, \\text { by }(2) \\\\\n\n& =\\left(a^{n-1}\\right)^n\\left(b^{n-1}\\right)^n \\\\\n\n& =\\left(a^{n-1} b^{n-1}\\right)^n, \\text { by }(1) \\\\\n\n& =(b a)^{n(n-1)}, \\text { by }(1) .\n\n\\end{aligned}\n\n$$\n\nThis completes our proof.\n\n\\end{proof}"}
{"id": "Herstein|exercise_2_3_16", "formal_statement": "theorem exercise_2_3_16 {G : Type*} [group G]\n  (hG : \u2200 H : subgroup G, H = \u22a4 \u2228 H = \u22a5) :\n  is_cyclic G \u2227 \u2203 (p : \u2115) (fin : fintype G), nat.prime p \u2227 @card G fin = p :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "If a group $G$ has no proper subgroups, prove that $G$ is cyclic of order $p$, where $p$ is a prime number.\n", "nl_proof": "\\begin{proof}\n\n    Case-1: $G=(e), e$ being the identity element in $G$. Then trivially $G$ is cyclic.\n\n    Case-2: $G \\neq(e)$. Then there exists an non-identity element in $G.$ Let us consider an non-identity element in $G$, say $a\\neq (e)$. Now look at the cyclic subgroup generated by $a$, that is, $\\langle a\\rangle$. Since\n\n    $a\\neq (e) \\in G,\\langle a\\rangle$ is a subgroup of $G$.\n\nIf $G \\neq\\langle a\\rangle$ then $\\langle a\\rangle$ is a proper non-trivial subgroup of $G$, which is an impossibility. Therfore we must have\n\n$$\n\nG=\\langle a\\rangle .\n\n$$\n\nThis implies, $G$ is a cyclic group generated by $a$. Then it follows that every non-identity element of $G$ is a generator of $G$. Now we claim that $G$ is finite.\n\n\\end{proof}"}
{"id": "Herstein|exercise_2_5_23", "formal_statement": "theorem exercise_2_5_23 {G : Type*} [group G] \n  (hG : \u2200 (H : subgroup G), H.normal) (a b : G) :\n  \u2203 (j : \u2124) , b*a = a^j * b:=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $G$ be a group such that all subgroups of $G$ are normal in $G$. If $a, b \\in G$, prove that $ba = a^jb$ for some $j$.\n", "nl_proof": "\\begin{proof}\n\n    Let $G$ be a group where each subgroup is normal in $G$. let $a, b \\in G$.\n\n$$\n\n\\begin{aligned}\n\n    \\langle a\\rangle\\triangleright  G  &\\Rightarrow b \\cdot\\langle a\\rangle=\\langle a\\rangle \\cdot b . \\\\\n\n& \\Rightarrow \\quad b \\cdot a=a^j \\cdot b \\text { for some } j \\in \\mathbb{Z}.\n\n\\end{aligned}\n\n$$\n\n(hence for $a_1 b \\in G \\quad a^j b=b \\cdot a$ ).\n\n\\end{proof}"}
{"id": "Herstein|exercise_2_5_31", "formal_statement": "theorem exercise_2_5_31 {G : Type*} [comm_group G] [fintype G]\n  {p m n : \u2115} (hp : nat.prime p) (hp1 : \u00ac p \u2223 m) (hG : card G = p^n*m)\n  {H : subgroup G} [fintype H] (hH : card H = p^n) : \n  characteristic H :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Suppose that $G$ is an abelian group of order $p^nm$ where $p \\nmid m$ is a prime.  If $H$ is a subgroup of $G$ of order $p^n$, prove that $H$ is a characteristic subgroup of $G$.\n", "nl_proof": "\\begin{proof}\n\n    Let $G$ be an abelian group of order $p^n m$, such that $p \\nmid m$. Now, Given that $H$ is a subgroup of order $p^n$. Since $G$ is abelian $H$ is normal. Now we want to prove that $H$ is a characterestic subgroup, that is $\\phi(H)=H$ for any automorphism $\\phi$ of $G$. Now consider $\\phi(H)$. Clearly $|\\phi(H)|=p^n$. Suppose $\\phi(H) \\neq H$, then $|H \\cap \\phi(H)|=p^s$, where $s<n$. Consider $H \\phi(H)$, this is a subgroup of $G$ as $H$ is normal. Also $|H \\phi(H)|=\\frac{|H||\\phi(H)|}{|H \\cap \\phi(H)|}=\\frac{p^{2 n}}{p^s}=p^{2 n-s}$, where $2 n-s>n$. By lagrange's theorem then $p^{2 n-s}\\left|p^n m \\Longrightarrow p^{n-s}\\right| m \\Longrightarrow p \\mid m$-contradiction. So $\\phi(H)=H$, and $H$ is characterestic subgroup of $G$.\n\n\\end{proof}"}
{"id": "Herstein|exercise_2_5_43", "formal_statement": "theorem exercise_2_5_43 (G : Type*) [group G] [fintype G]\n  (hG : card G = 9) :\n  comm_group G :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that a group of order 9 must be abelian.\n", "nl_proof": "\\begin{proof}\n\n    We use the result from problem 40 which is as follows: Suppose $G$ is a group, $H$ is a subgroup and $|G|=n$ and $n \\nmid\\left(i_G(H)\\right) !$. Then there exists a normal subgroup $\\$ K \\backslash$ neq $\\{$ e $\\} \\$$ and $K \\subseteq H$.\n\nSo, we have now a group $G$ of order 9. Suppose that $G$ is cyclic, then $G$ is abelian and there is nothing more to prove. Suppose that $G$ s not cyclic,then there exists an element $a$ of order 3 , and $A=\\langle a\\rangle$. Now $i_G(A)=3$, now $9 \\nmid 3$ !, hence by the above result there is a normal subgroup $K$, non-trivial and $K \\subseteq A$. But $|A|=3$, a prime order subgroup, hence has no non-trivial subgroup, so $K=A$. So $A$ is normal subgroup. Now since $G$ is not cyclic any non-identity element is of order 3.So Let $a(\\neq$ $e) \\in G$.Consider $A=\\langle a\\rangle$. As shown before $A$ is normal. $a$ commutes with any if its powers. Now Let $b \\in G$ such that $b \\notin A$. Then $b a b^{-1} \\in A$ and hence $b a b^{-1}=a^i$.This implies $a=b^3 a b^{-3}=a^{i^3} \\Longrightarrow a^{i^3-1}=e$. So, 3 divides $i^3-1$. Also by fermat's little theorem 3 divides $i^2-1$.So 3 divides $i-1$. But $0 \\leq i \\leq 2$. So $i=1$, is the only possibility and hence $a b=b a$. So $a \\in Z(G)$ as $b$ was arbitrary. Since $a$ was arbitrary $G=Z(G)$. Hence $G$ is abelian.\n\n\\end{proof}"}
{"id": "Herstein|exercise_2_5_52", "formal_statement": "theorem exercise_2_5_52 {G : Type*} [group G] [fintype G]\n  (\u03c6 : G \u2243* G) {I : finset G} (hI : \u2200 x \u2208 I, \u03c6 x = x\u207b\u00b9)\n  (hI1 : 0.75 * card G \u2264 card I) : \n  \u2200 x : G, \u03c6 x = x\u207b\u00b9 \u2227 \u2200 x y : G, x*y = y*x :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $G$ be a finite group and $\\varphi$ an automorphism of $G$ such that $\\varphi(x) = x^{-1}$ for more than three-fourths of the elements of $G$. Prove that $\\varphi(y) = y^{-1}$ for all $y \\in G$, and so $G$ is abelian.\n", "nl_proof": "\\begin{proof}\n\nLet us start with considering $b$ to be an arbitrary element in $A$. \n\n\n\n1. Show that $\\left|A \\cap\\left(b^{-1} A\\right)\\right|>\\frac{|G|}{2}$, where\n\n$$\n\nb^{-1} A=\\left\\{b^{-1} a \\mid a \\in A\\right\\}\n\n$$\n\nFirst notice that if we consider a map $f: A \\rightarrow b^{-1} A$ defined by $f(a)=b^{-1} a$, for all $a \\in A$, then $f$ is a 1-1 map and so $\\left|b^{-1} A\\right| \\geq|A|>\\frac{3}{4}|G|$. Now using inclusion-exclusion principle we have\n\n$$\n\n\\left|A \\cap\\left(b^{-1} A\\right)\\right|=|A|+\\left|b^{-1} A\\right|-\\left|A \\cup\\left(b^{-1} A\\right)\\right|>\\frac{3}{4}|G|+\\frac{3}{4}|G|-|G|=\\frac{1}{2}|G|\n\n$$\n\n2. Argue that $A \\cap\\left(b^{-1} A\\right) \\subseteq C(b)$, where $C(b)$ is the centralizer of $b$ in $G$.\n\n\n\nSuppose $x \\in A \\cap\\left(b^{-1} A\\right)$, that means, $x \\in A$ and $x \\in b^{-1} A$. Thus there exist an element $a \\in A$ such that $x=$ $b^{-1} a$, which gives us $x b=a \\in A$. Now notice that $x, b \\in A$ and $x b \\in A$, therefore we get\n\n$$\n\n\\phi(x b)=(x b)^{-1} \\Longrightarrow \\phi(x) \\phi(b)=(x b)^{-1} \\Longrightarrow x^{-1} b^{-1}=b^{-1} x^{-1} \\Longrightarrow x b=b x\n\n$$\n\nTherefore, we get $x b=b x$, for any $x \\in A \\cap\\left(b^{-1} A\\right)$, that means, $x \\in C(b)$.\n\n\n\n3. Argue that $C(b)=G$.\n\nWe know that centralizer of an element in a group $G$ is a subgroup (See Page 53). Therefore $C(b)$ is a subgroup of $G$. From statements $\\mathbf{1}$ and $\\mathbf{2}$, we have\n\n$$\n\n|C(b)| \\geq\\left|A \\cap\\left(b^{-1} A\\right)\\right|>\\frac{|G|}{2}\n\n$$\n\nWe need to use the following remark to argue $C(b)=G$ from the above step.\n\nRemark. Let $G$ be a finite group and $H$ be a subgroup with more then $|G| / 2$ elements then $H=G$.\n\n\n\nProof of Remark. Suppose $|H|=p$ Then by Lagrange Theorem, there exist an $n \\in \\mathbb{N}$, such that $|G|=n p$, as $|H|$ divide $|G|$. Now by hypothesis $p>\\frac{G]}{2}$ gives us,\n\n$$\n\np>\\frac{|G|}{2} \\Longrightarrow n p>\\frac{n|G|}{2} \\Longrightarrow n<2 \\Longrightarrow n=1\n\n$$\n\nTherefore we get $H=G$.\n\n\n\nNow notice that $C(b)$ is a subgroup of $G$ with $C(b)$ having more than $|G| / 2$ elements. Therefore, $C(b)=G$.\n\n\n\n4. Show that $A \\in Z(G)$.\n\n\n\nWe know that $x \\in Z(G)$ if and only if $C(a)=G$. Now notice that, for any $b \\in A$ we have $C(b)=G$. Therefore, every element of $A$ is in the center of $G$, that means, $A \\subseteq Z(G)$.\n\n\n\n5. 5how that $Z(G)=G$.\n\n\n\nAs it is given that $|A|>\\frac{3|G|}{4}$ and $A \\leq|Z(G)|$, therefore we get\n\n$$\n\n|Z(G)|>\\frac{3}{4}|G|>\\frac{1}{2}|G| .\n\n$$\n\nAs $Z(G)$ is a subgroup of $G$, so by the above Remark we have $Z(G)=G$. Hence $G$ is abelian.\n\n\n\n6. Finally show that $A=G$.\n\n\n\nFirst notice that $A$ is a subgroup of $G$. To show this let $p, q \\in A$. Then we have\n\n$$\n\n\\phi(p q)=\\phi(p) \\phi(q)=p^{-1} q^{-1}=(q p)^{-1}=(p q)^{-1}, \\quad \\text { As } G \\text { is abelian. }\n\n$$\n\nTherefore, $p q \\in A$ and so we have $A$ is a subgroup of $G$. Again by applying the above remark. we get $A=G$. Therefore we have\n\n$$\n\n\\phi(y)=y^{-1}, \\quad \\text { for all } y \\in G\n\n$$\n\n\n\n\\end{proof}"}
{"id": "Herstein|exercise_2_7_7", "formal_statement": "theorem exercise_2_7_7 {G : Type*} [group G] {G' : Type*} [group G']\n  (\u03c6 : G \u2192* G') (N : subgroup G) [N.normal] : \n  (map \u03c6 N).normal  :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "If $\\varphi$ is a homomorphism of $G$ onto $G'$ and $N \\triangleleft G$, show that $\\varphi(N) \\triangleleft G'$.\n", "nl_proof": "\\begin{proof}\n\nWe first claim that $\\varphi(N)$ is a subgroup of $G'$. To see this, note that since $N$ is a subgroup of $G$, the identity element $e_G$ of $G$ belongs to $N$. Therefore, the element $\\varphi(e_G) \\in \\varphi(N)$, so $\\varphi(N)$ is a non-empty subset of $G'$.\n\n\n\nNow, let $a', b' \\in \\varphi(N)$. Then there exist elements $a, b \\in N$ such that $\\varphi(a) = a'$ and $\\varphi(b) = b'$. Since $N$ is a subgroup of $G$, we have $a, b \\in N$, so $ab^{-1} \\in N$. Thus, we have\n\n$$\\varphi(ab^{-1}) = \\varphi(a) \\varphi(b^{-1}) = a'b'^{-1} \\in \\varphi(N),$$\n\nwhich shows that $a', b' \\in \\varphi(N)$ implies $a'b'^{-1} \\in \\varphi(N)$. Therefore, $\\varphi(N)$ is a subgroup of $G'$.\n\n\n\nNext, we will show that $\\varphi(N)$ is a normal subgroup of $G'$. Let $\\varphi(N) = N'$, a subgroup of $G'$. Let $x' \\in G'$ and $h' \\in N'$. Since $\\varphi$ is onto, there exist elements $x \\in G$ and $h \\in N$ such that $\\varphi(x) = x'$ and $\\varphi(h) = h'$.\n\n\n\nSince $N$ is a normal subgroup of $G$, we have $xhx^{-1} \\in N$. Thus,\n\n$$\\varphi(xhx^{-1}) = \\varphi(x)\\varphi(h)\\varphi(x^{-1}) = x'h'x'^{-1} \\in \\varphi(N),$$\n\nwhich shows that $x' \\in G'$ and $h' \\in N'$ implies $x'h'x'^{-1} \\in \\varphi(N)$. Therefore, $\\varphi(N)$ is a normal subgroup of $G'$. This completes the proof.\n\n\\end{proof}"}
{"id": "Herstein|exercise_2_8_15", "formal_statement": "theorem exercise_2_8_15 {G H: Type*} [fintype G] [group G] [fintype H]\n  [group H] {p q : \u2115} (hp : nat.prime p) (hq : nat.prime q) \n  (h : p > q) (h1 : q \u2223 p - 1) (hG : card G = p*q) (hH : card G = p*q) :\n  G \u2243* H :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that if $p > q$ are two primes such that $q \\mid p - 1$, then any two nonabelian groups of order $pq$ are isomorphic.\n", "nl_proof": "\\begin{proof}\n\n    For a nonabelian group of order $p q$, the structure of the group $G$ is set by determining the relation $a b a^{-1}=b^{k^{\\frac{p-1}{q}}}$ for some generator $k$ of the cyclic group. Here we are using the fact that $k^{\\frac{p-1}{q}}$ is a generator for the unique subgroup of order $q$ in $U_p$ (a cyclic group of order $m$ has a unique subgroup of order $d$ for each divisor $d$ of $m$ ). The other possible generators of this subgroup are $k^{\\frac{l(p-1)}{q}}$ for each $1 \\leq l \\leq q-1$, so these give potentially new group structures. Let $G^{\\prime}$ be a group with an element $c$ of order $q$, an element $d$ of order $p$ with structure defined by the relation $c d c^{-1}=d^{k^{\\frac{l(p-1)}{q}}}$. We may then define\n\n$$\n\n\\begin{aligned}\n\n\\phi: G^{\\prime} & \\rightarrow G \\\\\n\nc & \\mapsto a^l \\\\\n\nd & \\mapsto b\n\n\\end{aligned}\n\n$$\n\nsince $c$ and $a^l$ have the same order and $b$ and $d$ have the same order this is a well defined function.\n\nSince\n\n$$\n\n\\begin{aligned}\n\n\\phi(c) \\phi(d) \\phi(c)^{-1} & =a^l b a^{-l} \\\\\n\n& =b^{\\left(k^{\\frac{p-1}{q}}\\right)^l} \\\\\n\n& =b^{k^{\\frac{l(p-1)}{q}}} \\\\\n\n& =\\phi(d)^{k^{\\frac{l(p-1)}{q}}}\n\n\\end{aligned}\n\n$$\n\n$\\phi\\left(c^i d^j\\right)=a^{l i} b^j=e$ only if $i=j=0$, so $\\phi$ is 1-to-l. Therefore $G$ and $G^{\\prime}$ are isomorphic and so up to isomorphism there is only one nonabelian group of order $p q$.\n\n\\end{proof}"}
{"id": "Herstein|exercise_2_10_1", "formal_statement": "theorem exercise_2_10_1 {G : Type*} [group G] (A : subgroup G) \n  [A.normal] {b : G} (hp : nat.prime (order_of b)) :\n  A \u2293 (closure {b}) = \u22a5 :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $A$ be a normal subgroup of a group $G$, and suppose that $b \\in G$ is an element of prime order $p$, and that $b \\not\\in A$. Show that $A \\cap (b) = (e)$.\n", "nl_proof": "\\begin{proof}\n\nIf $b \\in G$ has order $p$, then $(b)$ is a cyclic group of order $p$. Since $A$ is a subgroup of $G$, we have $A \\cap (b)$ is a subgroup of $G$. Also, $A \\cap (b) \\subseteq (b)$. So $A \\cap (b)$ is a subgroup of $(b)$. Since $(b)$ is a cyclic group of order $p$, the only subgroups of $(b)$ are $(e)$ and $(b)$ itself.\n\n\n\nTherefore, either $A \\cap (b) = (e)$ or $A \\cap (b) = (b)$. If $A \\cap (b) = (e)$, then we are done. Otherwise, if $A \\cap (b) = (b)$, then $A \\subseteq (b)$. Since $A$ is a subgroup of $G$ and $A \\subseteq (b)$, it follows that $A$ is a subgroup of $(b)$.\n\n\n\nSince the only subgroups of $(b)$ are $(e)$ and $(b)$ itself, we have either $A = (e)$ or $A = (b)$. If $A = (e)$, then $A \\cap (b) = (e)$ and we are done. But if $A = (b)$, then $b \\in A$ as $b \\in (b)$, which contradicts our hypothesis that $b \\notin A$. So $A \\neq (b)$.\n\n\n\nHence $A \\cap (b) \\neq (b)$. Therefore, $A \\cap (b) = (e)$. This completes our proof.\n\n\\end{proof}"}
{"id": "Herstein|exercise_2_11_7", "formal_statement": "theorem exercise_2_11_7 {G : Type*} [group G] {p : \u2115} (hp : nat.prime p)\n  {P : sylow p G} (hP : P.normal) : \n  characteristic (P : subgroup G) :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "If $P \\triangleleft G$, $P$ a $p$-Sylow subgroup of $G$, prove that $\\varphi(P) = P$ for every automorphism $\\varphi$ of $G$.\n", "nl_proof": "\\begin{proof}\n\n    Let $\\phi$ be an automorphism of $G$. Let $P$ be a normal sylow p-subgroup. $\\phi(P)$ is also a sylow-p subgroup. But since $P$ is normal, it is unique. Hence $\\phi(P)=P$.\n\n\\end{proof}"}
{"id": "Herstein|exercise_3_2_21", "formal_statement": "theorem exercise_3_2_21 {\u03b1 : Type*} [fintype \u03b1] {\u03c3 \u03c4: equiv.perm \u03b1} \n  (h1 : \u2200 a : \u03b1, \u03c3 a = a \u2194 \u03c4 a \u2260 a) (h2 : \u03c4 \u2218 \u03c3 = id) : \n  \u03c3 = 1 \u2227 \u03c4 = 1 :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "If $\\sigma, \\tau$ are two permutations that disturb no common element and $\\sigma \\tau = e$, prove that $\\sigma = \\tau = e$.\n", "nl_proof": "\\begin{proof}\n\n    Note that $\\sigma \\tau=e$ can equivalentnly be phrased as $\\tau$ being the inverse of $\\sigma$. Our statement is then equivalent to the statement that an inverse of a nonidentity permutation disturbs at least one same element as that permutation. To prove this, let $\\sigma$ be a nonidentity permutation, then let $\\left(i_1 \\cdots i_n\\right)$ be a cycle in $\\sigma$. Then we have that\n\n$$\n\n\\sigma\\left(i_1\\right)=i_2, \\sigma\\left(i_2\\right)=i_2, \\ldots, \\sigma\\left(i_{n-1}\\right)=i_n, \\sigma\\left(i_n\\right)=i_1,\n\n$$\n\nbut then also\n\n$$\n\ni_1=\\tau\\left(i_2\\right), i_2=\\tau\\left(i_3\\right), \\ldots, i_{n-1}=\\tau\\left(i_n\\right), i_n=\\tau\\left(i_1\\right),\n\n$$\n\ni.e. its inverse disturbs $i_1, \\ldots, i_n$.\n\n\\end{proof}"}
{"id": "Herstein|exercise_4_1_34", "formal_statement": "theorem exercise_4_1_34 : equiv.perm (fin 3) \u2243* general_linear_group (fin 2) (zmod 2) :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $T$ be the group of $2\\times 2$ matrices $A$ with entries in the field $\\mathbb{Z}_2$ such that $\\det A$ is not equal to 0. Prove that $T$ is isomorphic to $S_3$, the symmetric group of degree 3.\n", "nl_proof": "\\begin{proof}\n\n    The order of $T$ is $2^4-2^3-2^2+2=6$; we now find those six matrices:\n\n$$\n\n\\begin{array}{ll}\n\nA_1=\\left(\\begin{array}{ll}\n\n1 & 0 \\\\\n\n0 & 1\n\n\\end{array}\\right), & A_2=\\left(\\begin{array}{ll}\n\n0 & 1 \\\\\n\n1 & 0\n\n\\end{array}\\right) \\\\\n\nA_3=\\left(\\begin{array}{ll}\n\n1 & 0 \\\\\n\n1 & 1\n\n\\end{array}\\right), & A_4=\\left(\\begin{array}{ll}\n\n1 & 1 \\\\\n\n0 & 1\n\n\\end{array}\\right) \\\\\n\nA_5=\\left(\\begin{array}{ll}\n\n0 & 1 \\\\\n\n1 & 1\n\n\\end{array}\\right), & A_6=\\left(\\begin{array}{ll}\n\n1 & 1 \\\\\n\n1 & 0\n\n\\end{array}\\right)\n\n\\end{array}\n\n$$\n\nwith orders $1,2,2,2,3,3$ respectively.\n\nNote that $S_3$ is composed of elements\n\n$$\n\n\\text{ id, (1 2), (1 3), (2 3), (1 2 3), (1 3 2)} \n\n$$\n\nwith orders 1, 2, 2, 2, 3, 3 respectively. Also note that, by Problem 17 of generate $S_3$. We also have that $\\left(\\begin{array}{llll}1 & 3 & 2\\end{array}\\right)=\\left(\\begin{array}{llll}1 & 2 & 3\\end{array}\\right)\\left(\\begin{array}{lll}1 & 2 & 3\\end{array}\\right)$, that $\\left(\\begin{array}{lll}1 & 3\\end{array}\\right)=\\left(\\begin{array}{lll}1 & 2 & 3\\end{array}\\right)\\left(\\begin{array}{ll}1 & 2\\end{array}\\right)$, $\\left(\\begin{array}{ll}1 & 2\\end{array}\\right)\\left(\\begin{array}{lll}1 & 2 & 3\\end{array}\\right)=\\left(\\begin{array}{ll}2 & 3\\end{array}\\right)$ and $\\left(\\begin{array}{lll}1 & 2\\end{array}\\right)\\left(\\begin{array}{ll}1 & 2\\end{array}\\right)=\\mathrm{id}$\n\n\n\nNow we can check that $\\tau\\left(A_2\\right)=\\left(\\begin{array}{ll}1 & 2\\end{array}\\right), \\tau\\left(A_5\\right)=\\left(\\begin{array}{lll}1 & 2 & 3\\end{array}\\right)$ induces an isomorphism. We compute\n\n$$\n\n\\begin{aligned}\n\n& \\tau\\left(A_1\\right)=\\tau\\left(A_2 A_2\\right)=\\tau\\left(A_2\\right) \\tau\\left(A_2\\right)=\\mathrm{id} \\\\\n\n& \\tau\\left(A_3\\right)=\\tau\\left(A_5 A_2\\right)=\\tau\\left(A_5\\right) \\tau\\left(A_2\\right)=\\left(\\begin{array}{llll}\n\n1 & 2 & 3\n\n\\end{array}\\right)\\left(\\begin{array}{lll}\n\n1 & 2\n\n\\end{array}\\right)=\\left(\\begin{array}{ll}\n\n1 & 3\n\n\\end{array}\\right) \\\\\n\n& \\tau\\left(A_4\\right)=\\tau\\left(A_2 A_5\\right)=\\tau\\left(A_2\\right) \\tau\\left(A_5\\right)=\\left(\\begin{array}{lll}\n\n1 & 2\n\n\\end{array}\\right)\\left(\\begin{array}{lll}\n\n1 & 2 & 3\n\n\\end{array}\\right)=\\left(\\begin{array}{ll}\n\n2 & 3\n\n\\end{array}\\right) \\\\\n\n& \\tau\\left(A_6\\right)=\\tau\\left(A_5 A_5\\right)=\\tau\\left(A_5\\right) \\tau\\left(A_5\\right)=\\left(\\begin{array}{lll}\n\n1 & 3 & 2\n\n\\end{array}\\right)\n\n\\end{aligned}\n\n$$\n\nThus we see that $\\tau$ extendeds to an isomorphism, since $A_2$ and $A_5$ generate $T$, so that $\\tau\\left(A_i A_j\\right)=\\tau\\left(A_i\\right) \\tau\\left(A_j\\right)$ follows from writing $A_i$ and $A_j$ in terms of $A_2$ and $A_5$ and using the equlities and relations shown above.\n\n\\end{proof}"}
{"id": "Herstein|exercise_4_2_6", "formal_statement": "theorem exercise_4_2_6 {R : Type*} [ring R] (a x : R) \n  (h : a ^ 2 = 0) : a * (a * x + x * a) = (x + x * a) * a :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "If $a^2 = 0$ in $R$, show that $ax + xa$ commutes with $a$.\n", "nl_proof": "\\begin{proof}\n\nWe need to show that\n\n$$\n\na(a x+x a)=(a x+x a) a \\text { for } a, x \\in R .\n\n$$\n\nNow,\n\n$$\n\n\\begin{gathered}\n\na(a x+x a)=a(a x)+a(x a) \\\\\n\n=a^2 x+a x a \\\\\n\n=0+a x a=a x a .\n\n\\end{gathered}\n\n$$\n\nAgain,\n\n$$\n\n\\begin{gathered}\n\n(a x+x a) a=(a x) a+(x a) a \\\\\n\n=a x a+x a^2 \\\\\n\n=a x a+0=a x a .\n\n\\end{gathered}\n\n$$\n\nIt follows that,\n\n$$\n\na(a x+x a)=(a x+x a) a, \\text { for } x, a \\in R .\n\n$$\n\nThis shows that $a x+x a$ commutes with $a$. This completes the proof.\n\n\\end{proof}"}
{"id": "Herstein|exercise_4_3_1", "formal_statement": "theorem exercise_4_3_1 {R : Type*} [comm_ring R] (a : R) :\n  \u2203 I : ideal R, {x : R | x*a=0} = I :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "If $R$ is a commutative ring and $a \\in R$, let $L(a) = \\{x \\in R \\mid xa = 0\\}$. Prove that $L(a)$ is an ideal of $R$.\n", "nl_proof": "\\begin{proof}\n\n    First, note that if $x \\in L(a)$ and $y \\in L(a)$ then $x a=0$ and $y a=0$, so that\n\n$$\n\n\\begin{aligned}\n\nx a-y a & =0 \\\\\n\n(x-y) a & =0,\n\n\\end{aligned}\n\n$$\n\ni.e. $L(a)$ is an additive subgroup of $R$. (We have used the criterion that $H$ is a subgroup of $G$ if for any $h_1, h_2 \\in H$ we have that $h_1 h_2^{-1} \\in H$. \n\n\n\nNow we prove the conclusion. Let $r \\in R$ and $b \\in L(a)$, then $b a=0$, and so $x b a=0$ which by associativity of multiplication in $R$ is equivalent to\n\n$$\n\n(x b) a=0,\n\n$$\n\nso that $x b \\in L(a)$. Since $R$ is commutative, (1) implies that $(bx)a=0$, so that $b x \\in L(a)$, which concludes the proof that $L(a)$ is an ideal.\n\n\\end{proof}"}
{"id": "Herstein|exercise_4_4_9", "formal_statement": "theorem exercise_4_4_9 (p : \u2115) (hp : nat.prime p) :\n  \u2203 S : finset (zmod p), S.card = (p-1)/2 \u2227 \u2203 x : zmod p, x^2 = p \u2227 \n  \u2203 S : finset (zmod p), S.card = (p-1)/2 \u2227 \u00ac \u2203 x : zmod p, x^2 = p :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Show that $(p - 1)/2$ of the numbers $1, 2, \\ldots, p - 1$ are quadratic residues and $(p - 1)/2$ are quadratic nonresidues $\\mod p$.\n", "nl_proof": "\\begin{proof}\n\n    To find all the quadratic residues $\\bmod p$ among the integers $1,2, \\ldots, p-1$, we compute the least positive residues modulo $p$ of the squares of the integers $1,2, \\ldots, p-1\\}$.\n\n\n\nSince there are $p-1$ squares to consider, and since each congruence $x^2 \\equiv a (\\bmod p)$ has either zero or two solutions, there must be exactly $\\frac{(p-1)}{2}$ quadratic residues mod $p$ among the integers $1,2, \\ldots, p-1$.\n\nThe remaining\n\n$$\n\n(p-1)-\\frac{(p-1)}{2}=\\frac{(p-1)}{2}\n\n$$\n\npositive integers less than $p-1$ are quadratic non-residues of $\\bmod p$.\n\n\\end{proof}"}
{"id": "Herstein|exercise_4_5_23", "formal_statement": "theorem exercise_4_5_23 {p q: polynomial (zmod 7)} \n  (hp : p = X^3 - 2) (hq : q = X^3 + 2) : \n  irreducible p \u2227 irreducible q \u2227 \n  (nonempty $ polynomial (zmod 7) \u29f8 ideal.span ({p} : set $ polynomial $ zmod 7) \u2243+*\n  polynomial (zmod 7) \u29f8 ideal.span ({q} : set $ polynomial $ zmod 7)) :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $F = \\mathbb{Z}_7$ and let $p(x) = x^3 - 2$ and $q(x) = x^3 + 2$ be in $F[x]$. Show that $p(x)$ and $q(x)$ are irreducible in $F[x]$ and that the fields $F[x]/(p(x))$ and $F[x]/(q(x))$ are isomorphic.\n", "nl_proof": "\\begin{proof}\n\n    We have that $p(x)$ and $q(x)$ are irreducible if they have no roots in $\\mathbb{Z}_7$, which can easily be checked. E.g. for $p(x)$ we have that $p(0)=5, p(1)=6, p(2)=6, p(3)=4, p(4)=6$, $p(5)=4, p(6)=4$, and similarly for $q(x)$.\n\n\n\nWe have that every element of $F[x] /(p(x))$ is equal to $a x^2+b x+c+(p(x))$, and likewise for $F[x] /(q(x))$. We consider a map $\\tau$ : $F[x] /(p(x)) \\rightarrow F[x] /(q(x))$ given by\n\n$$\n\n\\tau\\left(a x^2+b x+c+(p(x))\\right)=a x^2-b x+c+(q(x)) .\n\n$$\n\nThis map is obviously onto, and since $|F[x] /(p(x))|=|F[x] /(q(x))|=7^3$ by Problem 16, it is also one-to-one. We claim that it is a homomorphism. Additivity of $\\tau$ is immediate by the linearity of addition of polynomial coefficient, so we just have to check the multiplicativity; if $n=a x^2+b x+$ $c+(p(x))$ and $m=d x^2+e x+f+(p(x))$ then\n\n$$\n\n\\begin{aligned}\n\n\\tau(n m) & =\\tau\\left(a d x^4+(a e+b d) x^3+(a f+b e+c d) x^2+(b f+c e) x+c f+(p(x))\\right) \\\\\n\n& =\\tau\\left(2 a d x+2(a e+b d)+(a f+b e+c d) x^2+(b f+c e) x+c f+(p(x))\\right) \\\\\n\n& =\\tau\\left((a f+b e+c d) x^2+(b f+c e+2 a d) x+(c f+2 a e+2 b d)+(p(x))\\right) \\\\\n\n& =(a f+b e+c d) x^2-(b f+c e+2 a d) x+c f+2 a e+2 b d+(q(x)) \\\\\n\n& =a d x^4-(a e+b d) x^3+(a f+b e+c d) x^2-(b f+c e) x+c f+(q(x)) \\\\\n\n& =\\left(a x^2-b x+c+(q(x))\\right)\\left(d x^2-e x+f+(q(x))\\right) \\\\\n\n& =\\tau(n) \\tau(m) .\n\n\\end{aligned}\n\n$$\n\nwhere in the second equality we used that $x^3+p(x)=2+p(x)$ and in the fifth we used that $x^3+$ $q(x)=-2+q(x)$\n\n\\end{proof}"}
{"id": "Herstein|exercise_4_6_2", "formal_statement": "theorem exercise_4_6_2 : irreducible (X^3 + 3*X + 2 : polynomial \u211a) :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that $f(x) = x^3 + 3x + 2$ is irreducible in $Q[x]$.\n", "nl_proof": "\\begin{proof}    \n\nLet us assume that $f(x)$ is reducible over $\\mathbb{Q}[x]$.\n\nThen there exists a rational root of $f(x)$.\n\nLet $p / q$ be a rational root of $f(x)$, where $\\operatorname{gcd}(p, q)=1$.\n\nThen $f(p / q)=0$.\n\nNow,\n\n$$\n\n\\begin{aligned}\n\n& f(p / q)=(p / q)^3+3(p / q)+2 \\\\\n\n\\Longrightarrow & (p / q)^3+3(p / q)+2=0 \\\\\n\n\\Longrightarrow & p^3+3 p q^2=-2 q^3 \\\\\n\n\\Longrightarrow & p\\left(p^2+3 q^2\\right)=-q^3\n\n\\end{aligned}\n\n$$\n\nIt follows that, $p$ divides $q$ which is a contradiction to the fact that $\\operatorname{gcd}(p, q)=1$.\n\nThis implies that $f(x)$ has no rational root.\n\nNow we know that, a polynomial of degree two or three over a field $F$ is reducible if and only if it has a root in $F$.\n\nNow $f(x)$ is a 3 degree polynomial having no root in $\\mathbb{Q}$.\n\nSo, $f(x)$ is irreducible in $\\mathbb{Q}[x]$.\n\nThis completes the proof.\n\n\\end{proof}"}
{"id": "Herstein|exercise_5_1_8", "formal_statement": "theorem exercise_5_1_8 {p m n: \u2115} {F : Type*} [field F] \n  (hp : nat.prime p) (hF : char_p F p) (a b : F) (hm : m = p ^ n) : \n  (a + b) ^ m = a^m + b^m :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "If $F$ is a field of characteristic $p \\neq 0$, show that $(a + b)^m = a^m + b^m$, where $m = p^n$, for all $a, b \\in F$ and any positive integer $n$.\n", "nl_proof": "\\begin{proof}\n\n    Since $F$ is of characteristic $p$ and we have considered arbitrary two elements $a, b$ in $F$ we have\n\n$$\n\n\\begin{aligned}\n\n& p a=p b=0 \\\\\n\n& \\Longrightarrow p^n a=p^n b=0 \\\\\n\n& \\Longrightarrow m a=m b=0 \\text {. } \\\\\n\n&\n\n\\end{aligned}\n\n$$\n\nNow we know from Binomial Theorem that\n\n$$\n\n(a+b)^m=\\sum_{i=0}^m\\left(\\begin{array}{c}\n\nm \\\\\n\ni\n\n\\end{array}\\right) a^i b^{m-i}\n\n$$\n\nHere\n\n$$\n\n\\left(\\begin{array}{c}\n\nm \\\\\n\ni\n\n\\end{array}\\right)=\\frac{m !}{i !(m-i) !} .\n\n$$\n\nNow we know that for any integer $n$ and any integer $k$ satisfying $1 \\leq k<n, n$ always divides $\\left(\\begin{array}{l}n \\\\ k\\end{array}\\right)$. So in our case for $i$ in the range $1 \\leq i<m, m$ divides $\\left(\\begin{array}{c}m \\\\ i\\end{array}\\right)$. It follows that $p$ divides $\\left(\\begin{array}{c}m \\\\ i\\end{array}\\right)$, for $i$ satisfying $1 \\leq i<m$, since $m=p^n$ for any integer $n$. Therefore other than the terms $a^m$ and $b^m$ in the expansion $\\sum_{i=0}^m\\left(\\begin{array}{c}m \\\\ i\\end{array}\\right) a^i b^{m-i}$ will vanish due to char $p$ nature of $F$.\n\nHence we have\n\n$$\n\n\\sum_{i=0}^m\\left(\\begin{array}{c}\n\nm \\\\\n\ni\n\n\\end{array}\\right) a^i b^{m-i}=a^m+b^m\n\n$$\n\nThis follows that, for all $a, b \\in F$\n\n$$\n\n(a+b)^m=a^m+b^m .\n\n$$\n\nThis completes the proof.\n\n\\end{proof}"}
{"id": "Herstein|exercise_5_3_7", "formal_statement": "theorem exercise_5_3_7 {K : Type*} [field K] {F : subfield K} \n  {a : K} (ha : is_algebraic F (a ^ 2)) : is_algebraic F a :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "If $a \\in K$ is such that $a^2$ is algebraic over the subfield $F$ of $K$, show that a is algebraic over $F$.\n", "nl_proof": "\\begin{proof}\n\n    Since $a^2$ is algebraic over $F$, there exist a non-zero polynomial $f(x)$ in $F[x]$ such that $f\\left(a^2\\right)=0$. Consider a new polynomial $g(x)$ defined as $g(x)=f\\left(x^2\\right)$. Clearly $g(x) \\in F[x]$ and $g(a)=f\\left(a^2\\right)= 0$.\n\n\\end{proof}"}
{"id": "Herstein|exercise_5_4_3", "formal_statement": "theorem exercise_5_4_3 {a : \u2102} {p : \u2102 \u2192 \u2102} \n  (hp : p = \u03bb x, x^5 + real.sqrt 2 * x^3 + real.sqrt 5 * x^2 + \n  real.sqrt 7 * x + 11)\n  (ha : p a = 0) : \n  \u2203 p : polynomial \u2102 , p.degree < 80 \u2227 a \u2208 p.roots \u2227 \n  \u2200 n : p.support, \u2203 a b : \u2124, p.coeff n = a / b :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "If $a \\in C$ is such that $p(a) = 0$, where $p(x) = x^5 + \\sqrt{2}x^3 + \\sqrt{5}x^2 + \\sqrt{7}x + \\sqrt{11}$, show that $a$ is algebraic over $\\mathbb{Q}$ of degree at most 80.\n", "nl_proof": "\\begin{proof}\n\n    Given $a \\in \\mathbb{C}$ such that $p(a)=0$, where\n\n$$\n\np(x)=x^5+\\sqrt{2} x^3+\\sqrt{5} x^2+\\sqrt{7} x+\\sqrt{11}\n\n$$\n\nHere, we note that $p(x) \\in \\mathbb{Q}(\\sqrt{2}, \\sqrt{5}, \\sqrt{7}, \\sqrt{11})$ and\n\n$$\n\n\\begin{aligned}\n\n    {[Q(\\sqrt{2}, \\sqrt{5}, \\sqrt{7}, \\sqrt{11}): \\mathbb{Q}] } & =[Q(\\sqrt{2}, \\sqrt{5}, \\sqrt{7}, \\sqrt{11}): Q(\\sqrt{2}, \\sqrt{5}, \\sqrt{7})] \\cdot[\\mathbb{Q}(\\sqrt{2}, \\sqrt{5}, \\sqrt{7}): \\mathbb{Q}(\\sqrt{2}, \\sqrt{5})] \\\\\n\n& \\cdot[\\mathbb{Q}(\\sqrt{2}, \\sqrt{5}): \\mathbb{Q}(\\sqrt{2})] \\cdot[\\mathbb{Q}(\\sqrt{2}): \\mathbb{Q}] \\\\\n\n& =2 \\cdot 2 \\cdot 2 \\cdot 2 \\\\\n\n& =16\n\n\\end{aligned}\n\n$$\n\nHere, we note that $p(x)$ is of degree 5 over $\\mathbb{Q}(\\sqrt{2}, \\sqrt{5}, \\sqrt{7}, \\sqrt{11})$. If $a$ is root of $p(x)$, then\n\n$$\n\n[Q(\\sqrt{2}, \\sqrt{5}, \\sqrt{7}, \\sqrt{11}, a): \\mathbb{Q}]=[Q(\\sqrt{2}, \\sqrt{5}, \\sqrt{7}, \\sqrt{11}): Q(\\sqrt{2}, \\sqrt{5}, \\sqrt{7}, \\sqrt{11})] \\cdot 15\n\n$$\n\nand $[Q(\\sqrt{2}, \\sqrt{5}, \\sqrt{7}, \\sqrt{11}): Q(\\sqrt{2}, \\sqrt{5}, \\sqrt{7}, \\sqrt{11})] \\leq 5$. We get equality if $p(x)$ is irreducible over $Q(\\sqrt{2}, \\sqrt{5}, \\sqrt{7}, \\sqrt{11})$. This gives\n\n$$\n\n[Q(\\sqrt{2}, \\sqrt{5}, \\sqrt{7}, \\sqrt{11}, a): \\mathbb{Q}] \\leq 16 \\cdot 5=80\n\n$$\n\n\\end{proof}"}
{"id": "Herstein|exercise_5_6_14", "formal_statement": "theorem exercise_5_6_14 {p m n: \u2115} (hp : nat.prime p) {F : Type*} \n  [field F] [char_p F p] (hm : m = p ^ n) : \n  card (root_set (X ^ m - X : polynomial F) F) = m :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "If $F$ is of characteristic $p \\neq 0$, show that all the roots of $x^m - x$, where $m = p^n$, are distinct.\n", "nl_proof": "\\begin{proof}\n\n    Let us consider $f(x)=x^m-x$. Then $f \\in F[x]$.\n\nClaim: $f(x)$ has a multiple root in some extension of $F$ if and only if $f(x)$ is not relatively prime to its formal derivative, $f^{\\prime}(x)$. \n\n\n\nProof of the Claim: Let us assume that $f(x)$ has a multiple root in some extension of $F$. Let $y$ be a multiple root of $f(x)$. Then over a splitting field, we have\n\n$$\n\nf(x)=(x-y)^n g(x), \\text { for some integer } n \\geq 2 .\n\n$$\n\nHere $g(x)$ is a polynomial such that $g(y) \\neq 0$. Now taking derivative of $f$ we get\n\n$$\n\nf^{\\prime}(x)=n \\cdot(x-y)^{n-1} g(x)+(x-y)^n g^{\\prime}(x)\n\n$$\n\nhere $g^{\\prime}(x)$ implies derivative of $g$ with respect to $x$. Since we have $n \\geq 2$, this implies $(n-1) \\geq 1$. Hence, (1) shows that $f^{\\prime}(x)$ has $y$ as a root. Therefore, $f(x)$ is not relatively prime to $f^{\\prime}(x)$. We now prove the other direction.\n\nConversely, let us assume that $f(x)$ is not relatively prime to $f^{\\prime}(x)$. Let $y$ is a root of both $f(x)$ and $f^{\\prime}(x)$. Since $y$ is a root of $f(x)$, we can write\n\n$$\n\nf(x)=(x-y) \\cdot g(x)\n\n$$\n\nfor some polynomial $g(x)$. then taking derivative of $f(x)$ we have\n\n$$\n\nf^{\\prime}(x)=g(x)+(x-y) \\cdot g^{\\prime}(x)\n\n$$\n\nwhere $g^{\\prime}(x)$ is the derivative of $g(x)$ with respect to $x$. Since $y$ is a root of $f^{\\prime}(x)$ also we have\n\n$$\n\nf^{\\prime}(y)=0\n\n$$\n\nThen we have\n\n$$\n\n\\begin{aligned}\n\n& f^{\\prime}(y)=g(y)+(y-y) \\cdot g^{\\prime}(y) \\\\\n\n\\Longrightarrow & f^{\\prime}(y)=g(y) \\\\\n\n\\Longrightarrow & g(y)=0 .\n\n\\end{aligned}\n\n$$\n\nThis implies $y$ is a root of $g(x)$ also. Therefore we have\n\n$$\n\ng(x)=(x-y) \\cdot h(x)\n\n$$\n\nfor some polynomial $h(x)$. Now form (2) we have\n\n$$\n\nf(x)=(x-y)^2 \\cdot h(x) .\n\n$$\n\nThis follows that $y$ is a multiple root of $f(x)$. Therefore, $f(x)$ has a multiple root in some extension of the field $F$. This completes the proof of the Claim.\n\n\n\nIn our case, $f(x)=x^m-x$, where $m=p^n$. Now we calculate the derivative of $f$. That is\n\n$$\n\nf^{\\prime}(x)=m x^{m-1}-1=-1(\\bmod p) .\n\n$$\n\nBy the above condition it follows that, $f^{\\prime}$ has no root same as $f$, that is, $f(x)$ and $f^{\\prime}(x)$ are relatively prime. Hence, $f(x)$ has no multiple root in $F$. Since $f(x)=x^m-x$ is a polynomial of degree $m$, it follows that $f(x)$ has $m$ distinct roots in $F$, where $m=p^n$. This completes the proof.\n\n\\end{proof}"}
{"id": "Artin|exercise_2_3_2", "formal_statement": "theorem exercise_2_3_2 {G : Type*} [group G] (a b : G) :\n  \u2203 g : G, b* a = g * a * b * g\u207b\u00b9 :=", "src_header": "import .common \n\nopen function fintype subgroup ideal polynomial submodule zsqrtd char_p\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that the products $a b$ and $b a$ are conjugate elements in a group.\n", "nl_proof": "\\begin{proof}\n\n    We have that $(a^{-1})ab(a^{-1})^{-1} = ba$. \n\n\\end{proof}"}
{"id": "Artin|exercise_2_8_6", "formal_statement": "theorem exercise_2_8_6 {G H : Type*} [group G] [group H] :\n  center (G \u00d7 H) \u2243* (center G) \u00d7 (center H) :=", "src_header": "import .common \n\nopen function fintype subgroup ideal polynomial submodule zsqrtd char_p\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that the center of the product of two groups is the product of their centers.\n", "nl_proof": "\\begin{proof}\n\n    We have that $(g_1, g_2)\\cdot (h_1, h_2) = (h_1, h_2)\\cdot (g_1, g_2)$ if and only if $g_1h_1 = h_1g_1$ and $g_2h_2 = h_2g_2$. \n\n\\end{proof}"}
{"id": "Artin|exercise_3_2_7", "formal_statement": "theorem exercise_3_2_7 {F : Type*} [field F] {G : Type*} [field G]\n  (\u03c6 : F \u2192+* G) : injective \u03c6 :=", "src_header": "import .common \n\nopen function fintype subgroup ideal polynomial submodule zsqrtd char_p\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that every homomorphism of fields is injective.\n", "nl_proof": "\\begin{proof}\n\n    Suppose $f(a)=f(b)$, then $f(a-b)=0=f(0)$. If $u=(a-b) \\neq 0$, then $f(u) f\\left(u^{-1}\\right)=f(1)=1$, but that means that $0 f\\left(u^{-1}\\right)=1$, which is impossible. Hence $a-b=0$ and $a=b$.\n\n\\end{proof}"}
{"id": "Artin|exercise_3_7_2", "formal_statement": "theorem exercise_3_7_2 {K V : Type*} [field K] [add_comm_group V]\n  [module K V] {\u03b9 : Type*} [fintype \u03b9] (\u03b3 : \u03b9 \u2192 submodule K V) :\n  (\u22c2 (i : \u03b9), (\u03b3 i : set V)) \u2260 \u22a4 :=", "src_header": "import .common \n\nopen function fintype subgroup ideal polynomial submodule zsqrtd char_p\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $V$ be a vector space over an infinite field $F$. Prove that $V$ is not the union of finitely many proper subspaces.\n", "nl_proof": "\\begin{proof}\n\n    If $V$ is the set-theoretic union of $n$ proper subspaces $W_i$ ( $1 \\leq i \\leq n$ ), then $|F| \\leq n-1$.\n\nProof. We may suppose no $W_i$ is contained in the union of the other subspaces. Let $u \\in W_i, \\quad u \\notin \\bigcup_{j \\neq i} W_j$ and $v \\notin W_i$.\n\nThen $(v+F u) \\cap W_i=\\varnothing$ and $(v+F u) \\cap W_j(j \\neq i)$ contains at most one vector since otherwise $W_j$ would contain $u$. Hence\n\n$$\n\n|v+F u|=|F| \\leq n-1 .\n\n$$\n\nCorollary: Avoidance lemma for vector spaces.\n\nLet $E$ be a vector space over an infinite field. If a subspace is contained in a finite union of subspaces, it is contained in one of them.\n\n\\end{proof}"}
{"id": "Artin|exercise_6_4_2", "formal_statement": "theorem exercise_6_4_2 {G : Type*} [group G] [fintype G] {p q : \u2115}\n  (hp : prime p) (hq : prime q) (hG : card G = p*q) :\n  is_simple_group G \u2192 false :=", "src_header": "import .common \n\nopen function fintype subgroup ideal polynomial submodule zsqrtd char_p\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that no group of order $p q$, where $p$ and $q$ are prime, is simple.\n", "nl_proof": "\\begin{proof}\n\n    If $|G|=n=p q$ then the only two Sylow subgroups are of order $p$ and $q$.\n\nFrom Sylow's third theorem we know that $n_p \\mid q$ which means that $n_p=1$ or $n_p=q$.\n\nIf $n_p=1$ then we are done (by a corollary of Sylow's theorem)\n\nIf $n_p=q$ then we have accounted for $q(p-1)=p q-q$ elements of $G$ and so there is only one group of order $q$ and again we are done.\n\n\\end{proof}"}
{"id": "Artin|exercise_6_4_12", "formal_statement": "theorem exercise_6_4_12 {G : Type*} [group G] [fintype G]\n  (hG : card G = 224) :\n  is_simple_group G \u2192 false :=", "src_header": "import .common \n\nopen function fintype subgroup ideal polynomial submodule zsqrtd char_p\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that no group of order 224 is simple.\n", "nl_proof": "\\begin{proof}\n\n    The following proves there must exist a normal Sylow 2 -subgroup of order 32 ,\n\nSuppose there are $n_2=7$ Sylow 2 -subgroups in $G$. Making $G$ act on the set of these Sylow subgroups by conjugation (Mitt wrote about this but on the set of the other Sylow subgroups, which gives no contradiction), we get a homomorphism $G \\rightarrow S_7$ which must be injective if $G$ is simple (why?).\n\n\n\nBut this cannot be since then we would embed $G$ into $S_7$, which is impossible since $|G| \\nmid 7 !=\\left|S_7\\right|$ (why?)\n\n\\end{proof}"}
{"id": "Artin|exercise_10_1_13", "formal_statement": "theorem exercise_10_1_13 {R : Type*} [ring R] {x : R}\n  (hx : is_nilpotent x) : is_unit (1 + x) :=", "src_header": "import .common \n\nopen function fintype subgroup ideal polynomial submodule zsqrtd char_p\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "An element $x$ of a ring $R$ is called nilpotent if some power of $x$ is zero. Prove that if $x$ is nilpotent, then $1+x$ is a unit in $R$.\n", "nl_proof": "\\begin{proof}\n\n    If $x^n=0$, then\n\n$$\n\n(1+x)\\left(\\sum_{k=0}^{n-1}(-1)^k x^k\\right)=1+(-1)^{n-1} x^n=1 .\n\n$$\n\n\\end{proof}"}
{"id": "Artin|exercise_10_6_7", "formal_statement": "theorem exercise_10_6_7 {I : ideal gaussian_int}\n  (hI : I \u2260 \u22a5) : \u2203 (z : I), z \u2260 0 \u2227 (z : gaussian_int).im = 0 :=", "src_header": "import .common \n\nopen function fintype subgroup ideal polynomial submodule zsqrtd char_p\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that every nonzero ideal in the ring of Gauss integers contains a nonzero integer.\n", "nl_proof": "\\begin{proof}\n\n    Let $I$ be some nonzero ideal. Then there exists some $z \\in \\mathbb{Z}[i], z \\neq 0$ such that $z \\in I$. We know that $z=a+b i$, for some $a, b \\in \\mathbb{Z}$. We consider three cases:\n\n1. If $b=0$, then $z=a$, so $z \\in \\mathbb{Z} \\cap I$, and $z \\neq 0$, so the statement of the exercise holds.\n\n2. If $a=0$, then $z=i b$. Since $z \\neq 0$, we conclude that $b \\neq 0$. Since $I$ is an ideal in $\\mathbb{Z}[i]$, and $i \\in \\mathbb{Z}[i]$, we conclude that $i z \\in I$. Furthermore, $i z=-b \\in \\mathbb{Z}$. Thus, $i z$ is a nonzero integer which is in $I$.\n\n3. Let $a \\neq 0$ and $b \\neq 0$. Since $I$ is an ideal and $z \\in I$, we conclude that $z^2 \\in I$; that is,\n\n$$\n\n(a+b i)^2=a^2-b^2+2 a b i \\in I\n\n$$\n\nFurthermore, since $-2 a \\in \\mathbb{Z}[i]$, and $z \\in I$ and $I$ is an ideal, $-2 a z \\in I$; that is,\n\n$$\n\n-2 a z=-2 a(a+b i)=-2 a^2-2 a b i \\in I\n\n$$\n\nSince $I$ is closed under addition,\n\n$$\n\n\\left(a^2-b^2+2 a b i\\right)+\\left(-2 a^2-2 a b i\\right) \\in I \\Longrightarrow-a^2-b^2 \\in I\n\n$$\n\nNotice that $-a^2-b^2 \\neq 0$ since $a^2>0$ and $b^2>0$, so $-a^2-b^2<0$. Furthermore, it is an integer. Thus, we have found a nonzero integer in $I$.\n\n\\end{proof}"}
{"id": "Artin|exercise_10_4_7a", "formal_statement": "theorem exercise_10_4_7a {R : Type*} [comm_ring R] [no_zero_divisors R]\n  (I J : ideal R) (hIJ : I + J = \u22a4) : I * J = I \u2293 J :=", "src_header": "import .common \n\nopen function fintype subgroup ideal polynomial submodule zsqrtd char_p\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $I, J$ be ideals of a ring $R$ such that $I+J=R$. Prove that $I J=I \\cap J$.\n", "nl_proof": "\\begin{proof}\n\n    We have seen that $IJ \\subset I \\cap J$, so it remains to show that $I \\cap J \\subset IJ$.  Since $I+J = (1)$, there are elements $i \\in I$ and $j \\in J$ such that $i+j = 1$.  Let $k \\in I \\cap J$, and multiply $i+j=1$ through by $k$ to get $ki+kj = k$.  Write this more suggestively as\n\n\\[ k = ik+kj. \\]\n\nThe first term is in $IJ$ because $k \\in J$, and the second term is in $IJ$ because $k \\in I$, so $k \\in IJ$ as desired.\n\n\\end{proof}"}
{"id": "Artin|exercise_11_2_13", "formal_statement": "theorem exercise_11_2_13 (a b : \u2124) :\n  (of_int a : gaussian_int) \u2223 of_int b \u2192 a \u2223 b :=", "src_header": "import .common \n\nopen function fintype subgroup ideal polynomial submodule zsqrtd char_p\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "If $a, b$ are integers and if $a$ divides $b$ in the ring of Gauss integers, then $a$ divides $b$ in $\\mathbb{Z}$.\n", "nl_proof": "\\begin{proof}\n\n    Suppose $a|b$ in $\\mathbb{Z}[i]$ and $a,b\\in\\mathbb{Z}$. Then $a(x+yi)=b$ for $x,y\\in\\mathbb{Z}$. Expanding this we get $ax+ayi=b$, and equating imaginary parts gives us $ay=0$, implying $y=0$. \n\n\\end{proof}"}
{"id": "Artin|exercise_11_4_6a", "formal_statement": "theorem exercise_11_4_6a {F : Type*} [field F] [fintype F] (hF : card F = 7) :\n  irreducible (X ^ 2 + 1 : polynomial F) :=", "src_header": "import .common \n\nopen function fintype subgroup ideal polynomial submodule zsqrtd char_p\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that $x^2+x+1$ is irreducible in the field $\\mathbb{F}_2$.\n", "nl_proof": "\\begin{proof}\n\n    If $x^2+x+1$ were reducible in $\\mathbb{F}_2$, its factors must be linear. But we neither have that $0^2+0+1=$ nor $1^2+1+1=0$, therefore $x^2+x+1$ is irreducible.  \n\n\\end{proof}"}
{"id": "Artin|exercise_11_4_6c", "formal_statement": "theorem exercise_11_4_6c : irreducible (X^3 - 9 : polynomial (zmod 31)) :=", "src_header": "import .common \n\nopen function fintype subgroup ideal polynomial submodule zsqrtd char_p\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that $x^3 - 9$ is irreducible in $\\mathbb{F}_{31}$.\n", "nl_proof": "\\begin{proof}\n\n    If $p(x) = x^3-9$ were reducible, it would have a linear factor, since it either has a linear factor and a quadratic factor or three linear factors. We can then verify by brute force that $p(x)\\neq 0$ for $x \\in \\mathbb{F}_31$. \n\n\\end{proof}"}
{"id": "Artin|exercise_11_13_3", "formal_statement": "theorem exercise_11_13_3 (N : \u2115):\n  \u2203 p \u2265 N, nat.prime p \u2227 p + 1 \u2261 0 [MOD 4] :=", "src_header": "import .common \n\nopen function fintype subgroup ideal polynomial submodule zsqrtd char_p\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that there are infinitely many primes congruent to $-1$ (modulo $4$).\n", "nl_proof": "\\begin{proof}\n\n    First we show a lemma: if $a \\equiv 3(\\bmod 4)$ then there exists a prime $p$ such that $p \\mid a$ and $p \\equiv 3(\\bmod 4)$.\n\n\n\n    Clearly, all primes dividing $a$ are odd. Suppose all of them would be $\\equiv 1(\\bmod 4)$. Then their product would also be $a \\equiv 1(\\bmod 4)$, which is a contradiction.\n\n\n\nTo prove the main claim, suppose that $p_1, \\ldots, p_n$ would be all such primes. (In particular, we have $p_1=3$.) Consider $a=4 p_2 \\cdots p_n+3$. (Or you can take $a=4 p_2 \\cdots p_n-1$.) Show that $p_i \\nmid a$ for $i=1, \\ldots, n$. (The case $3 \\nmid a$ is solved differently than the other primes - this is the reason for omitting $p_1$ in the definition of $a$.) Then use the above lemma to get a contradiction.\n\n\\end{proof}"}
{"id": "Artin|exercise_13_6_10", "formal_statement": "theorem exercise_13_6_10 {K : Type*} [field K] [fintype K\u02e3] :\n  \u220f (x : K\u02e3), x = -1 :=", "src_header": "import .common \n\nopen function fintype subgroup ideal polynomial submodule zsqrtd char_p\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $K$ be a finite field. Prove that the product of the nonzero elements of $K$ is $-1$.\n", "nl_proof": "\\begin{proof}\n\n    Since we are working with a finite field with $q$ elements, anyone of them is a root of the following polynomial\n\n$$\n\nx^q-x=0 .\n\n$$\n\nIn particular if we rule out the 0 element, any $a_i \\neq 0$ is a root of\n\n$$\n\nx^{q-1}-1=0 .\n\n$$\n\nThis polynomial splits completely in $\\mathbb{F}_q$ so we find\n\n$$\n\n\\left(x-a_1\\right) \\cdots\\left(x-a_{q-1}\\right)=0\n\n$$\n\nin particular\n\n$$\n\nx^{q-1}-1=\\left(x-a_1\\right) \\cdots\\left(x-a_{q-1}\\right)\n\n$$\n\nThus $a_1 \\cdots a_{q-1}=-1$.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_1_1_3", "formal_statement": "theorem exercise_1_1_3 (n : \u2124) : \n  \u2200 (a b c : \u2124), (a+b)+c \u2261 a+(b+c) [ZMOD n] :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that the addition of residue classes $\\mathbb{Z}/n\\mathbb{Z}$ is associative.\n", "nl_proof": "\\begin{proof}\n\n    We have\n\n$$\n\n\\begin{aligned}\n\n(\\bar{a}+\\bar{b})+\\bar{c} &=\\overline{a+b}+\\bar{c} \\\\\n\n&=\\overline{(a+b)+c} \\\\\n\n&=\\overline{a+(b+c)} \\\\\n\n&=\\bar{a}+\\overline{b+c} \\\\\n\n&=\\bar{a}+(\\bar{b}+\\bar{c})\n\n\\end{aligned}\n\n$$\n\nsince integer addition is associative.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_1_1_5", "formal_statement": "theorem exercise_1_1_5 (n : \u2115) (hn : 1 < n) : \n  is_empty (group (zmod n)) :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that for all $n>1$ that $\\mathbb{Z}/n\\mathbb{Z}$ is not a group under multiplication of residue classes.\n", "nl_proof": "\\begin{proof}\n\n    Note that since $n>1, \\overline{1} \\neq \\overline{0}$. Now suppose $\\mathbb{Z} /(n)$ contains a multiplicative identity element $\\bar{e}$. Then in particular,\n\n$$\n\n\\bar{e} \\cdot \\overline{1}=\\overline{1}\n\n$$\n\nso that $\\bar{e}=\\overline{1}$. Note, however, that\n\n$$\n\n\\overline{0} \\cdot \\bar{k}=\\overline{0}\n\n$$\n\nfor all k, so that $\\overline{0}$ does not have a multiplicative inverse. Hence $\\mathbb{Z} /(n)$ is not a group under multiplication.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_1_1_16", "formal_statement": "theorem exercise_1_1_16 {G : Type*} [group G] \n  (x : G) (hx : x ^ 2 = 1) :\n  order_of x = 1 \u2228 order_of x = 2 :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $x$ be an element of $G$. Prove that $x^2=1$ if and only if $|x|$ is either $1$ or $2$.\n", "nl_proof": "\\begin{proof}\n\n    $(\\Rightarrow)$ Suppose $x^2=1$. Then we have $0<|x| \\leq 2$, i.e., $|x|$ is either 1 or 2 .\n\n( $\\Leftarrow$ ) If $|x|=1$, then we have $x=1$ so that $x^2=1$. If $|x|=2$ then $x^2=1$ by definition. So if $|x|$ is 1 or 2 , we have $x^2=1$.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_1_1_18", "formal_statement": "theorem exercise_1_1_18 {G : Type*} [group G]\n  (x y : G) : x * y = y * x \u2194 y\u207b\u00b9 * x * y = x \u2194 x\u207b\u00b9 * y\u207b\u00b9 * x * y = 1 :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $x$ and $y$ be elements of $G$. Prove that $xy=yx$ if and only if $y^{-1}xy=x$ if and only if $x^{-1}y^{-1}xy=1$.\n", "nl_proof": "\\begin{proof}\n\nIf $x y=y x$, then $y^{-1} x y=y^{-1} y x=1 x=x$. Multiplying by $x^{-1}$ then gives $x^{-1} y^{-1} x y=1$.\n\n\n\nOn the other hand, if $x^{-1} y^{-1} x y=1$, then we may multiply on the left by $x$ to get $y^{-1} x y=x$. Then multiplying on the left by $y$ gives $x y=y x$ as desired.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_1_1_22a", "formal_statement": "theorem exercise_1_1_22a {G : Type*} [group G] (x g : G) :\n  order_of x = order_of (g\u207b\u00b9 * x * g) :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "If $x$ and $g$ are elements of the group $G$, prove that $|x|=\\left|g^{-1} x g\\right|$.\n", "nl_proof": "\\begin{proof}\n\n    First we prove a technical lemma:\n\n\n\n    {\\bf Lemma.} For all $a, b \\in G$ and $n \\in \\mathbb{Z},\\left(b^{-1} a b\\right)^n=b^{-1} a^n b$.\n\nThe statement is clear for $n=0$. We prove the case $n>0$ by induction; the base case $n=1$ is clear. Now suppose $\\left(b^{-1} a b\\right)^n=b^{-1} a^n b$ for some $n \\geq 1$; then\n\n$$\n\n\\left(b^{-1} a b\\right)^{n+1}=\\left(b^{-1} a b\\right)\\left(b^{-1} a b\\right)^n=b^{-1} a b b^{-1} a^n b=b^{-1} a^{n+1} b .\n\n$$\n\nBy induction the statement holds for all positive $n$.\n\nNow suppose $n<0$; we have\n\n$$\n\n\\left(b^{-1} a b\\right)^n=\\left(\\left(b^{-1} a b\\right)^{-n}\\right)^{-1}=\\left(b^{-1} a^{-n} b\\right)^{-1}=b^{-1} a^n b .\n\n$$\n\nHence, the statement holds for all integers $n$.\n\nNow to the main result. Suppose first that $|x|$ is infinity and that $\\left|g^{-1} x g\\right|=n$ for some positive integer $n$. Then we have\n\n$$\n\n\\left(g^{-1} x g\\right)^n=g^{-1} x^n g=1,\n\n$$\n\nand multiplying on the left by $g$ and on the right by $g^{-1}$ gives us that $x^n=1$, a contradiction. Thus if $|x|$ is infinity, so is $\\left|g^{-1} x g\\right|$. Similarly, if $\\left|g^{-1} x g\\right|$ is infinite and $|x|=n$, we have\n\n$$\n\n\\left(g^{-1} x g\\right)^n=g^{-1} x^n g=g^{-1} g=1,\n\n$$\n\na contradiction. Hence if $\\left|g^{-1} x g\\right|$ is infinite, so is $|x|$.\n\nSuppose now that $|x|=n$ and $\\left|g^{-1} x g\\right|=m$ for some positive integers $n$ and $m$. We have\n\n$$\n\n\\left(g^{-1} x g\\right)^n=g^{-1} x^n g=g^{-1} g=1,\n\n$$\n\nSo that $m \\leq n$, and\n\n$$\n\n\\left(g^{-1} x g\\right)^m=g^{-1} x^m g=1,\n\n$$\n\nso that $x^m=1$ and $n \\leq m$. Thus $n=m$.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_1_1_25", "formal_statement": "theorem exercise_1_1_25 {G : Type*} [group G] \n  (h : \u2200 x : G, x ^ 2 = 1) : \u2200 a b : G, a*b = b*a :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that if $x^{2}=1$ for all $x \\in G$ then $G$ is abelian.\n", "nl_proof": "\\begin{proof}\n\n    Solution: Note that since $x^2=1$ for all $x \\in G$, we have $x^{-1}=x$. Now let $a, b \\in G$. We have\n\n$$\n\na b=(a b)^{-1}=b^{-1} a^{-1}=b a .\n\n$$\n\nThus $G$ is abelian.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_1_1_34", "formal_statement": "theorem exercise_1_1_34 {G : Type*} [group G] {x : G} \n  (hx_inf : order_of x = 0) (n m : \u2124) :\n  x ^ n \u2260 x ^ m :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "If $x$ is an element of infinite order in $G$, prove that the elements $x^{n}, n \\in \\mathbb{Z}$ are all distinct.\n", "nl_proof": "\\begin{proof}\n\n    Solution: Suppose to the contrary that $x^a=x^b$ for some $0 \\leq a<b \\leq n-1$. Then we have $x^{b-a}=1$, with $1 \\leq b-a<n$. However, recall that $n$ is by definition the least integer $k$ such that $x^k=1$, so we have a contradiction. Thus all the $x^i$, $0 \\leq i \\leq n-1$, are distinct. In particular, we have\n\n$$\n\n\\left\\{x^i \\mid 0 \\leq i \\leq n-1\\right\\} \\subseteq G,\n\n$$\n\nso that $|x|=n \\leq|G|$\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_1_6_4", "formal_statement": "theorem exercise_1_6_4 : \n  is_empty (multiplicative \u211d \u2243* multiplicative \u2102) :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that the multiplicative groups $\\mathbb{R}-\\{0\\}$ and $\\mathbb{C}-\\{0\\}$ are not isomorphic.\n", "nl_proof": "\\begin{proof}\n\n    Isomorphic groups necessarily have the same number of elements of order $n$ for all finite $n$.\n\n\n\nNow let $x \\in \\mathbb{R}^{\\times}$. If $x=1$ then $|x|=1$, and if $x=-1$ then $|x|=2$. If (with bars denoting absolute value) $|x|<1$, then we have\n\n$$\n\n1>|x|>\\left|x^2\\right|>\\cdots,\n\n$$\n\nand in particular, $1>\\left|x^n\\right|$ for all $n$. So $x$ has infinite order in $\\mathbb{R}^{\\times}$.\n\nSimilarly, if $|x|>1$ (absolute value) then $x$ has infinite order in $\\mathbb{R}^{\\times}$. So $\\mathbb{R}^{\\times}$has 1 element of order 1,1 element of order 2 , and all other elements have infinite order.\n\nIn $\\mathbb{C}^{\\times}$, on the other hand, $i$ has order 4 . Thus $\\mathbb{R}^{\\times}$and $\\mathbb{C}^{\\times}$are not isomorphic.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_1_6_17", "formal_statement": "theorem exercise_1_6_17 {G : Type*} [group G] (f : G \u2192 G) \n  (hf : f = \u03bb g, g\u207b\u00b9) :\n  \u2200 x y : G, f x * f y = f (x*y) \u2194 \u2200 x y : G, x*y = y*x :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $G$ be any group. Prove that the map from $G$ to itself defined by $g \\mapsto g^{-1}$ is a homomorphism if and only if $G$ is abelian.\n", "nl_proof": "\\begin{proof}\n\n    $(\\Rightarrow)$ Suppose $G$ is abelian. Then\n\n$$\n\n\\varphi(a b)=(a b)^{-1}=b^{-1} a^{-1}=a^{-1} b^{-1}=\\varphi(a) \\varphi(b),\n\n$$\n\nso that $\\varphi$ is a homomorphism.\n\n$(\\Leftarrow)$ Suppose $\\varphi$ is a homomorphism, and let $a, b \\in G$. Then\n\n$$\n\na b=\\left(b^{-1} a^{-1}\\right)^{-1}=\\varphi\\left(b^{-1} a^{-1}\\right)=\\varphi\\left(b^{-1}\\right) \\varphi\\left(a^{-1}\\right)=\\left(b^{-1}\\right)^{-1}\\left(a^{-1}\\right)^{-1}=b a,\n\n$$\n\nso that $G$ is abelian.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_2_1_5", "formal_statement": "theorem exercise_2_1_5 {G : Type*} [group G] [fintype G] \n  (hG : card G > 2) (H : subgroup G) [fintype H] : \n  card H \u2260 card G - 1 :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that $G$ cannot have a subgroup $H$ with $|H|=n-1$, where $n=|G|>2$.\n", "nl_proof": "\\begin{proof}\n\n    Solution: Under these conditions, there exists a nonidentity element $x \\in H$ and an element $y \\notin H$. Consider the product $x y$. If $x y \\in H$, then since $x^{-1} \\in H$ and $H$ is a subgroup, $y \\in H$, a contradiction. If $x y \\notin H$, then we have $x y=y$. Thus $x=1$, a contradiction. Thus no such subgroup exists.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_2_4_4", "formal_statement": "theorem exercise_2_4_4 {G : Type*} [group G] (H : subgroup G) : \n  subgroup.closure ((H : set G) \\ {1}) = \u22a4 :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that if $H$ is a subgroup of $G$ then $H$ is generated by the set $H-\\{1\\}$.\n", "nl_proof": "\\begin{proof}\n\n    If $H=\\{1\\}$ then $H-\\{1\\}$ is the empty set which indeed generates the trivial subgroup $H$. So suppose $|H|>1$ and pick a nonidentity element $h \\in H$. Since $1=h h^{-1} \\in\\langle H-\\{1\\}\\rangle$ (Proposition 9), we see that $H \\leq\\langle H-\\{1\\}\\rangle$. By minimality of $\\langle H-\\{1\\}\\rangle$, the reverse inclusion also holds so that $\\langle H-\\{1\\}\\rangle=$ $H$.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_2_4_16b", "formal_statement": "theorem exercise_2_4_16b {n : \u2115} {hn : n \u2260 0} \n  {R : subgroup (dihedral_group n)} \n  (hR : R = subgroup.closure {dihedral_group.r 1}) : \n  R \u2260 \u22a4 \u2227 \n  \u2200 K : subgroup (dihedral_group n), R \u2264 K \u2192 K = R \u2228 K = \u22a4 :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Show that the subgroup of all rotations in a dihedral group is a maximal subgroup.\n", "nl_proof": "\\begin{proof}\n\n    Fix a positive integer $n>1$ and let $H \\leq D_{2 n}$ consist of the rotations of $D_{2 n}$. That is, $H=\\langle r\\rangle$. Now, this subgroup is proper since it does not contain $s$. If $H$ is not maximal, then by the previous proof we know there is a maximal subset $K$ containing $H$. Then $K$ must contain a reflection $s r^k$ for $k \\in\\{0,1, \\ldots, n-1\\}$. Then since $s r^k \\in K$ and $r^{n-k} \\in K$, it follows by closure that\n\n$$\n\ns=\\left(s r^k\\right)\\left(r^{n-k}\\right) \\in K .\n\n$$\n\nBut $D_{2 n}=\\langle r, s\\rangle$, so this shows that $K=D_{2 n}$, which is a contradiction. Therefore $H$ must be maximal.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_3_1_3a", "formal_statement": "theorem exercise_3_1_3a {A : Type*} [comm_group A] (B : subgroup A) :\n  \u2200 a b : A \u29f8 B, a*b = b*a :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $A$ be an abelian group and let $B$ be a subgroup of $A$. Prove that $A / B$ is abelian.\n", "nl_proof": "\\begin{proof}\n\n    Lemma: Let $G$ be a group. If $|G|=2$, then $G \\cong Z_2$.\n\nProof: Since $G=\\{e a\\}$ has an identity element, say $e$, we know that $e e=e, e a=a$, and $a e=a$. If $a^2=a$, we have $a=e$, a contradiction. Thus $a^2=e$. We can easily see that $G \\cong Z_2$.\n\n\n\nIf $A$ is abelian, every subgroup of $A$ is normal; in particular, $B$ is normal, so $A / B$ is a group. Now let $x B, y B \\in A / B$. Then\n\n$$\n\n(x B)(y B)=(x y) B=(y x) B=(y B)(x B) .\n\n$$\n\nHence $A / B$ is abelian.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_3_1_22b", "formal_statement": "theorem exercise_3_1_22b {G : Type*} [group G] (I : Type*)\n  (H : I \u2192 subgroup G) (hH : \u2200 i : I, subgroup.normal (H i)) : \n  subgroup.normal (\u2a05 (i : I), H i):=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that the intersection of an arbitrary nonempty collection of normal subgroups of a group is a normal subgroup (do not assume the collection is countable).\n", "nl_proof": "\\begin{proof}\n\nLet $\\left\\{H_i \\mid i \\in I\\right\\}$ be an arbitrary collection of normal subgroups of $G$ and consider the intersection\n\n$$\n\n\\bigcap_{i \\in I} H_i\n\n$$\n\nTake an element $a$ in the intersection and an arbitrary element $g \\in G$. Then $g a g^{-1} \\in H_i$ because $H_i$ is normal for any $i \\in H$\n\nBy the definition of the intersection, this shows that $g a g^{-1} \\in \\bigcap_{i \\in I} H_i$ and therefore it is a normal subgroup.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_3_2_11", "formal_statement": "theorem exercise_3_2_11 {G : Type*} [group G] {H K : subgroup G}\n  (hHK : H \u2264 K) : \n  H.index = K.index * H.relindex K :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $H \\leq K \\leq G$. Prove that $|G: H|=|G: K| \\cdot|K: H|$ (do not assume $G$ is finite).\n", "nl_proof": "\\begin{proof}\n\n    Proof. Let $G$ be a group and let $I$ be a nonempty set of indices, not necessarily countable. Consider the collection of subgroups $\\left\\{N_\\alpha \\mid \\alpha \\in I\\right\\}$, where $N_\\alpha \\unlhd G$ for each $\\alpha \\in I$. Let\n\n$$\n\nN=\\bigcap_{\\alpha \\in I} N_\\alpha .\n\n$$\n\nWe know $N$ is a subgroup of $G$. \n\nFor any $g \\in G$ and any $n \\in N$, we must have $n \\in N_\\alpha$ for each $\\alpha$. And since $N_\\alpha \\unlhd G$, we have $g n g^{-1} \\in N_\\alpha$ for each $\\alpha$. Therefore $g n g^{-1} \\in N$, which shows that $g N g^{-1} \\subseteq N$ for each $g \\in G$. As before, this is enough to complete the proof.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_3_2_21a", "formal_statement": "theorem exercise_3_2_21a (H : add_subgroup \u211a) (hH : H \u2260 \u22a4) : H.index = 0 :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that $\\mathbb{Q}$ has no proper subgroups of finite index.\n", "nl_proof": "\\begin{proof}\n\n    Solution: We begin with a lemma.\n\nLemma: If $D$ is a divisible abelian group, then no proper subgroup of $D$ has finite index.\n\nProof: We saw previously that no finite group is divisible and that every proper quotient $D / A$ of a divisible group is divisible; thus no proper quotient of a divisible group is finite. Equivalently, $[D: A]$ is not finite.\n\nBecause $\\mathbb{Q}$ and $\\mathbb{Q} / \\mathbb{Z}$ are divisible, the conclusion follows.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_3_4_1", "formal_statement": "theorem exercise_3_4_1 (G : Type*) [comm_group G] [is_simple_group G] :\n    is_cyclic G \u2227 \u2203 G_fin : fintype G, nat.prime (@card G G_fin) :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that if $G$ is an abelian simple group then $G \\cong Z_{p}$ for some prime $p$ (do not assume $G$ is a finite group).\n", "nl_proof": "\\begin{proof}\n\n    Solution: Let $G$ be an abelian simple group.\n\nSuppose $G$ is infinite. If $x \\in G$ is a nonidentity element of finite order, then $\\langle x\\rangle<G$ is a nontrivial normal subgroup, hence $G$ is not simple. If $x \\in G$ is an element of infinite order, then $\\left\\langle x^2\\right\\rangle$ is a nontrivial normal subgroup, so $G$ is not simple.\n\n\n\nSuppose $G$ is finite; say $|G|=n$. If $n$ is composite, say $n=p m$ for some prime $p$ with $m \\neq 1$, then by Cauchy's Theorem $G$ contains an element $x$ of order $p$ and $\\langle x\\rangle$ is a nontrivial normal subgroup. Hence $G$ is not simple. Thus if $G$ is an abelian simple group, then $|G|=p$ is prime. We saw previously that the only such group up to isomorphism is $\\mathbb{Z} /(p)$, so that $G \\cong \\mathbb{Z} /(p)$. Moreover, these groups are indeed simple.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_3_4_5a", "formal_statement": "theorem exercise_3_4_5a {G : Type*} [group G] \n  (H : subgroup G) [is_solvable G] : is_solvable H :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that subgroups of a solvable group are solvable.\n", "nl_proof": "\\begin{proof}\n\n    Let $G$ be a solvable group and let $H \\leq G$. Since $G$ is solvable, we may find a chain of subgroups\n\n$$\n\n1=G_0 \\unlhd G_1 \\unlhd G_2 \\unlhd \\cdots \\unlhd G_n=G\n\n$$\n\nso that each quotient $G_{i+1} / G_i$ is abelian. For each $i$, define\n\n$$\n\nH_i=G_i \\cap H, \\quad 0 \\leq i \\leq n .\n\n$$\n\nThen $H_i \\leq H_{i+1}$ for each $i$. Moreover, if $g \\in H_{i+1}$ and $x \\in H_i$, then in particular $g \\in G_{i+1}$ and $x \\in G_i$, so that\n\n$$\n\ng x g^{-1} \\in G_i\n\n$$\n\nbecause $G_i \\unlhd G_{i+1}$. But $g$ and $x$ also belong to $H$, so\n\n$$\n\ng x g^{-1} \\in H_i,\n\n$$\n\nwhich shows that $H_i \\unlhd H_{i+1}$ for each $i$.\n\nNext, note that\n\n$$\n\nH_i=G_i \\cap H=\\left(G_i \\cap G_{i+1}\\right) \\cap H=G_i \\cap H_{i+1} .\n\n$$\n\nBy the Second Isomorphism Theorem, we then have\n\n$$\n\nH_{i+1} / H_i=H_{i+1} /\\left(H_{i+1} \\cap G_i\\right) \\cong H_{i+1} G_i / G_i \\leq G_{i+1} / G_i .\n\n$$\n\nSince $H_{i+1} / H_i$ is isomorphic to a subgroup of the abelian group $G_{i+1} / G_i$, it follows that $H_{i+1} / H_i$ is also abelian. This completes the proof that $H$ is solvable.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_3_4_11", "formal_statement": "theorem exercise_3_4_11 {G : Type*} [group G] [is_solvable G] \n  {H : subgroup G} (hH : H \u2260 \u22a5) [H.normal] : \n  \u2203 A \u2264 H, A.normal \u2227 \u2200 a b : A, a*b = b*a :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that if $H$ is a nontrivial normal subgroup of the solvable group $G$ then there is a nontrivial subgroup $A$ of $H$ with $A \\unlhd G$ and $A$ abelian.\n", "nl_proof": "\\begin{proof}\n\n    Suppose $H$ is a nontrivial normal subgroup of the solvable group $G$.\n\nFirst, notice that $H$, being a subgroup of a solvable group, is itself solvable. By exercise $8, H$ has a chain of subgroups\n\n$$\n\n1 \\leq H_1 \\leq \\ldots \\leq H\n\n$$\n\nsuch that each $H_i$ is a normal subgroup of $H$ itself and $H_{i+1} / H_i$ is abelian. But then the first group in the series\n\n$$\n\nH_1 / 1 \\cong H\n\n$$\n\nis an abelian subgroup of $H$.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_4_3_26", "formal_statement": "theorem exercise_4_3_26 {\u03b1 : Type*} [fintype \u03b1] (ha : fintype.card \u03b1 > 1)\n  (h_tran : \u2200 a b: \u03b1, \u2203 \u03c3 : equiv.perm \u03b1, \u03c3 a = b) : \n  \u2203 \u03c3 : equiv.perm \u03b1, \u2200 a : \u03b1, \u03c3 a \u2260 a :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $G$ be a transitive permutation group on the finite set $A$ with $|A|>1$. Show that there is some $\\sigma \\in G$ such that $\\sigma(a) \\neq a$ for all $a \\in A$.\n", "nl_proof": "\\begin{proof}\n\n    Let $G$ be a transitive permutation group on the finite set $A,|A|>1$. We want to find an element $\\sigma$ which doesn't stabilize anything, that is, we want a $\\sigma$ such that\n\n$$\n\n\\sigma \\notin G_a\n\n$$\n\nfor all $a \\in A$.\n\nSince the group is transitive, there is always a $g \\in G$ such that $b=g \\cdot a$. Let us see in what relationship the stabilizers of $a$ and $b$ are. We find\n\n$$\n\n\\begin{aligned}\n\nG_b & =\\{h \\in G \\mid h \\cdot b=b\\} \\\\\n\n& =\\{h \\in G \\mid h g \\cdot a=g \\cdot a\\} \\\\\n\n& =\\left\\{h \\in G \\mid g^{-1} h g \\cdot a=a\\right\\}\n\n\\end{aligned}\n\n$$\n\nPutting $h^{\\prime}=g^{-1} h g$, we have $h=g h^{\\prime} g^{-1}$ and\n\n$$\n\n\\begin{aligned}\n\nG_b & =g\\left\\{h^{\\prime} \\in H \\mid h^{\\prime} \\cdot a=a\\right\\} g^{-1} \\\\\n\n& =g G_a g^{-1}\n\n\\end{aligned}\n\n$$\n\nBy the above, the stabilizer subgroup of any element is conjugate to some other stabilizer subgroup. Now, the stabilizer cannot be all of $G$ (else $\\{a\\}$ would be a orbit). Thus it is a proper subgroup of $G$. By the previous exercise, we have\n\n$$\n\n\\bigcup_{a \\in A} G_a=\\bigcup_{g \\in G} g G_a g^{-1} \\subset G\n\n$$\n\n(the union of conjugates of a proper subgroup can never be all of $G$ ). This shows there is an element $\\sigma$ which is not in any stabilizer of any element of $A$. Then $\\sigma(a) \\neq a$ for all $a \\in A$, as we wanted to show.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_4_2_14", "formal_statement": "theorem exercise_4_2_14 {G : Type*} [fintype G] [group G] \n  (hG : \u00ac (card G).prime) (hG1 : \u2200 k \u2223 card G, \n  \u2203 (H : subgroup G) (fH : fintype H), @card H fH = k) : \n  \u00ac is_simple_group G :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $G$ be a finite group of composite order $n$ with the property that $G$ has a subgroup of order $k$ for each positive integer $k$ dividing $n$. Prove that $G$ is not simple.\n", "nl_proof": "\\begin{proof}\n\n    Solution: Let $p$ be the smallest prime dividing $n$, and write $n=p m$. Now $G$ has a subgroup $H$ of order $m$, and $H$ has index $p$. Then $H$ is normal in $G$.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_4_4_6a", "formal_statement": "theorem exercise_4_4_6a {G : Type*} [group G] (H : subgroup G)\n  [subgroup.characteristic H] : subgroup.normal H  :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that characteristic subgroups are normal.\n", "nl_proof": "\\begin{proof}\n\n    Let $H$ be a characterestic subgroup of $G$. By definition $\\alpha(H) \\subset H$ for every $\\alpha \\in \\operatorname{Aut}(G)$. So, $H$ is in particular invariant under the inner automorphism. Let $\\phi_g$ denote the conjugation automorphism by $g$. Then $\\phi_g(H) \\subset H \\Longrightarrow$ $g H g^{-1} \\subset H$. So, $H$ is normal. \n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_4_4_7", "formal_statement": "theorem exercise_4_4_7 {G : Type*} [group G] {H : subgroup G} [fintype H]\n  (hH : \u2200 (K : subgroup G) (fK : fintype K), card H = @card K fK \u2192 H = K) : \n  H.characteristic :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "If $H$ is the unique subgroup of a given order in a group $G$ prove $H$ is characteristic in $G$.\n", "nl_proof": "\\begin{proof}\n\n    Let $G$ be group and $H$ be the unique subgroup of order $n$. Now, let $\\sigma \\in \\operatorname{Aut}(G)$. Now Clearly $|\\sigma(G)|=n$, because $\\sigma$ is a one-one onto map. But then as $H$ is the only subgroup of order $n$, and because of the fact that a automorphism maps subgroups to subgroups, we have $\\sigma(H)=$ $H$ for every $\\sigma \\in \\operatorname{Aut}(G)$. Hence, $H$ is a characterestic subgroup of $G$.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_4_5_1a", "formal_statement": "theorem exercise_4_5_1a {p : \u2115} {G : Type*} [group G] \n  {P : subgroup G} (hP : is_p_group p P) (H : subgroup G) \n  (hH : P \u2264 H) : is_p_group p H :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that if $P \\in \\operatorname{Syl}_{p}(G)$ and $H$ is a subgroup of $G$ containing $P$ then $P \\in \\operatorname{Syl}_{p}(H)$.\n", "nl_proof": "\\begin{proof}\n\nIf $P \\leq H \\leq G$ is a Sylow $p$-subgroup of $G$, then $p$ does not divide $[G: P]$. Now $[G: P]=[G: H][H: P]$, so that $p$ does not divide $[H: P]$; hence $P$ is a Sylow $p$-subgroup of $H$.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_4_5_14", "formal_statement": "theorem exercise_4_5_14 {G : Type*} [group G] [fintype G]\n  (hG : card G = 312) :\n  \u2203 (p : \u2115) (P : sylow p G), P.normal :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that a group of order 312 has a normal Sylow $p$-subgroup for some prime $p$ dividing its order.\n", "nl_proof": "\\begin{proof}\n\n    Since $|G|=351=3^{2}.13$, $G$ has $3-$Sylow subgroup of order $9$, as well as $13-$Sylow subgroup of order $13$. Now, we count the number of such subgroups. Let $n_{13}$ be the number of $13-$Sylow subgroup and $n_{3}$ be the number of  $3-$Sylow subgroup. Now $n_{13}=1+13k$ where $1+13k|9$. The choices for $k$ is $0$. Hence, there is a unique $13-$Sylow subgroup and hence is normal.\n\n\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_4_5_16", "formal_statement": "theorem exercise_4_5_16 {p q r : \u2115} {G : Type*} [group G] \n  [fintype G]  (hpqr : p < q \u2227 q < r) \n  (hpqr1 : p.prime \u2227 q.prime \u2227 r.prime)(hG : card G = p*q*r) : \n  nonempty (sylow p G) \u2228 nonempty(sylow q G) \u2228 nonempty(sylow r G) :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $|G|=p q r$, where $p, q$ and $r$ are primes with $p<q<r$. Prove that $G$ has a normal Sylow subgroup for either $p, q$ or $r$.\n", "nl_proof": "\\begin{proof}\n\n    Let $|G|=p q r$. We also assume $p<q<r$. We prove that $G$ has a normal Sylow subgroup of $p$, $q$ or $r$. Now, Let $n_p, n_q, n_r$ be the number of Sylow-p subgroup, Sylow-q subgroup, Sylow-r subgroup resp. So, we have $n_r=1+r k$ such that $1+r k \\mid p q$. So, in this case as $r$ is greatest $n_r$ can be 1 or $p q$. We assume $n_r=p q$. Now we have $n_q=1+q k$ such that $1+q k \\mid p r$. Now,as $p<q<r, n_q$ can be 1 or $r$, or $p r$. Assume that $n_q=r$. Now we turn to $n_p$. Again my similar method we can conclude $n_p$ can be $1, q, r$, or $q r$. We assume that $n_p$ is $q$. Now we count the number of elements of order $p, q, r$. Since $n_r=p q$, the number of elements of order $r$ is $p q(r-1)$. Since $n_q=r$, the number of elements of order $q$ is $(q-1) r$. And as $n_p=q$, the number of elements of order $p$ is $(p-1) q$. So, in total we get $p q(r-1)+(q-1) r+(p-$ 1) $q=p q r+q r-r-q=p q r+r(q-1)-r$. But observe that as $q>1, r(q-1)-r>$ 0 . So the number of elements exceeds $p q r$. So, it proves that atleast $n_p$ or $n_q$ or $n_r$ is 1, which ultimately proves the result, because a unique Sylow-p subgroup is always normal.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_4_5_18", "formal_statement": "theorem exercise_4_5_18 {G : Type*} [fintype G] [group G] \n  (hG : card G = 200) : \n  \u2203 N : sylow 5 G, N.normal :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that a group of order 200 has a normal Sylow 5-subgroup.\n", "nl_proof": "\\begin{proof}\n\n    Let $G$ be a group of order $200=5^2 \\cdot 8$. Note that 5 is a prime not dividing 8 . Let $P \\in$ $S y l_5(G)$. [We know $P$ exists since $S y l_5(G) \\neq \\emptyset$ by Sylow's Theorem]\n\n\n\nThe number of Sylow 5-subgroups of $G$ is of the form $1+k \\cdot 5$, i.e., $n_5 \\equiv 1(\\bmod 5)$ and $n_5$ divides 8 . The only such number that divides 8 and equals $1 (\\bmod 5)$ is 1 so $n_5=1$. Hence $P$ is the unique Sylow 5-subgroup.\n\nSince $P$ is the unique Sylow 5-subgroup, this implies that $P$ is normal in $G$.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_4_5_20", "formal_statement": "theorem exercise_4_5_20 {G : Type*} [fintype G] [group G]\n  (hG : card G = 1365) : \u00ac is_simple_group G :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that if $|G|=1365$ then $G$ is not simple.\n", "nl_proof": "\\begin{proof}    \n\nSince $|G|=1365=3.5.7.13$, $G$ has $13-$Sylow subgroup of order $13$. Now, we count the number of such subgroups. Let $n_{13}$ be the number of $13-$Sylow subgroup. Now $n_{13}=1+13k$ where $1+13k|3.5.7$. The choices for $k$ is $0$. Hence, there is a unique $13-$Sylow subgroup and hence is normal. so $G$ is not simple.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_4_5_22", "formal_statement": "theorem exercise_4_5_22 {G : Type*} [fintype G] [group G]\n  (hG : card G = 132) : \u00ac is_simple_group G :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that if $|G|=132$ then $G$ is not simple.\n", "nl_proof": "\\begin{proof}    \n\nSince $|G|=132=2^{2}.3.11$, $G$ has $2-$Sylow subgroup of order $4$, as well as $11-$Sylow subgroup of order $11$, and $3-$Sylow subgroup of order $3$. Now, we count the number of such subgroups. Let $n_{11}$ be the number of  $11-$Sylow subgroup and $n_{3}$ be the number of  $3-$Sylow subgroup. Now $n_{11}=1+11k$ where $1+11k|12$. The choices for $k$ are $0$ or $1$. If $k=0$, there is only one $11-$Sylow subgroup and hence normal. So, assume now, that there are $12$ $11-$Sylow subgroup(for $k=1$). Now we look at $3-$ Sylow subgroups. $n_{3}=1+3k| 44$. So choice for $k$ are $0$, $1$, and $7$. If $k=0$, there is only one $3-$Sylow subgroup and hence normal. So, assume now, that there are $4$ $2-$Sylow subgroup (for $k=3$). Now we claim that simultaneously, there cannot be $12$ $11-$Sylow subgroup and $4$ $3-$Sylow subgroups provided there is more than one $2-$Sylow subgroups. So, either $2-$Sylow subgroup is normal or if not, then, either $11-$Sylow subgroup is normal being unique, or  the $3-$Sylow subgroup is normal(We don't consider the possibility of $22$ $3-$Sylow subgroup because of obvious reason). Now, to prove the claim, we observe that there are $120$ elements of order $11$. Also there are $8$ elements of order $3$. So we already get $120+8+1=129$ distinct elements in the group. Let us count the number of $2-$Sylow subgroups in $G$. $n_{2}=1+2k|33$. The possibilities for $k$ are $0$, $1$, $5$, $16$. Now, assume there is more than one $2-$Sylow subgroups. Let $H_{1}$ and $H_{2}$ be two distinct  $2-$Sylow subgroup. Now $|H_{1}|=4$. So we already get $129+3=132$ distinct elements in the group. Now $H_{2}$ being distinct from $H_{1}$, has at least one element which is not in $H_{1}$. This adds one more element in the group, at the least. Now already we have number of elements in the group exceeding the number of element in $G$. This gives a contradiction and proves the claim.\n\nHence $G$ is not simple.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_4_5_28", "formal_statement": "theorem exercise_4_5_28 {G : Type*} [group G] [fintype G] \n  (hG : card G = 105) (P : sylow 3 G) [hP : P.normal] : \n  comm_group G :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $G$ be a group of order 105. Prove that if a Sylow 3-subgroup of $G$ is normal then $G$ is abelian.\n", "nl_proof": "\\begin{proof}\n\n    Given that $G$ is a group of order $1575=3^2 .5^2 .7$. Now, Let $n_p$ be the number of Sylow-p subgroups. It is given that Sylow-3 subgroup is normal and hence is unique, so $n_3=1$. First we prove that both Sylow-5 subgroup and Sylow 7-subgroup are normal. Let $P$ be the Sylow3 subgroup. Now, Consider $G / P$, which has order $5^2 .7$. Now, the number of Sylow $-5$ subgroup of $G / P$ is given by $1+5 k$, where $1+5 k \\mid 7$. Clearly $k=0$ is the only choice and hence there is a unique Sylow-5 subgroup of $G / P$, and hence normal. In the same way Sylow-7 subgroup of $G / P$ is also unique and hence normal. Consider now the canonical map $\\pi: G \\rightarrow G / P$. The inverse image of Sylow-5 subgroup of $G / P$ under $\\pi$, call it $H$, is a normal subgroup of $G$, and $|H|=3^2 .5^2$. Similarly, the inverse image of Sylow-7 subgroup of $G / P$ under $\\pi$ call it $K$ is also normal in $G$ and $|K|=3^2 .7$. Now, consider $H$. Observe first that the number of Sylow-5 subgroup in $H$ is $1+5 k$ such that $1+5 k \\mid 9$. Again $k=0$ and hence $H$ has a unique Sylow-5 subgroup, call it $P_1$. But, it is easy to see that $P_1$ is also a Sylow-5 subgroup of $G$, because $\\left|P_1\\right|=25$. But now any other Sylow 5 subgroup of $G$ is of the form $g P_1 g^{-1}$ for some $g \\in G$. But observe that since $P_1 \\subset H$ and $H$ is normal in $G$, so $g P_1 g^1 \\subset H$, and $g P_1 g^{-1}$ is also Sylow-5 subgroup of $H$. But, then as Sylow-5 subgroup of $H$ is unique we have $g P_1 g^{-1}=P_1$. This shows that Sylow-5 subgroup of $G$ is unique and hence normal in $G$.\n\n\n\nSimilarly, one can argue the same for $K$ and deduce that Sylow-7 subgroup of $G$ is unique and hence normal. So, the first part of the problem is done.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_5_4_2", "formal_statement": "theorem exercise_5_4_2 {G : Type*} [group G] (H : subgroup G) : \n  H.normal \u2194 \u2045(\u22a4 : subgroup G), H\u2046 \u2264 H :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that a subgroup $H$ of $G$ is normal if and only if $[G, H] \\leq H$.\n", "nl_proof": "\\begin{proof}\n\n    $H \\unlhd G$ is equivalent to $g^{-1} h g \\in H, \\forall g \\in G, \\forall h \\in H$. We claim that holds if and only if $h^{-1} g^{-1} h g \\in H, \\forall g \\in G, \\forall h \\in H$, i.e., $\\left\\{h^{-1} g^{-1} h g: h \\in H, g \\in G\\right\\} \\subseteq H$. That holds by the following argument:\n\nIf $g^{-1} h g \\in H, \\forall g \\in G, \\forall h \\in H$, note that $h^{-1} \\in H$, so multiplying them, we also obtain an element of $H$.\n\nOn the other hand, if $h^{-1} g^{-1} h g \\in H, \\forall g \\in G, \\forall h \\in H$, then\n\n$$\n\nh h^{-1} g^{-1} h g=g^{-1} h g \\in H, \\forall g \\in G, \\forall h \\in H .\n\n$$\n\nSince $\\left\\{h^{-1} g^{-1} h g: h \\in H, g \\in G\\right\\} \\subseteq H \\Leftrightarrow\\left\\langle\\left\\{h^{-1} g^{-1} h g: h \\in H, g \\in G\\right\\}\\right\\rangle \\leq H$, we've solved the exercise by definition of $[H, G]$.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_7_1_11", "formal_statement": "theorem exercise_7_1_11 {R : Type*} [ring R] [is_domain R] \n  {x : R} (hx : x^2 = 1) : x = 1 \u2228 x = -1 :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that if $R$ is an integral domain and $x^{2}=1$ for some $x \\in R$ then $x=\\pm 1$.\n", "nl_proof": "\\begin{proof}\n\n    Solution: If $x^2=1$, then $x^2-1=0$. Evidently, then,\n\n$$\n\n(x-1)(x+1)=0 .\n\n$$\n\nSince $R$ is an integral domain, we must have $x-1=0$ or $x+1=0$; thus $x=1$ or $x=-1$.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_7_1_15", "formal_statement": "theorem exercise_7_1_15 {R : Type*} [ring R] (hR : \u2200 a : R, a^2 = a) :\n  comm_ring R :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "A ring $R$ is called a Boolean ring if $a^{2}=a$ for all $a \\in R$. Prove that every Boolean ring is commutative.\n", "nl_proof": "\\begin{proof}\n\n    Solution: Note first that for all $a \\in R$,\n\n$$\n\n-a=(-a)^2=(-1)^2 a^2=a^2=a .\n\n$$\n\nNow if $a, b \\in R$, we have\n\n$$\n\na+b=(a+b)^2=a^2+a b+b a+b^2=a+a b+b a+b .\n\n$$\n\nThus $a b+b a=0$, and we have $a b=-b a$. But then $a b=b a$. Thus $R$ is commutative.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_7_2_12", "formal_statement": "theorem exercise_7_2_12 {R G : Type*} [ring R] [group G] [fintype G] : \n  \u2211 g : G, monoid_algebra.of R G g \u2208 center (monoid_algebra R G) :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $G=\\left\\{g_{1}, \\ldots, g_{n}\\right\\}$ be a finite group. Prove that the element $N=g_{1}+g_{2}+\\ldots+g_{n}$ is in the center of the group ring $R G$.\n", "nl_proof": "\\begin{proof}\n\n    Let $M=\\sum_{i=1}^n r_i g_i$ be an element of $R[G]$. Note that for each $g_i \\in G$, the action of $g_i$ on $G$ by conjugation permutes the subscripts. Then we have the following.\n\n$$\n\n\\begin{aligned}\n\nN M &=\\left(\\sum_{i=1}^n g_i\\right)\\left(\\sum_{j=1}^n r_j g_j\\right) \\\\\n\n&=\\sum_{j=1}^n \\sum_{i=1}^n r_j g_i g_j \\\\\n\n&=\\sum_{j=1}^n \\sum_{i=1}^n r_j g_j g_j^{-1} g_i g_j \\\\\n\n&=\\sum_{j=1}^n r_j g_j\\left(\\sum_{i=1}^n g_j^{-1} g_i g_j\\right) \\\\\n\n&=\\sum_{j=1}^n r_j g_j\\left(\\sum_{i=1}^n g_i\\right) \\\\\n\n&=\\left(\\sum_{j=1}^n r_j g_j\\right)\\left(\\sum_{i=1}^n g_i\\right) \\\\\n\n&=M N .\n\n\\end{aligned}\n\n$$\n\nThus $N \\in Z(R[G])$.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_7_3_37", "formal_statement": "theorem exercise_7_3_37 {R : Type*} {p m : \u2115} (hp : p.prime) \n  (N : ideal $ zmod $ p^m) : \n  is_nilpotent N \u2194  is_nilpotent (ideal.span ({p} : set $ zmod $ p^m)) :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "An ideal $N$ is called nilpotent if $N^{n}$ is the zero ideal for some $n \\geq 1$. Prove that the ideal $p \\mathbb{Z} / p^{m} \\mathbb{Z}$ is a nilpotent ideal in the ring $\\mathbb{Z} / p^{m} \\mathbb{Z}$.\n", "nl_proof": "\\begin{proof}\n\n    First we prove a lemma.\n\nLemma: Let $R$ be a ring, and let $I_1, I_2, J \\subseteq R$ be ideals such that $J \\subseteq I_1, I_2$. Then $\\left(I_1 / J\\right)\\left(I_2 / J\\right)=I_1 I_2 / J$.\n\nProof: ( $\\subseteq$ ) Let\n\n$$\n\n\\alpha=\\sum\\left(x_i+J\\right)\\left(y_i+J\\right) \\in\\left(I_1 / J\\right)\\left(I_2 / J\\right) .\n\n$$\n\nThen\n\n$$\n\n\\alpha=\\sum\\left(x_i y_i+J\\right)=\\left(\\sum x_i y_i\\right)+J \\in\\left(I_1 I_2\\right) / J .\n\n$$\n\nNow let $\\alpha=\\left(\\sum x_i y_i\\right)+J \\in\\left(I_1 I_2\\right) / J$. Then\n\n$$\n\n\\alpha=\\sum\\left(x_i+J\\right)\\left(y_i+J\\right) \\in\\left(I_1 / J\\right)\\left(I_2 / J\\right) .\n\n$$\n\nFrom this lemma and the lemma to Exercise 7.3.36, it follows by an easy induction that\n\n$$\n\n\\left(p \\mathbb{Z} / p^m \\mathbb{Z}\\right)^m=(p \\mathbb{Z})^m / p^m \\mathbb{Z}=p^m \\mathbb{Z} / p^m \\mathbb{Z} \\cong 0 .\n\n$$\n\nThus $p \\mathbb{Z} / p^m \\mathbb{Z}$ is nilpotent in $\\mathbb{Z} / p^m \\mathbb{Z}$.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_8_1_12", "formal_statement": "theorem exercise_8_1_12 {N : \u2115} (hN : N > 0) {M M': \u2124} {d : \u2115}\n  (hMN : M.gcd N = 1) (hMd : d.gcd N.totient = 1) \n  (hM' : M' \u2261 M^d [ZMOD N]) : \n  \u2203 d' : \u2115, d' * d \u2261 1 [ZMOD N.totient] \u2227 \n  M \u2261 M'^d' [ZMOD N] :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $N$ be a positive integer. Let $M$ be an integer relatively prime to $N$ and let $d$ be an integer relatively prime to $\\varphi(N)$, where $\\varphi$ denotes Euler's $\\varphi$-function. Prove that if $M_{1} \\equiv M^{d} \\pmod N$ then $M \\equiv M_{1}^{d^{\\prime}} \\pmod N$ where $d^{\\prime}$ is the inverse of $d \\bmod \\varphi(N)$: $d d^{\\prime} \\equiv 1 \\pmod {\\varphi(N)}$.\n", "nl_proof": "\\begin{proof}\n\n    Note that there is some $k \\in \\mathbb{Z}$ such that $M^{d d^{\\prime}} \\equiv M^{k \\varphi(N)+1} \\equiv\\left(M^{\\varphi(N)}\\right)^k \\cdot M \\bmod N$. By Euler's Theorem we have $M^{\\varphi(N)} \\equiv 1 \\bmod N$, so that $M_1^{d^{\\prime}} \\equiv M \\bmod N$.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_8_3_4", "formal_statement": "theorem exercise_8_3_4 {R : Type*} {n : \u2124} {r s : \u211a} \n  (h : r^2 + s^2 = n) : \n  \u2203 a b : \u2124, a^2 + b^2 = n :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that if an integer is the sum of two rational squares, then it is the sum of two integer squares.\n", "nl_proof": "\\begin{proof}\n\n    Let $n=\\frac{a^2}{b^2}+\\frac{c^2}{d^2}$, or, equivalently, $n(b d)^2=a^2 d^2+c^2 b^2$. From this, we see that $n(b d)^2$ can be written as a sum of two squared integers. Therefore, if $q \\equiv 3(\\bmod 4)$ and $q^i$ appears in the prime power factorization of $n, i$ must be even. Let $j \\in \\mathbb{N} \\cup\\{0\\}$ such that $q^j$ divides $b d$. Then $q^{i-2 j}$ divides $n$. But since $i$ is even, $i-2 j$ is even as well. Consequently, $n$ can be written as a sum of two squared integers.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_8_3_6a", "formal_statement": "theorem exercise_8_3_6a {R : Type*} [ring R]\n  (hR : R = (gaussian_int \u29f8 ideal.span ({\u27e80, 1\u27e9} : set gaussian_int))) :\n  is_field R \u2227 \u2203 finR : fintype R, @card R finR = 2 :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that the quotient ring $\\mathbb{Z}[i] /(1+i)$ is a field of order 2.\n", "nl_proof": "\\begin{proof}\n\n    Let $a+b i \\in \\mathbb{Z}[i]$. If $a \\equiv b \\bmod 2$, then $a+b$ and $b-a$ are even and $(1+i)\\left(\\frac{a+b}{2}+\\frac{b-a}{2} i\\right)=a+b i \\in\\langle 1+i\\rangle$. If $a \\not \\equiv b \\bmod 2$ then $a-1+b i \\in\\langle 1+i\\rangle$. Therefore every element of $\\mathbb{Z}[i]$ is in either $\\langle 1+i\\rangle$ or $1+\\langle 1+i\\rangle$, so $\\mathbb{Z}[i] /\\langle 1+i\\rangle$ is a finite ring of order 2 , which must be a field.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_9_1_6", "formal_statement": "theorem exercise_9_1_6 : \u00ac is_principal \n  (ideal.span ({X 0, X 1} : set (mv_polynomial (fin 2) \u211a))) :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that $(x, y)$ is not a principal ideal in $\\mathbb{Q}[x, y]$.\n", "nl_proof": "\\begin{proof}\n\n    Suppose, to the contrary, that $(x, y)=p$ for some polynomial $p \\in \\mathbb{Q}[x, y]$. From $x, y \\in$ $(x, y)=(p)$ there are $s, t \\in \\mathbb{Q}[x, y]$ such that $x=s p$ and $y=t p$.\n\nThen:\n\n$$\n\n\\begin{aligned}\n\n& 0=\\operatorname{deg}_y(x)=\\operatorname{deg}_y(s)+\\operatorname{deg}_y(p) \\text { so } \\\\\n\n& 0=\\operatorname{deg}_y(p) \\\\\n\n& 0=\\operatorname{deg}_x(y)=\\operatorname{deg}_x(s)+\\operatorname{deg}_x(p) \\text { so } \\\\\n\n& 0=\\operatorname{deg}_x(p) \\text { so }\n\n\\end{aligned}\n\n$$\n\nFrom : $\\quad 0=\\operatorname{deg}_y(p)=\\operatorname{deg}_x(p)$ we get $\\operatorname{deg}(p)=0$ and $p \\in \\mathbb{Q}$.\n\nBut $p \\in(p)=(x, y)$ so $p=a x+b y$ for some $a, b \\in \\mathbb{Q}[x, y]$\n\n$$\n\n\\begin{aligned}\n\n\\operatorname{deg}(p) & =\\operatorname{deg}(a x+b y) \\\\\n\n& =\\min (\\operatorname{deg}(a)+\\operatorname{deg}(x), \\operatorname{deg}(b)+\\operatorname{deg}(y)) \\\\\n\n& =\\min (\\operatorname{deg}(a)+1, \\operatorname{deg}(b)+1) \\geqslant 1\n\n\\end{aligned}\n\n$$\n\nwhich contradicts $\\operatorname{deg}(p)=0$.\n\nSo we conclude that $(x, y)$ is not principal ideal in $\\mathbb{Q}[x, y]$\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_9_3_2", "formal_statement": "theorem exercise_9_3_2 {f g : polynomial \u211a} (i j : \u2115)\n  (hfg : \u2200 n : \u2115, \u2203 a : \u2124, (f*g).coeff = a) :\n  \u2203 a : \u2124, f.coeff i * g.coeff j = a :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that if $f(x)$ and $g(x)$ are polynomials with rational coefficients whose product $f(x) g(x)$ has integer coefficients, then the product of any coefficient of $g(x)$ with any coefficient of $f(x)$ is an integer.\n", "nl_proof": "\\begin{proof}\n\n    Let $f(x), g(x) \\in \\mathbb{Q}[x]$ be such that $f(x) g(x) \\in \\mathbb{Z}[x]$.\n\nBy Gauss' Lemma there exists $r, s \\in \\mathbb{Q}$ such that $r f(x), s g(x) \\in \\mathbb{Z}[x]$, and $(r f(x))(s g(x))=r s f(x) g(x)=f(x) g(x)$. From this last relation we can conclude that $s=r^{-1}$.\n\n\n\nTherefore for any coefficient $f_i$ of $f(x)$ and $g_j$ of $g(x)$ we have that $r f_i, r^{-1} g_j \\in$ $\\mathbb{Z}$ and by multiplicative closure and commutativity of $\\mathbb{Z}$ we have that $r f_i r^{-1} g_j=$ $f_i g_j \\in \\mathbb{Z}$\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_9_4_2b", "formal_statement": "theorem exercise_9_4_2b : irreducible \n  (X^6 + 30*X^5 - 15*X^3 + 6*X - 120 : polynomial \u2124) :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that $x^6+30x^5-15x^3 + 6x-120$ is irreducible in $\\mathbb{Z}[x]$.\n", "nl_proof": "\\begin{proof}\n\n    $$\n\nx^6+30 x^5-15 x^3+6 x-120\n\n$$\n\nThe coefficients of the low order.: $30,-15,0,6,-120$\n\nThey are divisible by the prime 3 , but $3^2=9$ doesn 't divide $-120$. So this polynomial is irreducible over $\\mathbb{Z}$. \n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_9_4_2d", "formal_statement": "theorem exercise_9_4_2d {p : \u2115} (hp : p.prime \u2227 p > 2) \n  {f : polynomial \u2124} (hf : f = (X + 2)^p): \n  irreducible (\u2211 n in (f.support - {0}), (f.coeff n) * X ^ (n-1) : \n  polynomial \u2124) :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that $\\frac{(x+2)^p-2^p}{x}$, where $p$ is an odd prime, is irreducible in $\\mathbb{Z}[x]$.\n", "nl_proof": "\\begin{proof}\n\n$\\frac{(x+2)^p-2^p}{x} \\quad \\quad p$ is on add pprime $Z[x]$\n\n$$\n\n\\frac{(x+2)^p-2^p}{x} \\quad \\text { as a polynomial we expand }(x+2)^p\n\n$$\n\n$2^p$ cancels with $-2^p$, every remaining term has $x$ as $a$ factor\n\n$$\n\n\\begin{aligned}\n\n& x^{p-1}+2\\left(\\begin{array}{l}\n\np \\\\\n\n1\n\n\\end{array}\\right) x^{p-2}+2^2\\left(\\begin{array}{l}\n\np \\\\\n\n2\n\n\\end{array}\\right) x^{p-3}+\\ldots+2^{p-1}\\left(\\begin{array}{c}\n\np \\\\\n\np-1\n\n\\end{array}\\right) \\\\\n\n& 2^k\\left(\\begin{array}{l}\n\np \\\\\n\nk\n\n\\end{array}\\right) x^{p-k-1}=2^k \\cdot p \\cdot(p-1) \\ldots(p-k-1), \\quad 0<k<p\n\n\\end{aligned}\n\n$$\n\nEvery lower order coef. has $p$ as a factor but doesnt have $\\$ \\mathrm{p}^{\\wedge} 2 \\$$ as a fuction so the polynomial is irreducible by Eisensteins Criterion.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_9_4_11", "formal_statement": "theorem exercise_9_4_11 : \n  irreducible ((X 0)^2 + (X 1)^2 - 1 : mv_polynomial (fin 2) \u211a) :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that $x^2+y^2-1$ is irreducible in $\\mathbb{Q}[x,y]$.\n", "nl_proof": "\\begin{proof}\n\n$$\n\np(x)=x^2+y^2-1 \\in Q[y][x] \\cong Q[y, x]\n\n$$\n\nWe have that $y+1 \\in Q[y]$ is prime and $Q[y]$ is an UFD, since $p(x)=x^2+y^2-1=x^2+$ $(y+1)(y-1)$ by the Eisenstein criterion $x^2+y^2-1$ is irreducibile in $Q[x, y]$.\n\n\\end{proof}"}