hcju/proofwikiqa huggingface dataset:
Consider the difference {{begin-eqn}} {{eqn | l = \Delta F_{y'}       | r = \map F {x + \Delta x, y + \Delta y, y' + \Delta y'} - \map F {x, y, y'}       | c =  }} {{eqn | r = \Delta x \overline F_{y' x} + \Delta y \overline F_{y'y} + \Delta y' \overline F_{y'y'}       | c = Multivariate Mean Value Theorem }} {{end-eqn}} Overbar indicates that derivatives are evaluated along certain intermediate curves. {{Explain|intermediate curves}} Divide $\Delta F_{y'} $ by $\Delta x$ and consider the limit $\Delta x\to 0$: :$\displaystyle \lim_{\Delta x \mathop \to 0} \frac {\Delta F_{y'} } {\Delta x} = \lim_{\Delta x \mathop \to 0} \paren {\overline F_{y'x} + \frac {\Delta y} {\Delta x} \overline F_{y' y} + \frac {\Delta y'} {\Delta x} \overline F_{y'y'} }$ Existence of second derivatives and continuity of $F$ is guaranteed by conditions of the theorem: :$\displaystyle \lim_{\Delta x \mathop \to 0} \frac {\Delta F_{y'} } {\Delta x} = F_{y' x}$ :$\displaystyle \lim_{\Delta x \mathop \to 0} \overline F_{y' x} = F_{y' x}$ :$\displaystyle \lim_{\Delta x \mathop \to 0} \overline F_{y' y} = F_{y' y}$ :$\displaystyle \lim_{\Delta x \mathop \to 0} \overline F_{y' y} = F_{y'y'}$ Similarly: :$\displaystyle \lim_{\Delta x \mathop \to 0} \frac {\Delta y} {\Delta x} =  y'$ By Product Rule for Limits of Functions, it follows that: :$\displaystyle \lim_{\Delta x \mathop \to 0} \frac {\Delta y'} {\Delta x} = y''$ Hence $y''$ exists wherever $F_{y' y'} \ne 0$. Euler's equation and continuity of necessary derivatives of $F$ and $y$ implies that $y''$ is continuous. {{Explain|explain how barred functionals become unbarred}}
Let $x \in S$. By definition of closure operator/inflationary: :$x \preceq \map c x$ By definition of upper closure of element: :$\map c x \in x^\succeq$ By definition of image of mapping: :$\map c x \in c \sqbrk S$ By definition of intersection: :$\map c x \in x^\succeq \cap c \sqbrk S$ By definitions of infimum and lower bound: :$\map {\inf_L} {x^\succeq \cap c \sqbrk S} \preceq \map c x$ We will prove that :$\map c x$ is lower bound for $x^\succeq \cap c \sqbrk S$ Let $y \in x^\succeq \cap c \sqbrk S$ By definition of intersection: :$y \in x^\succeq$ and $y \in c \sqbrk S$ By definition of image of mapping: :$\exists z \in S: y = \map c z$ By definition of closure operator/idempotent: :$y = \map c y$ By definition of upper closure of element: :$x \preceq y$ Thus by definition of closure operator/increasing: :$\map c x \preceq y$ {{qed|lemma}} By definition of infimum: :$\map c x \preceq \map {\inf_L} {x^\succeq \cap c \sqbrk S}$ By definition of antisymmetry: :$\map c x = \map {\inf_L} {x^\succeq \cap c \sqbrk S}$ Thus by definition of operator generated by $\struct {c \sqbrk S, \precsim}$: :$\map {\map {\operatorname {operator} } {\struct {c \sqbrk S, \precsim} } } x = \map c x$ Hence by Equality of Mappings: :$\map {\operatorname {operator} } {\struct {c \sqbrk S, \precsim} } = c$ {{qed}}
{{AimForCont}} $\alpha \left({x, y}\right)$ is nonzero at some point in $R$. Then $\alpha \left({x, y}\right)$ is also nonzero in some disk $D$ such that: :$\left({x - x_0}\right)^2 + \left({y - y_0}\right)^2 \le \epsilon^2$ Suppose: :$\left({x, y}\right) = \operatorname {sgn} \left({\alpha \left({x, y}\right)}\right) \left({\epsilon^2 - \left({x - x_0}\right)^2 + \left({y - y_0}\right)^2}\right)^3$  in this disk and $0$ elsewhere. Thus $h \left({x, y}\right)$ satisfies conditions of the theorem. However: :$\displaystyle \int \int_R \alpha \left({x, y}\right) h \left({x, y}\right) \rd x \rd y = \int \int_D \left\vert{\alpha \left({x, y}\right)}\right\vert \left({\epsilon^2 - \left({x - x_0}\right)^2 + \left({y - y_0}\right)^2}\right)^3 \ge 0$ Hence the result, by Proof by Contradiction. {{qed}}
From Continued Fraction Expansion of $\sqrt {13}$: :$\sqrt {13} = \sqbrk {3, \sequence {1, 1, 1, 1, 6} }$ The cycle is of length is $5$. By the solution of Pell's Equation, the only solutions of $x^2 - 13 y^2 = 1$ are: :${p_{5 r} }^2 - 13 {q_{5 r} }^2 = \paren {-1}^{5 r}$ for $r = 1, 2, 3, \ldots$ When $r = 1$ this gives: :${p_5}^2 - 13 {q_5}^2 = -1$ which is not the solution required. When $r = 2$ this gives: :${p_{10} }^2 - 13 {q_{10} }^2 = 1$ From Convergents of Continued Fraction Expansion of $\sqrt {13}$: :$p_{10} = 649$ :$q_{10} = 180$ although on that page the numbering goes from $p_0$ to $p_9$, and $q_0$ to $q_9$. {{qed}}
From Continued Fraction Expansion of $\sqrt {29}$: :$\sqrt {29} = \sqbrk {5, \sequence {2, 1, 1, 2, 10} }$ The cycle is of length is $5$. By the solution of Pell's Equation, the only solutions of $x^2 - 29 y^2 = 1$ are: :${p_{5 r} }^2 - 29 {q_{5 r} }^2 = \paren {-1}^{5 r}$ for $r = 1, 2, 3, \ldots$ When $r = 1$ this gives: :${p_5}^2 - 29 {q_5}^2 = -1$ which is not the solution required. When $r = 2$ this gives: :${p_{10} }^2 - 29 {q_{10} }^2 = 1$ From Convergents of Continued Fraction Expansion of $\sqrt {29}$: :$p_{10} = 9801$ :$q_{10} = 1820$ although on that page the numbering goes from $p_0$ to $p_9$, and $q_0$ to $q_9$. {{qed}}
From Condition for Differentiable Functional to have Extremum we have :$\bigvalueat {\delta J \sqbrk {y; h} } {y \mathop = \hat y} = 0$ The variation exists if $J$ is a differentiable functional. We will start from the increment of a functional: {{explain|make the above link point to a page dedicated to the appropriate definition}} {{begin-eqn}} {{eqn | l = \Delta J \sqbrk {y; h}       | r = J \sqbrk {y + h} - J \sqbrk y       | c = definition }} {{eqn | r = \int_a^b \map F {x, y + h, y' + h'} \rd x - \int_a^b \map F {x, y, y'} \rd x       | c = $(1)$ }} {{eqn | r = \int_a^b \paren {\map F {x, y + h, y' + h'} - \map F {x, y, y'} } \rd x }} {{end-eqn}} {{explain|"definition" in the above -- point to what it is a definition of}} Using multivariate Taylor's theorem, one can expand $\map F {x, y + h, y' + h'}$ with respect to $h$ and $h'$: :$\map F {x, y + h, y' + h'} = \bigvalueat {\map F {x, y + h, y' + h'} } {h \mathop = 0, \, h' \mathop = 0} + \valueat {\dfrac {\partial \map F {x, y + h, y' + h'} } {\partial y} } {h \mathop = 0, \, h' \mathop = 0} h + \valueat {\dfrac {\partial {\map F {x, y + h, y' + h'} } } {\partial y'} } {h \mathop = 0, \, h' \mathop = 0} h' + \map \OO {h^2, h h', h'^2}$ Substitute this back into the integral.  Note that the first term in the expansion and the negative one in the integral will cancel out: :$\ds \Delta J \paren {y; h} = \int_a^b \paren {\map F {x, y, y'}_y h + \map F {x, y, y'}_{y'} h' + \map \OO {h^2, h h', h'^2} } \rd x$ {{explain|What do the subscripted $y$ and $y'$ mean in the above?}} Terms in $\map \OO {h^2, h'^2}$ represent terms of order higher than 1 with respect to $h$ and $h'$. Now we expand: :$\ds \int_a^b\map \OO {h^2, h h', h'^2} \rd x$ Every term in this expansion will be of the form: :$\ds \int_a^b \map A {m, n} \frac {\partial^{m + n} \map F {x, y, y'} } {\partial y^m \partial {y'}^n} h^m h'^n \rd x$ where $m, n \in \N$ and $m + n \ge 2$ {{Explain|How to convert powers of $h'$ into $h$? Integration by parts is the only obvious candidate, but how precisely? Also check, if this is necessary}} By definition, the integral not counting in $\map \OO {h^2, h h', h'^2}$ is a variation of functional. :$\ds \delta J \sqbrk {y; h} = \int_a^b \paren {F_y h + F_{y'} h'} \rd x$ Now, integrate by parts and note that $\map h x$ does not necessarily vanish at the endpoints: {{begin-eqn}} {{eqn | l = \delta J \sqbrk {y; h}       | r = \int_a^b \paren {F_y - \frac \d {\d x} F_{y'} } \map h x \rd x + \bigintlimits {F_{y'} \map h x} {x \mathop = a} {x \mathop = b} }} {{eqn | r = \int_a^b \paren {F_y - \frac \d {\d x} F_{y'} } \map h x \rd x + \bigvalueat {F_{y'} } {x \mathop = b} \map h b - \bigvalueat {F_{y'} } {x \mathop = a} \map h a }} {{end-eqn}} Then, for arbitrary $\map h x$, $J$ has an extremum if: :$ \begin {cases} F_y - \dfrac \d {\d x} F_{y'} = 0 \\ \bigvalueat {F_{y'} } {x \mathop = a} = 0\\ \bigvalueat {F_{y'} } {x \mathop = b} = 0 \end {cases}$ {{qed}}
By definition of ordered set of closure operators: :$\operatorname{Closure}\left({L}\right) = \left({X, \preceq'}\right)$ where :$X$ is the set of all closure operators on $L$, :for all closure operators $f, g$ on $L$: $f \preceq' g \iff f \preceq g$ :$\preceq$ denotes the ordering on mappings. By definition of ordered set of closure systems: :$\operatorname{ClSystems}\left({L}\right) = \left({Y, \precsim}\right)$ where :$Y$ is the set of all closure systems on $L$, :for all closure systems $s_1 = \left({T_1, \preceq_1}\right)$, $s_2 = \left({T_2, \preceq_2}\right)$ on $L$: $s_1 \precsim s_2 \iff T_1 \subseteq T_2$ Define $F:X \to Y$ by Image of Closure Operator Inherits Infima: :for all closure operators $f$ on $L$: $F\left({f}\right) = \left({f\left[{S}\right], \mathord\preceq \cap \left({f\left[{S}\right] \times f\left[{S}\right]}\right)}\right)$ By Operator Generated by Closure System is Closure Operator and Image of Operator Generated by Closure System is Set of Closure System: :$F$ is a surjection. By Ordering on Closure Operators iff Images are Including: :$F$ is an order embedding. Thue by definition: :$F$ is an order isomorphism. {{qed}}
This follows by mutatis mutandis of the proof of Infimum in Ordered Subset. {{qed}}
=== Sufficient Condition === Assume that :$\operatorname{operator}\left({S}\right)$ preserves directed suprema. Let $Z$ be directed subset of $Y$ such that :$Z$ admits a supremum in $L$. By Image of Operator Generated by Closure System is Set of Closure System: :$\operatorname{operator}\left({S}\right)\left[{X}\right] = Y$ By Operator Generated by Closure System is Closure Operator: :$\operatorname{operator}\left({S}\right)$ is closure operator. By definition of closure operator/idempotent: :$\operatorname{operator}\left({S}\right)\left[{Z}\right] = Z$ By definitions of Definition:Mapping Preserves Supremum/Directed and Definition:Mapping Preserves Supremum/Subset: :$\sup_L Z = \operatorname{operator}\left({S}\right)\left({\sup_L Z}\right)$ Thus by definition of image of mapping: :$\sup_L Z \in Y$ {{qed|lemma}} === Necessary Condition === Assume that :$S$ inherits directed suprema. Define $f := \operatorname{operator}\left({S}\right)$. By Image of Operator Generated by Closure System is Set of Closure System: :$f\left[{X}\right] = Y$ Let $A$ be a directed subset of $X$ such that :$A$ admits a supremum in $L$. Thus by definition of complete lattice: :$f\left[{A}\right]$ admits a supremum in $L$. We will prove that :$f\left({\sup_L A}\right)$ is upper bound for $f\left[{A}\right]$ Let $x \in f\left[{A}\right]$. By definition of image of subset: :$\exists y \in A: x = f\left({y}\right)$ By definitions of supremum and upper bound: :$y \preceq \sup_L A$ Thus by definition of closure operator/increasing: :$x \preceq f\left({\sup_L A}\right)$ {{qed|lemma}} By definition of supremum: :$\sup_L\left({f\left[{A}\right]}\right) \preceq f\left({\sup_L A}\right)$ We will prove that :$\sup_L\left({f\left[{A}\right]}\right)$ is upper bound for $A$ Let $x \in A$. By definition of image of set: :$f \left({x}\right) \in f \left[{A}\right]$ By definitions of supremum and upper bound: :$f \left({x}\right) \preceq \sup_L \left({ f \left[{A}\right]}\right)$ By definition of closure operator/inflationary: :$x \preceq f \left({x}\right)$ Thus by definition of transitivity: :$x \preceq \sup_L \left({ f \left[{A}\right]}\right)$ {{qed|lemma}} By definition of supremum: :$\sup_L A \preceq \sup_L \left({ f \left[{A}\right]}\right)$ By Image of Directed Subset under Increasing Mapping is Directed: :$f \left[{A}\right]$ is directed in $L$. By definition of ordered subset: :$f \left[{A}\right]$ is directed in $S$. By definition of directed suprema inheriting: :$\sup_L \left({ f \left[{A}\right]}\right) \in Y$ By definition of closure operator/idempotent: :$f \left({\sup_L \left({ f \left[{A}\right]}\right)}\right) = \sup_L \left({ f \left[{A}\right]}\right)$ By definition of closure operator/increasing: :$f \left({\sup_L A}\right) \preceq \sup_L \left({ f \left[{A}\right]}\right)$ Thus by definition of antisymmetry: :$f \left({\sup_L A}\right) = \sup_L \left({ f \left[{A}\right]}\right)$ {{qed}}
By Operator Generated by Image of Closure Operator is Closure Operator: :$\operatorname{operator}\left({\left({h \left[{S}\right], \precsim}\right)}\right) = h$ where $\operatorname{operator}\left({\left({h \left[{S}\right], \precsim}\right)}\right)$ denotes the operator generated by $\left({h \left[{S}\right], \precsim}\right)$ Hence the result holds by Operator Generated by Closure System Preserves Directed Suprema iff Closure System Inherits Directed Suprema. {{qed}}
We will prove that :$\inf X$ is lower bound for $Y$. Let $x \in Y$. By definition of coarser subset: :$\exists y \in X: y \preceq x$ By definitions of infimum and lower bound: :$\inf X \preceq y$ Thus by definition of transitivity: :$\inf X \preceq x$ {{qed|lemma}} Hence by definition of infimum: :$\inf X \preceq \inf Y$ {{qed}}
We will prove that :$\sup Y$ is upper bound for $X$. Let $x \in X$. By definition of finer subset: :$\exists y \in Y: x \preceq y$ By definitions of supremum and upper bound: :$y \preceq \sup Y$ Thus by definition of transitivity: :$x \preceq \sup Y$ {{qed|lemma}} Hence by definition of supremum: :$\sup X \preceq \sup Y$ {{qed}}
By definition, the Catalan polyhedra are the dual polyhedra of the Archimedean polyhedra. There are $13$ Archimedean polyhedra, and so there are $13$ Catalan polyhedra. {{ProofWanted|Include links to specific results linking the Archimedean polyhedra with their duals.}}
In the functional: :$\displaystyle \JJ \sqbrk y = \int_{x_0}^{x_1} \map F {x, y, y'} \rd x$ let the argument $y$ stand for a curve which is given in a parametric form.  In other words, let the curve be described by the vector $\tuple {\map y t,\map x t}$ rather than $\paren {\map y x, x}$. Then the functional can be rewritten as: {{begin-eqn}} {{eqn | l = \JJ \sqbrk y       | r = \int_{t_0}^{t_1} F \sqbrk {\map x t, \map y t, \frac {\d \, \map y t} {\d \, \map x t} } \rd \map x t }} {{eqn | r = \int_{t_0}^{t_1} F \sqbrk {\map x t, \map y t, \frac {\frac {\d y} {\d t} } {\frac {\d x} {\d t} } } \frac {\d x} {\d t} \rd t }} {{eqn | r = \int_{t_0}^{t_1} F \sqbrk {\map x t, \map y t, \frac {\map {\dot y} t} {\map {\dot x} t} } \map {\dot x} t \rd t }} {{eqn | r = \int_{t_0}^{t_1} \map \Phi {x, y, \dot x, \dot y} \rd t }} {{end-eqn}} The function on the {{RHS}} does not involve $t$ explicitly. Suppose that it is positive-homogeneous of degree 1 in $\map {\dot x} t$ and $\map {\dot y} t$: :$\map \Phi {x, y, \lambda \dot x, \lambda \dot y} = \lambda \map \Phi {x, y, \dot x, \dot y}$ for all $\lambda > 0$. Now we will show that the value of such functional depends only on the curve in the $x y$-plane defined by the parametric equations $x = \map x t$ and $y = \map y t$, and not on the functions $\map x t$, $\map y t$ themselves.  Let a new parameter $\tau$ be chosen such that $t = \map t \tau$, where $\dfrac {\d t} {\d \tau} > 0$. Let the interval $\closedint {t_0} {t_1}$ be mapped onto $\closedint {\tau_0} {\tau_1}$. Since $\Phi$ is positive-homogeneous of degree $1$ in $\dot x$ and $\dot y$, it follows that: {{begin-eqn}} {{eqn | l = \int_{\tau_0}^{\tau_1} \map \Phi {x, y, \frac {\d x} {\d \tau}, \frac {\d y} {\d \tau} } \rd \tau       | r = \int_{\tau_0}^{\tau_1} \map \Phi {x, y, \dot x \frac {\d t} {\d \tau}, \dot y \frac {\d t} {\d \tau} } \rd \tau       | c = Chain Rule for Derivatives: $\dfrac {\d x} {\d \tau} = \dfrac {\d x} {\d t} \dfrac {\d t} {\d \tau}$ }} {{eqn | r = \int_{\tau_0}^{\tau_1} \map \Phi {x, y, \dot x, \dot y} \frac {\d t} {\d \tau} \rd \tau }} {{eqn | r = \int_{t_0}^{t_1} \map \Phi {x, y, \dot x, \dot y} \rd t }} {{end-eqn}} {{qed}} {{proof wanted|seems that only one condition is proven; check for the other one}}
Thus by assumption: :$A \cap B$ is a non-empty set. Let $x, y \in A \cap B$. By definition of intersection: :$x, y \in A$ and $x, y \in B$ By definition of directed subset: :$\exists z \in B: x \preceq z \land y \preceq z$ By definition of upper set: :$z \in A$ Thus by definition of intersection: :$x \in A \cap B$ Thus :$x \preceq z$ and $y \preceq z$ {{qed}}
Let $x_0 \in A$. Define $H : A \times \closedint 0 1 \to A$ by: :$\map H {x, t} = t x_0 + \paren {1 - t} x$ This yields a homotopy between the identity map $I_A$ and the constant map $x_0$. Thanks to the assumption of convexity for $A$, $H$ takes values in $A$. $H$ is a continuous function, since it is polynomial separately in $x, t$, and: :$\map H {-, 0} = I_A$ :$\map H {-, 1} \equiv x_0$ (the constant function on $x_0$)  This proves that $H: I_A \simeq c_{x_0}$. {{qed}} Category:Vector Spaces Category:Topology b92d0ryhoj78u9ja7ei7z55f2zfbcrd
By Right and Left Regular Representations in Topological Group are Homeomorphisms, the left cosets of $H$ are open. By Left Coset Space forms Partition, the complement of $H$ is a union of left cosets of $H$. Because the complement of $H$ is open, $H$ is closed. {{qed}} Category:Topology Category:Group Theory Category:Topological Groups ppam4e1on2bq4v6hnwnnxn4ast6zz5o
=== Necessary Condition === Suppose $G$ is Hausdorff. By T2 Space is T1 Space, $\left\{ {e}\right\}$ is closed. {{qed|lemma}} === Sufficient Condition === Suppose $\left\{ {e}\right\}$ is closed. Let $f: G \times G \to G$ be defined as: :$f \left({g h}\right) = g h^{-1}$ Because $f$ is continuous, $f^{-1} \left({e}\right)$ is closed. We have that $f^{-1} \left({e}\right)$ is the diagonal set on $G$. By Hausdorff Space iff Diagonal Set on Product is Closed, $G$ is Hausdorff. {{qed}}
=== Necessary Condition === Follows directly from $T_2$ Space is $T_1$ Space. {{qed|lemma}} === Sufficient Condition === Let $G$ be a $T_1$ space. Then $\left\{ {e}\right\}$ is closed. By Topological Group is Hausdorff iff Identity is Closed, $G$ is Hausdorff. {{qed}} Category:Topological Groups Category:T1 Spaces Category:Hausdorff Spaces cjvgmlo7cljoijr1ke0sljwnk0ylcly
Let $D$ be a directed subset of $S$ such that :$D \subseteq \complement_S\left({X}\right)$ Aiming for a contradiction suppose that :$\sup D \notin \complement_S\left({X}\right)$ By definition of relative complement: :$\sup D \in X$ By definition of property (S): :$\exists y \in D: \forall x \in D: y \preceq x \implies x \in X$ By definition of reflexivity: :$y \in X$ By definitions of intersection and non-empty: :$D \cap X \ne \varnothing$ Thus this by Empty Intersection iff Subset of Complement contradicts :$D \subseteq \complement_S\left({X}\right)$ {{qed}}
This follows from the Lifting The Exponent Lemma with $y$ replaced by $-y$. {{qed}}
Let $k = \map {\nu_2} n$. Then $n = 2^k m$ with $2 \nmid m$. By P-adic Valuation of Difference of Powers with Coprime Exponent: :$\map {\nu_2} {x^n - y^n} = \map {\nu_2} {x^{2^k} - y^{2^k} }$ Note that: :$x^{2^k} - y^{2^k} = \paren {x - y} \cdot \displaystyle \prod_{i \mathop = 0}^{k - 1} \paren {x^{2^i} + y^{2^i} }$ By Square Modulo 4: :$\map {\nu_2} {x^{2^i} + y^{2^i} } = 1$ for $i > 0$. Because $4 \divides \paren {x - y}$, $4 \nmid \paren {x + y}$. Thus: :$\map {\nu_2} {x + y} = 1$ Thus: {{begin-eqn}} {{eqn | l = \map {\nu_2} {x^{2^k} - y^{2^k} }       | r = \map {\nu_2} {x - y} + \sum_{i \mathop = 0}^{k - 1} \map {\nu_2} {x^{2^i} + y^{2^i} }       | c = p-adic Valuation is Additive }} {{eqn | r = \map {\nu_2} {x - y} + k }} {{end-eqn}} {{qed}}
The Platonic solids are the convex polyhedra all of whose faces are congruent and whose vertices are regular. From Five Platonic Solids, there are $5$ of these. The regular prisms are made from two regular polygons of an arbitrary number of sides forming the bases, separated by as many squares as there are sides of the two bases, forming the lateral faces. From Regular Prisms are Countably Infinite, they form a countably infinite set. Note that when the bases are themselves squares, the regular prism is then a cube, and so has already been counted amongst the Platonic solids. The regular antiprisms are made from two regular polygons of an arbitrary number of sides forming the bases, separated by twice as many equilateral triangles as there are sides of the two bases, forming the lateral faces. From Regular Antiprisms are Countably Infinite, they form a countably infinite set. Note that when the bases are themselves equilateral triangles, the regular antiprism is then a regular octahedron, and so has already been counted amongst the Platonic solids. The Archimedean polyhedra consist of all vertex-transitive convex polyhedra whose faces are all regular polygons, excluding: :the Platonic solids :the regular prisms :the regular antiprisms. From Thirteen Archimedean Polyhedra, there are $13$ of these. The Johnson polyhedra are all the other convex polyhedra whose faces are all regular polygons. From 92 Johnson Polyhedra, there are $92$ of these. {{qed}}
=== Existence === We prove this by induction on $k$. ==== Basis for the induction ==== Let $k = 2$. Let $d = \gcd \left({n_1, n_2}\right)$. By Bézout's Lemma, there exist $s, t \in \Z$ such that $d = s n_1 + t n_2$. Let $\left({q_1, r}\right)$ and $\left({q_2, r}\right)$ be the quotient and remainder of $b_1$ and $b_2$ upon division by $d$. Then $x = q_1 s n_2 + q_2 t n_2 + r$ is a solution. ==== Induction step ==== Let $n = \lcm \left({n_2, \ldots, n_k}\right)$. By the induction hypothesis, the last $k - 1$ congruences are equivalent to a single congruence: :$x \equiv b \pmod n$ for a certain $b \in \Z$ which satisfies $b \equiv b_i \pmod {n_i}$ for all $i \in \left\{ {2, \ldots, k}\right\}$. It remains to be shown that :$\lcm \left({n, n_1}\right) = \lcm \left({n_1, \ldots, n_k}\right)$ :$b \equiv b_1 \pmod {\gcd \left({n, n_1}\right)}$ to apply the case $k = 2$ and conclude. The first statement follows from Lowest Common Multiple is Associative. From GCD and LCM Distribute Over Each Other: :$\gcd \left({n, n_1}\right) = \lcm \left({\gcd \left({n_1, n_2}\right), \ldots, \gcd \left({n_1, n_k}\right)}\right)$ For all $i \in \left\{ {2, \ldots, k}\right\}$: : $b \equiv b_i \pmod {n_i}$ and: : $b_i \equiv b_1 \pmod {\gcd \left({n_1, n_i}\right)}$ Thus: : $b \equiv b_1 \pmod {\gcd \left({n_1, n_i}\right)}$ Therefore: :$b \equiv b_1 \pmod {\lcm \left({\gcd \left({n_1, n_2}\right), \ldots, \gcd \left({n_1, n_k}\right)}\right)}$ {{qed|lemma}} === Uniqueness=== {{ProofWanted}} Category:Chinese Remainder Theorem rblaaqhqiu6lqz5u8gvqms58tp3ug1e
=== $(1)$ implies $(2)$ === Suppose $\displaystyle \prod_{n \mathop = 1}^\infty z_n$ converges to $z$. By Convergence of Series of Complex Numbers by Real and Imaginary Part, it suffices to show that: :$\displaystyle \sum_{n \mathop = 1}^\infty \Re \log z_n = \Re \log z$ :$\displaystyle \sum_{n \mathop = 1}^\infty \Im \log z_n = \Im \log z + 2 k \pi$ ==== Real Part ==== Let $P_n$ denote the $n$th partial product. Then $P_n \to z$. By Modulus of Limit: :$\cmod {P_n} \to \cmod z$ By Natural Logarithm Function is Continuous; :$\log \cmod {P_n} \to \log \cmod z$ By definition of complex natural logarithm and Real Part of Sum of Complex Numbers: :$\displaystyle \log \cmod {P_n} = \sum_{j \mathop = 1}^n \Re \log z_j$ ==== Imaginary Part ==== It remains to show that: :$\displaystyle \sum_{j \mathop = 1}^n \Im \log z_j \to \arg z + 2 k \pi$ for some $k \in \Z$. Let $\theta = \arg z$ be the argument of $z$. Let $\theta_n = \arg P_n$ be the argument of $P_n$ in the half-open interval $\hointl {\theta - \pi} {\theta + \pi}$. By the corollary to Convergence of Complex Sequence in Polar Form: :$\theta_n \to \theta$ Let $k_n \in \Z$ be such that: :$\displaystyle \sum_{j \mathop = 1}^n \Im \log z_j = \theta_n + 2 k_n \pi$ We show that $k_n$ is eventually constant. By the Triangle Inequality: :$2 \pi \cmod {k_{n + 1} - k_n} \le \cmod {\theta_{n + 1} - \theta_n} + \cmod {\Im \log z_{n + 1} }$ By Factors in Convergent Product Converge to One, $z_n \to 1$. By Complex Logarithm is Continuous Outside Branch, $\log z_n \to 0$. Thus $2 \pi \cmod {k_{n + 1} - k_n} \to 0$. So $k_n$ is eventually constant, say equal to $k$. Then: :$\displaystyle \sum_{j \mathop = 1}^n \Im \log z_j \to \theta + 2 k \pi$ {{qed|lemma}} === $(2)$ implies $(1)$ === Suppose $\displaystyle \sum_{n \mathop = 1}^\infty \log z_n = \log z + 2 k \pi i$. By Exponential of Series Equals Infinite Product: :$\displaystyle \prod_{n \mathop = 1}^\infty z_n = z$ Hence the result. {{qed}}