hcju/proofwikiqa huggingface dataset:
Let $y \in \map {B_r} x$. Let $a \in \map {B_r} y$. By the definition of an open ball, then: :$\norm {a - y} < r$ :$\norm {y - x} < r$ Hence: {{begin-eqn}} {{eqn | l = \norm {a - x}       | r = \norm {a - y + y - x} }} {{eqn | o = \le       | r = \max \set {\norm {a - y}, \norm{y - x} }        | c = {{Defof|Non-Archimedean Division Ring Norm}} }} {{eqn | o = <       | r = r       | c =  }} {{end-eqn}} By the definition of an open ball:  :$a \in \map {B_r} x$ Hence: :$\map {B_r} y \subseteq \map {B_r} x$ By Norm of Negative: :$\norm {x - y} < r$ By the definition of an open ball: :$x \in \map {B_r} y$ Similarly it follows that: :$\map {B_r} x \subseteq \map {B_r} y$ By set equality: :$\map {B_r} y = \map {B_r} x$ {{qed}}
Let $y \in \map { {B_r}^-} x$. Let $a \in \map { {B_r}^-} y$. By the definition of an closed ball, then: :$\norm {a - y} \le r$ :$\norm {y - x} \le r$ Hence: {{begin-eqn}} {{eqn | l = \norm {a - x}       | r = \norm {a - y + y - x} }} {{eqn | r = \max \set {\norm {a - y}, \norm {y - x} }        | o = \le       | c = {{Defof|Non-Archimedean Division Ring Norm}} }} {{eqn | r = r       | o = \le       | c =  }} {{end-eqn}} By the definition of a closed ball, then:  :$a \in \map { {B_r}^-} x$. Hence: :$\map { {B_r}^-} y \subseteq \map { {B_r}^-} x$ By Norm of Negative then: :$\norm {x - y} \le r$ By the definition of a closed ball, then: :$x \in \map { {B_r}^-} y$ Similarly it follows that: :$\map { {B_r}^-} x \subseteq \map { {B_r}^-} y$ By set equality: :$\map { {B_r}^-} x = \map { {B_r}^-} y$ {{qed}}
Let $d$ be the metric induced by the norm $\norm {\,\cdot\,}$. By the definition of an open ball in $\norm {\,\cdot\,}$: :$\map {B_r} x$ is an open ball in the metric space $\struct {R, d}$. By Open Ball of Metric Space is Open Set then $\map {B_r} x$ is open in $\struct {R, d}$. So it remains to show that $\map {B_r} x$ is closed in $\struct {R, d}$. Let $\map \cl {\map {B_r} x}$ denote the closure of $\map {B_r} x$. Let $y \in \map \cl {\map {B_r} x}$. By the definition of the closure of $\map {B_r} x$ then: :$\forall s > 0: \map {B_s} y \cap \map {B_r} x \ne \O$ In particular: :$\map {B_r} y \cap \map {B_r} x \ne \O$ Let $z \in \map {B_r} y \cap \map {B_r} x$. By Centers of Open Balls: :$\map {B_r} y = B_r \paren{z} = \map {B_r} x$ By the definition of an open ball: :$y \in \map {B_r} y = \map {B_r} x$. Hence: :$\map \cl {\map {B_r} x} \subseteq \map {B_r} x$ By Subset of Metric Space is Subset of its Closure then: :$\map {B_r} x \subseteq \map \cl {\map {B_r} x}$ So by definition of set equality: :$\map \cl {\map {B_r} x} = \map {B_r} x$ By Set is Closed iff Equals Topological Closure then $\map {B_r} x$ is closed. {{qed}}
Let $d$ be the metric induced by the norm $\norm {\,\cdot\,}$. By the definition of a closed ball in $\norm {\,\cdot\,}$ then: :$\map { {B_r}^-} x$ is a closed ball in the metric space $\struct {R, d}$. By Closed Ball is Closed in Metric Space then $\map { {B_r}^-} c$ is closed in $d$. So it remains to show that $\map { {B_r}^-} x$ is open in $d$. Let $y \in \map { {B_r}^-} x$. By Centers of Closed Balls then: :$\map { {B_r}^-} y = \map { {B_r}^-} x$ By the definition of an open ball then: :$y \in \map {B_r} y \subseteq \map { {B_r}^-} y = \map { {B_r}^-} x$ By the definition of an open set in a metric space, $\map { {B_r}^-} x$ is open. {{qed}}
=== Necessary Condition === Let $z \in \map {B_r} x \cap \map {B_s} y$. If $r \le s$ then: {{begin-eqn}} {{eqn| l = \map {B_r} x      | r = \map {B_r} z      | c = Every element in an open ball is the center }} {{eqn| o = \subseteq       | r = \map {B_s} z      | c = as $r \le s$ }} {{eqn| r = \map {B_s} y      | c = Every element in an open ball is the center }} {{end-eqn}} Similarly, if $s \le r$ then: {{begin-eqn}} {{eqn| l = \map {B_s} y       | o = \subseteq       | r = \map {B_r} x }} {{end-eqn}} {{qed|lemma}} === Sufficient Condition === Let  :$\map {B_r} x \subseteq \map {B_s} y$  or  :$\map {B_s} y \subseteq \map {B_r} x$ By the definition of an open ball then: :$x \in \map {B_r} x \ne \O$ :$y \in \map {B_s} y \ne \O$ The result follows. {{qed}}
=== Necessary Condition === Let $z \in \map { {B_r}^-} x \cap \map { {B_s}^-} y$. If $r \le s$ then: {{begin-eqn}} {{eqn| l = \map { {B_r}^-} x       | r = \map { {B_r}^-} z      | c = Every element in an open ball is the center }} {{eqn| o = \subseteq      | r = \map { {B_s}^-} z      | c = as $r \le s$ }} {{eqn| r = \map { {B_s}^-} y      | c = Every element in an open ball is the center }} {{end-eqn}} Similarly, if $s \le r$ then: {{begin-eqn}} {{eqn| l = \map { {B_s}^-} y       | o = \subseteq      | r = \map { {B_r}^-} x }} {{end-eqn}} {{qed|lemma}} === Sufficient Condition === Let: :$\map { {B_r}^-} x \subseteq \map { {B_s}^-} y$  or: :$\map { {B_s}^-} y \subseteq \map { {B_r}^-} x$ By the definition of an open ball then: :$x \in \map { {B_r}^-} x \ne \O$ :$y \in \map { {B_s}^-} y \ne \O$ The result follows. {{qed}}
{{begin-eqn}} {{eqn	| l = \pi \, \map \coth {\pi z} 	| r = \pi i \, \map \cot {\pi i z} 	| c = Hyperbolic Cotangent in terms of Cotangent }} {{eqn	| r = i \paren {\frac 1 {i z} + 2 i \sum_{n \mathop = 1}^\infty \frac z {\paren {i z}^2 - n^2} } 	| c = Mittag-Leffler Expansion for Cotangent Function }} {{eqn	| r = \frac 1 z - 2 \sum_{n \mathop = 1}^\infty \frac z {-z^2 - n^2} 	| c = $i^2 = -1$ }} {{eqn	| r = \frac 1 z + 2 \sum_{n \mathop = 1}^\infty \frac z {z^2 + n^2} }} {{end-eqn}} {{qed}} {{Namedfor|Magnus Gustaf Mittag-Leffler|cat = Mittag-Leffler}}
Supposed there exists an identity $e$ in one of the standard number systems $\F$. {{begin-eqn}} {{eqn | lo= \forall x \in \F:       | l = x       | r = x - e       | c =  }} {{eqn | r = e - x       | c =  }} {{eqn | ll= \leadsto       | l = x + \paren {-e}       | r = e + \paren {-x}       | c =  }} {{eqn | ll= \leadsto       | l = x + x       | r = e + e       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = e       | c =  }} {{end-eqn}} That is: :$\forall x \in \F: x = e$ from which it follows that $\F$ has no such $e$. {{qed}}
This is demonstrated by showing that: :$n \times 0 = 0 = 0 \times n$ for all $n$ in all standard number systems. {{qed}}
Suppose $z$ is a zero element for addition in a standard number system $\F$. Then: {{begin-eqn}} {{eqn | ll= \forall n \in \F       | l = n + z       | r = z       | c =  }} {{eqn | ll= \leadsto       | l = n       | r = 0       | c = subtracting $z$ from both sides }} {{end-eqn}} As $n$ is arbitrary, and therefore not always $0$, it follows there can be no such $z$. {{qed}}
We have that Subtraction on Numbers is Not Associative. Hence $\struct {\R, -}$ is not a semigroup by definition. {{qed}}
From Group has Latin Square Property, we have that: {{begin-eqn}} {{eqn | l = a + g       | r = b       | c =  }} {{eqn | ll= \leadsto       | l = g       | r = \paren {-a} + b       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = h + a       | r = b       | c =  }} {{eqn | ll= \leadsto       | l = h       | r = b + \paren {-a}       | c =  }} {{end-eqn}} {{qed}}
Let $\zeta \in \C \setminus X$.  Then:  :$\displaystyle \frac {\map f z} {z - \zeta}$  has simple poles for $z \in X \cup \set \zeta$.  Let $X_N$ be the set of poles contained within $C_N$.  Then:  {{begin-eqn}} {{eqn	| l = \frac 1 {2 \pi i} \oint_{\partial C_N} \frac {\map f z} {z - \zeta} \rd z 	| r = \Res {\frac {\map f z} {z - \zeta} } \zeta + \sum_{n \mathop \in X_N} \Res {\frac {\map f z} {z - \zeta} } n 	| c = Residue Theorem }} {{eqn	| r = \lim_{z \mathop \to \zeta} \paren {\frac {\paren {z - \zeta} \map f z} {z - \zeta} } + \sum_{n \mathop \in X_N} \paren {\lim_{z \mathop \to n} \frac {\paren {z - n} \map f z} {z - \zeta} } 	| c = Residue at Simple Pole }} {{eqn	| r = \lim_{z \mathop \to \zeta} \map f z + \sum_{n \mathop \in X_N} \paren {\lim_{z \mathop \to n} \paren {\paren {z - n} \map f z } \cdot \lim_{z \mathop \to n} \frac 1 {z - \zeta} } 	| c = Product Rule for Limits  }} {{eqn	| r = \map f \zeta + \sum_{n \mathop \in X_N} \frac {\Res f n} {n - \zeta} 	| c = $f$ is continuous at $\zeta$, Residue at Simple Pole }} {{end-eqn}} Setting $\zeta = 0$, we obtain:  :$\displaystyle \frac 1 {2 \pi i} \oint_{\partial C_N} \frac {\map f z} z \rd z = \map f 0 + \sum_{n \mathop \in X_N} \frac {\Res f n} n$  So:  :$\displaystyle \frac 1 {2 \pi i} \oint_{\partial C_N} \map f z \paren {\frac 1 {z - \zeta} - \frac 1 z} \rd z = \map f \zeta + \sum_{n \in X_N} {\Res f n} \paren {\frac 1 {n - \zeta} - \frac 1 n} - \map f 0$ It remains to show that the integral on the left hand side vanishes as $N \to \infty$. We have: {{begin-eqn}} {{eqn	| l = \cmod {\frac 1 {2 \pi i} \oint_{\partial C_N} \map f z \paren {\frac 1 {z - \zeta} - \frac 1 z} \rd z} 	| r = \cmod {\frac \zeta {2 \pi i} \oint_{\partial C_N} \frac {\map f z} {z \paren {z - \zeta} } \rd z} }} {{eqn	| o = \le 	| r = \frac {\cmod \zeta} {2 \pi} \cdot \frac M {R_N \paren {R_N - \cmod \zeta} } \cdot 2 \pi R_N 	| c = Triangle Inequality for Contour Integrals, Reverse Triangle Inequality, noting that $\cmod z$ for $z \in \partial C_N$   }} {{eqn	| o = \sim 	| r = \frac 1 {R_N} }} {{eqn	| o = \to 	| r = 0 	| c = as $N \to \infty$, $R_N \to \infty$ }} {{end-eqn}} Letting $N \to \infty$ gives:  :$\displaystyle 0 = \map f \zeta + \sum_{n \in X} {\Res f n} \paren {\frac 1 {n - \zeta} - \frac 1 n} - \map f 0$ Giving: :$\displaystyle \map f \zeta = \map f 0 + \sum_{n \mathop \in X} \Res f n \paren {\frac 1 {\zeta - n} + \frac 1 n}$  {{qed}} {{Namedfor|Magnus Gustaf Mittag-Leffler|cat = Mittag-Leffler}}
By $p$-Product Metric Induces Product Topology and Continuous Mapping is Continuous on Induced Topological Spaces, it suffices to consider the case $p = \infty$. Let $\tuple {x_0, y_0} \in R \times R$. Let $\epsilon > 0$ be given. Let $\tuple {x, y} \in R \times R$ such that: :$\map {d_\infty} {\tuple {x, y},\tuple{x_0, y_0} } < \dfrac \epsilon 2$ By the definition of the product metric $d_\infty$ then: :$\max \set {\map d {x, x_0}, \map d {y, y_0} } < \dfrac \epsilon 2$ or equivalently: :$\map d {x, x_0} < \dfrac \epsilon 2$ :$\map d {y, y_0} < \dfrac \epsilon 2$ Then: {{begin-eqn}} {{eqn | l = \map d {x_0 + y_0, x + y}       | r = \norm {\paren {x_0 + y_0} - \paren {x+y} }       | c = {{Defof|Metric Induced by Norm on Division Ring}} }} {{eqn | r = \norm {\paren {x_0 - x} + \paren {y_0 - y} }       | c =  }} {{eqn | r = \norm {x_0 - x } + \norm {y_0 - y}       | o = \le       | c = Norm Axiom (N3) (Triangle Inequality) }} {{eqn | r = \map d {x_0, x} + \map d {y_0, y}       | o = \le       | c = {{Defof|Metric Induced by Norm on Division Ring}} }} {{eqn | r = \dfrac \epsilon 2 + \dfrac \epsilon 2       | o = <       | c =  }} {{eqn | r = \epsilon }} {{end-eqn}} Since $\tuple {x_0, y_0}$ and $\epsilon$ were arbitrary, by the definition of continuity then the mapping: :$+ : \struct {R \times R, d_\infty} \to \struct {R, d}$  is continuous. {{qed}}
Let $x_0 \in R$. Let $\epsilon > 0$ be given. Let $x \in R$ such that: :$\map d {x, x_0} < \epsilon$ Then: {{begin-eqn}} {{eqn | l = \map d {-x, -x_0}       | r = \norm {-x - \paren {-x_0} }       | c = {{Defof|Metric Induced by Norm on Division Ring}} }} {{eqn | r = \norm {-x + x_0} }} {{eqn | r = \norm {x_0 - x}       | c = $+$ is Commutative. }} {{eqn | r = \map d {x_0, x}       | c = {{Defof|Metric Induced by Norm on Division Ring}} }} {{eqn | r = \map d {x, x_0}       | c = Metric Space Axiom $(\text M 2)$ }} {{eqn | r = \epsilon       | o = < }} {{end-eqn}} Since $x_0$ and $\epsilon$ were arbitrary, by the definition of continuity then the mapping: :$\eta: \struct {R, d} \to \struct {R, d} : \map \eta x = -x$  is continuous. {{qed}}
By $p$-Product Metric Induces Product Topology and Continuous Mapping is Continuous on Induced Topological Spaces, it suffices to consider the case $p = \infty$. Let $\tuple {x_0, y_0} \in R \times R$. Let $\epsilon > 0$ be given. Let $\delta = \min \set {\dfrac \epsilon {1 + \norm {y_0} + \norm {x_0} }, 1}$ Since $1 + \norm {y_0} + \norm {x_0} > 0$ then $\delta > 0$ Let $\tuple {x,y} \in R \times R$ such that: :$\map {d_\infty} {\tuple {x, y}, \tuple{x_0, y_0} } < \delta$ By the definition of the $p$-product metric $d_\infty$: :$\max \set {\map d {x, x_0}, \map d {y, y_0}} < \delta$ or equivalently: :$\map d {x, x_0} < \delta$ :$\map d {y, y_0} < \delta$ Then: {{begin-eqn}} {{eqn | l = \norm y       | r = \norm {y - y_0 + y_0}       | c = }} {{eqn | o = \le       | r = \norm {y - y_0}  + \norm {y_0}       | c = {{NormAxiom|3}} }} {{eqn | o = \le       | r = \map d {y, y_0}  + \norm {y_0}       | c = {{Defof|Metric Induced by Norm on Division Ring}} }} {{eqn | o = <       | r = \delta  + \norm {y_0}       | c =  }} {{eqn | o = \le       | r = 1  + \norm {y_0}       | c =  }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \map d {x y, x_0 y_0}       | r = \norm {x y - x_0 y_0}       | c = {{Defof|Metric Induced by Norm on Division Ring}} }} {{eqn | r = \norm {x y - x_0 y + x_0 y - x_0 y_0}       | c = }} {{eqn | o = \le       | r = \norm {x y - x_0 y} + \norm {x_0 y - x_0 y_0}       | c = {{NormAxiom|3}} }} {{eqn | o = \le       | r = \norm {x - x_0} \norm y + \norm {x_0} \norm {y - y_0}       | c = {{NormAxiom|2}} }} {{eqn | o = \le       | r = \map d {x, x_0} \norm y + \norm {x_0} \map d {y, y_0}       | c = {{Defof|Metric Induced by Norm on Division Ring}} }} {{eqn | o = <       | r = \delta \norm y + \norm {x_0} \delta       | c =  }} {{eqn | o = \le       | r = \delta \paren {\norm y + \norm {x_0} }       | c = }} {{eqn | o = <       | r = \delta \paren {1 + \norm {y_0} + \norm {x_0} }       | c =  }} {{eqn | o = \le       | r = \dfrac \epsilon {1 + \norm {y_0} + \norm {x_0} } \paren {1 + \norm {y_0} + \norm {x_0} }       | c =  }} {{eqn | r = \epsilon }} {{end-eqn}} We have that $\tuple {x_0, y_0}$ and $\epsilon$ are arbitrary. Hence, by the definition of continuity, the mapping: :$* : \struct {R \times R, d_\infty} \to \struct {R, d}$  is continuous. {{qed}}
Let $x_0 \in R^*$. Let $\epsilon \gt 0$ be given. Let $\delta = \min \set {\dfrac {\norm {x_0} } 2, \dfrac {\norm {x_0}^2 \epsilon} 2 }$ Let $x \in R^*$ such that: :$\map {d^*} {x, x_0} < \delta$ By the definition of the subspace metric on $R^*$ and the definition of the metric induced by the norm on $R$: :$\map {d^*} {x, x_0} = \map d {x, x_0} = \norm {x - x_0} < \delta$ Then: {{begin-eqn}} {{eqn | l = \norm {x_0}       | o = \le       | r = \norm {x - x_0} + \norm x       | c = {{NormAxiom|3}} }} {{eqn | o = }} {{eqn | o = <       | r = \delta + \norm x       | c = by assumption: $\norm {x - x_0} < \delta$ }} {{eqn | o = }} {{eqn | o = \le       | r = \dfrac {\norm {x_0} } 2 + \norm x       | c = by assumption: $\delta \le \dfrac {\norm {x_0} } 2$ }} {{eqn | o = }} {{eqn | ll= \leadsto       | l = \dfrac {\norm {x_0} } 2       | o = <       | r = \norm x       | c = subtracting $\dfrac {\norm {x_0} } 2$ from both sides }} {{eqn | o = }} {{eqn | ll= \leadsto       | l = \dfrac 2 {\norm {x_0} }       | o = >       | r = \dfrac 1 {\norm x}       | c = inverting both sides of the equation }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \map d {x^{-1}, x_0^{-1} }       | r = \norm {x^{-1} - x_0^{-1} }       | c = {{Defof|Metric Induced by Norm on Division Ring}} }} {{eqn | o = }} {{eqn | r = \dfrac 1 {\norm x} \paren {\norm x \norm {x^{-1} - x_0^{-1} } \norm {x_0} } \dfrac 1 {\norm {x_0} }        | c =  }} {{eqn | o = }} {{eqn | r = \dfrac 1 {\norm x \norm {x_0} } \paren {\norm {x \paren {x^{-1} - x_0^{-1} } x_0} }       | c = {{NormAxiom|2}} }} {{eqn | o = }} {{eqn | r = \dfrac 1 {\norm x \norm {x_0} } \paren {\norm {x x^{-1} x_0 - x x_0^{-1} x_0} }       | c = Ring Axiom $(\text D)$: Product is Distributive over Addition }} {{eqn | o = }} {{eqn | r = \dfrac 1 {\norm x \norm {x_0} } \paren {\norm {x_0 - x} }       | c = {{Defof|Division Ring}} }} {{eqn | o = }} {{eqn | o = <       | r = \dfrac 2 {\norm {x_0}^2} \paren {\norm {x_0 - x} }       | c = from $\dfrac 1 {\norm x} < \dfrac 2 {\norm {x_0} }$ above }} {{eqn | o = }} {{eqn | o = <       | r = \dfrac {2 \delta} {\norm {x_0}^2}       | c = by assumption: $\norm {x - x_0} < \delta$ }} {{eqn | o = }} {{eqn | o = \le       | r = \dfrac 2 {\norm {x_0}^2} \paren {\dfrac {\norm {x_0}^2 \epsilon} 2}       | c = by assumption: $\delta \le \dfrac {\norm {x_0}^2 \epsilon} 2$ }} {{eqn | o = }} {{eqn | r = \epsilon       | c = cancelling terms }} {{end-eqn}} Since $x_0$ and $\epsilon$ were arbitrary, by the definition of continuity then the mapping: :$\iota: \struct {R^*, d^*} \to \struct {R, d} : \map \iota x = x^{-1}$  is continuous. {{qed}}
Let $d_\infty$ be the Chebyshev distance metric on $R \times R$. Let $\tau^\times$ be the product topology on $R \times R$. By $p$-Product Metric Induces Product Topology, $\tau^\times$ is the topology induced by the metric $d_\infty$. Let $R^* = R \setminus \set 0$. Let $d^*$ be the restriction of $d$ to $R^*$. Let $\tau^*$ be the subspace topology on $R^*$. By Metric Subspace Induces Subspace Topology then $\tau^*$ is the topology induced by the metric $d^*$ By Normed Division Ring Operations are Continuous and Continuous Mapping is Continuous on Induced Topological Spaces, the mappings: ::$\phi : \struct {R \times R, \tau^\times} \to \struct {R, \tau} : \map \phi {x, y} = x + y$  ::$\theta : \struct {R ,\tau} \to \struct {R, \tau} : \map \theta x = -x$  ::$\psi : \struct {R \times R, \tau^\times} \to \struct {R, \tau} : \map \psi {x, y} = x y$  ::$\xi : \struct {R^* ,\tau^*} \to \struct {R, \tau} : \map \xi x = x^{-1}$  are continuous. By the definition of a topological division ring then the result follows. {{qed}}
Let $x \in S$. Then {{begin-eqn}} {{eqn | l = x       | r = x \circ z       | c = {{Defof|Identity Element}} }} {{eqn | r = z       | c = {{Defof|Zero Element}} }} {{end-eqn}} and so there is no other element of $S$ but $z$. {{qed}}
We verify the group axioms, in the following order (for convenience): === G0: Closure === Let $\theta_{a, b}$ and $\theta_{c, d}$ be elements of $G$. From Composition of Linear Real Functions: :$\theta_{c, d} \circ \theta_{a, b} = \theta_{a c, b c + d}$ As $a \ne 0$ and $c \ne 0$ it follows that $a c \ne 0$. Thus $\theta_{a c, b c + d} \in G$ This proves closure of $\circ$. {{qed|lemma}} === G1: Associativity === We have that Composition of Mappings is Associative. {{qed|lemma}} === G2: Identity === We assert that $\theta_{1, 0}$ is the identity with respect to $\circ$. {{begin-eqn}} {{eqn | l = \theta_{a, b} \circ \theta_{1, 0}       | r = \theta_{1 \times a, \, 0 \times a + b}       | c = Composition of Linear Real Functions }} {{eqn | r = \theta_{a, b}       | c =  }} {{eqn | r = \theta_{a \times 1, \, b \times 1 + 0}       | c =  }} {{eqn | r = \theta_{1, 0} \circ \theta_{a, b}       | c = Composition of Linear Real Functions }} {{end-eqn}} So indeed $\theta_{1, 0}$ is the identity for $\circ$. {{qed|lemma}} === G3: Inverses === For any element $\theta_{a, b} \in S$, we claim that $\theta_{1 / a, \, -b / a}$ is the inverse for $\theta_{a, b}$. From Inverse of Linear Function on Real Numbers: :$\theta^{-1}_{a, b} = \theta_{1 / a, \, -b / a}$ Verifying this: {{begin-eqn}} {{eqn | l = \theta_{1 / a, \, -b / a} \circ \theta_{a, b}       | r = \theta_{a \times 1 / a, \, b \times 1 / a + \paren {-b / a} }       | c = Composition of Linear Real Functions }} {{eqn | r = \theta_{1, 0}       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = \theta_{a, b} \circ \theta_{1 / a, \, -b / a}       | r = \theta_{\paren {1 / a} \times a, \, \paren {-b / a} \times a + b}       | c = Composition of Linear Real Functions }} {{eqn | r = \theta_{1, 0}       | c =  }} {{end-eqn}} We conclude that $\theta_{1 / a, \, -b / a}$ is indeed the inverse for $\theta_{a, b}$. {{qed|lemma}} It follows that $\struct {G, \circ}$ is indeed a group. {{qed|lemma}} === Non-Abelian === Let $\theta_{a, b}$ and $\theta_{c, d}$ be elements of $G$. From Condition for Composition of Linear Real Functions to be Commutative, it is not true in general that $\theta_{c, d} \circ \theta_{a, b} = \theta_{a, b} \circ \theta_{c, d}$. Thus $\struct {G, \circ}$ is non-abelian by definition. {{qed}}
Clearly $0 \in n \Z$ so $n \Z \ne \O$. Now suppose $x, y \in n \Z$. Then $\exists r, s \in \Z: x = n r, y = n s$. Also, $-y = - n s$. So $x - y = n \paren {r - s}$. As $r - s \in \Z$ it follows that $x - y \in n \Z$. So by the One-Step Subgroup Test it follows that $\struct {n \Z, +}$ is a subgroup of the additive group of integers $\struct {\Z, +}$. {{qed}}
By definition of generated subgroup, $\gen H$ is the smallest subgroup of $H$ containing $H$. Hence the result. {{qed}}
By definition of generated subgroup, $\gen G$ is the smallest subgroup of $G$ containing $G$. Hence the result by Group is Subgroup of Itself. {{qed}}
=== $(1)$ implies $(2)$ === Let $G = \gen g$ be an infinite cyclic group by definition 1. Then by definition: :$(1): \quad \forall n \in \Z_{> 0}: n > 0 \implies \nexists a \in G, a \ne e: a^n = e$ {{AimForCont}}: :$\exists a \in G: \exists m, n \in \Z, m \ne n: a^m = a^n$ Then by Equal Powers of Finite Order Element: :$\exists k \in \Z_{> 0}: k \divides \paren {m - n}$ such that: :$a^k = e$ As $m \ne n$, we have that $k \ge 1$. This contradicts the condition $(1)$ for $G$ to be an infinite cyclic group. From this contradiction we deduce that: :$\forall a \in G: \forall m, n \in \Z: m \ne n \implies a^m \ne a^n$ Thus $G$ is an infinite cyclic group by definition 2. {{qed|lemma}} === $(2)$ implies $(1)$ === Let $G = \gen g$ be an infinite cyclic group by definition 2. Then by definition: :$\forall a \in G: \forall m, n \in \Z: m \ne n \implies a^m \ne a^n$ {{AimForCont}}: :$\exists a \in G, a \ne e: \exists n \in \Z: a^n = e$ Then: :$a^{2 n} = a^n \circ a^n = e^2 = e = a^n$ That is: :$\exists n, 2 n \in \Z: n \ne 2 n: a^n = a^{2 n}$ and so it is not the case that: :$\forall a \in G: \forall m, n \in \Z: m \ne n \implies a^m \ne a^n$ From this contradiction we deduce that: :$\forall n \in \Z_{> 0}: n > 0 \implies \nexists a \in G, a \ne e: a^n = e$ Thus $G$ is an infinite cyclic group by definition 1. {{qed}}
Let $A, B \in \mathscr S$. We have that: {{begin-eqn}} {{eqn | l = T       | o = \subseteq       | r = A       | c = Definition of $\mathscr S$ }} {{eqn | l = T       | o = \subseteq       | r = B       | c = Definition of $\mathscr S$ }} {{eqn | n = 1       | ll= \leadsto       | l = T       | o = \subseteq       | r = A \cup B       | c = Set is Subset of Union }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = A       | o = \subseteq       | r = S       | c = {{Defof|Power Set}} }} {{eqn | l = B       | o = \subseteq       | r = S       | c = {{Defof|Power Set}} }} {{eqn | ll= \leadsto       | l = A \cup B       | o = \subseteq       | r = S       | c = Union is Smallest Superset }} {{eqn | n = 2       | ll= \leadsto       | l = A \cup B       | o = \in       | r = \powerset S       | c = {{Defof|Power Set}} }} {{end-eqn}} Thus we have: {{begin-eqn}} {{eqn | l = T       | o = \subseteq       | r = A \cup B       | c = from $(1)$ }} {{eqn | l = A \cup B       | o = \in       | r = \powerset S       | c = from $(2)$ }} {{eqn | ll= \leadsto       | l = A \cup B       | o = \in       | r = \mathscr S       | c = Definition of $\mathscr S$ }} {{end-eqn}} Hence the result by definition of closed algebraic structure. {{qed}}