hcju/proofwikiqa huggingface dataset:
Let $\order G = m p^n$. Let $\mathbb S = \set {S \subseteq G: \order S = p^n}$, that is, the set of all of subsets of $G$ which have exactly $p^n$ elements. Let $N = \order {\mathbb S}$. Now $N$ is the number of ways $p^n$ elements can be chosen from a set containing $p^n k$ elements.   From Cardinality of Set of Subsets, this is given by: :$N = \dbinom {p^n m} {p^n} = \dfrac {\paren {p^n m} \paren {p^n m - 1} \cdots \paren {p^n m - i} \cdots \paren {p^n m - p^n + 1} } {\paren {p^n} \paren {p^n - 1} \cdots \paren {p^n - i} \cdots \paren 1}$ For $0 < i < p^n$, the highest power of $p$ which divides $m p^n - 1$ is the same as that dividing $p^n - 1$. {{Explain|why the above holds}} Thus the power of $p$ in the numerator and denominator of $N$ cancel out, except those of $m$. Thus $m$ and $N$ have the same number of divisors which are power of $p$. Let $p^r$ be the largest power of $p$ which divides $m$ and $N$. Now let $G$ act on $\mathbb S$ by the rule: :$\forall S \in \mathbb S: g * S = g S = \set {x \in G: x = g s: s \in S}$ That is, $g * S$ is the left coset of $S$ by $g$. From Group Action on Sets with k Elements, this is a group action. If every orbit under $*$ were divisible by $p^{r - 1}$, then $p^{r + 1} \divides N$. {{explain|why the above holds}} But we have stated that $p^r$ is the largest power of $p$ which divides $N$. Therefore there is at least one orbit under $*$ which can be represented as: :$\set {S_1, S_2, \ldots, S_k}$ for which $p^{r + 1} \nmid k$. Let $H = \Stab {S_1}$ denote the stabilizer of $S_1$. From the Orbit-Stabilizer Theorem, we have that: :$k = \index G H = \dfrac {\order g} {\order H}$ As $p^r \divides m$, we have: :$p^{n + r} \divides p^n m$ But: :$p^n m = \order G = k \order H$ and so: :$p^{n + r} \divides k \order H$ But we also have that: :$p^{r + 1} \nmid k$ which means: :$p^n \divides \order H$ and that: :$p^n \le \order H$ Now because $H = \Stab {S_1}$, we haveL :$H g \subseteq S_1$ for all $g \in \S_1$ Thus: :$\order H = \card {H g} \le \card {S_1} = p^n$ So we have $p^n \le \order H$ and $\order H \le p^n$ which means: :$\order H = p^n$ and so $H$ is the subgroup of $G$ whose existence we were to demonstrate. {{qed}}
From Set Difference with Superset is Empty Set: :$T \subseteq S \iff T \setminus S = \O$ From Set Difference and Intersection form Partition: :$S = \paren {S \setminus T} \cup T$ Thus from Cardinality of Set Union: :$\card S = \card T + \card {S \setminus T} - \card {T \cap \paren {S \setminus T} }$ But from Set Difference Intersection with Second Set is Empty Set: :$T \cap \paren {S \setminus T} = 0$ Hence the result. {{qed}}
By Hankel Representation of Riemann Zeta Function:  :$\displaystyle \map \zeta {-n} = \frac {i \map \Gamma {1 + n} } {2 \pi} \oint_C \frac 1 {z^{n + 1} \paren {e^z - 1} } \rd z$  where $C$ is the Hankel contour. Note that the integrand is meromorphic, with a pole at $z = 0$ lying inside the contour. So:  {{begin-eqn}} {{eqn	| l = \map \zeta {-n} 	| r = \frac {i n!} {2 \pi} \oint_C \frac 1 {\paren {-z}^{n + 2} } \cdot \frac {-z} {e^z - 1} \rd z 	| c = Gamma Function Extends Factorial }} {{eqn	| r = -\frac {i n!} {2 \pi} \cdot 2 \pi i \Res {\frac 1 {\paren {-z}^{n + 2} } \cdot \frac {-z} {e^z - 1} } 0 	| c = Residue Theorem  }} {{eqn	| r = n! \Res {\paren {-1}^2 \paren {-1}^n \sum _{k \mathop = 0}^\infty \frac {B_k} {k!} z^{k - n - 2} } 0 	| c = {{Defof|Bernoulli Numbers/Generating Function|Bernoulli Numbers}}, $i^2 = -1$ }} {{eqn	| r = n! \paren {-1}^n \Res {\sum _{k \mathop = 0}^\infty \frac {B_k} {k!} z^{k - n - 2} } 0 }} {{end-eqn}} By the definition of a residue, the residue at $0$ is given by the coefficient of the $\dfrac 1 z$ term. This is the term where $k = n + 1$, so: {{begin-eqn}} {{eqn	| l = \map \zeta {-n} 	| r = n! \paren {-1}^n \frac {B_{n + 1} } {\paren {n + 1}!}  }} {{eqn	| r = \paren {-1}^n \frac {B_{n + 1} } {n + 1} }} {{end-eqn}} {{qed}} Category:Riemann Zeta Function svniiy7i2jmd86hhra1u1ymikmr4he0
The decimal expansion of $\sqrt 5$ is: :$\sqrt 5 \approx 2 \cdotp 23606 \, 79774 \, 99789 \, 6964 \ldots$ Thus: :$2 \le \sqrt 5 < 3$ Hence $2$ is the floor of $\sqrt 5$ by definition. {{qed}}
We have that $3$ is an integer. Thus this is a specific example of Real Number is Integer iff equals Floor: $\floor x = x \iff x \in \Z$ {{qed}}
We first prove that $100 \times 10^n = 4 \pmod {12}$, where $n \in \N$. Proof by induction: For all $n \in \N$, let $P \paren n$ be the proposition: :$100 \times 10^n = 4 \pmod {12}$ === Basis for the Induction === $P(0)$ says $100 = 4 \pmod {12}$, which is true as $100 = 8 \times 12 + 4$. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \paren k$ is true, where $k \ge 0$, then it logically follows that $P \paren{k+1}$ is true. So this is our induction hypothesis: :$100 \times 10^k = 4 \pmod {12}$ Then we need to show: :$100 \times 10^{k+1} = 4 \pmod {12}$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = 100 \times 10^{k+1}       | r = 10 \times 100 \times 10^k \pmod {12} }} {{eqn | r = 10 \times 4 \pmod {12}       | c = Induction Hypothesis }} {{eqn | r = 4 \pmod {12}       | c = $10 \times 4 = 40 = 3 \times 12 + 4$ }} {{end-eqn}} So $P \paren k \implies P \paren{k+1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$100 \times 10^n = 4 \pmod {12}$ And then: {{begin-eqn}} {{eqn | l = N       | r = a_0 + a_1 10 + \sum^n_{r \mathop = 2} a_r 10^r       | c =  }} {{eqn | r = a_0 + a_1 10 + \sum^n_{r \mathop = 2} a_r \paren {100 \times 10^{r-2} } \pmod {12}       | c =  }} {{eqn | r = a_0 + a_1 10 + 4 \sum^n_{r \mathop = 2} a_r \pmod {12}       | c = Applying earlier result }} {{eqn | r = a_0 - 2 a_1 + 4 \sum^n_{r \mathop = 2} a_r \pmod {12}       | c = $-2 = -1 \times 12 + 10$ }} {{end-eqn}} Therefore: :$N = 0 \pmod {12} \iff a_0 - 2 a_1 + 4 \paren {\displaystyle \sum_{r \mathop = 2}^n a_r} = 0 \pmod {12}$ The result follows from Congruent to Zero if Modulo is Divisor {{qed}}
<onlyinclude> From Exponential is Strictly Increasing: :$\exp$ is strictly increasing on $\R$. From Strictly Monotone Mapping with Totally Ordered Domain is Injective: :$\exp$ is an injection. {{qed}}
We have Exponential on Real Numbers is Injection. Let $y \in \R_{> 0}$. Then $\exists x \in \R: x = \map \ln y$ That is: :$\exp x = y$ and so $\exp: \R \to \R_{>0}$ is a surjection. Hence the result. {{qed}}
=== Case $1$: $l = m$ === Let $l = m$. Then: :$\max \set {l, m} = l = m$ Let $\epsilon > 0$ be given. By definition of the limit of a real sequence, we can find $N_1$ such that: :$\forall n > N_1: \size {x_n - l} < \epsilon$ Similarly we can find $N_2$ such that: :$\forall n > N_2: \size {y_n - m} < \epsilon$ Let $N = \max \set {N_1, N_2}$. Then if $n > N$, both the above inequalities will be true: :$n > N_1$ :$n > N_2$ Thus $\forall n > N$: :if $y_n < x_n$ then: ::$\size {\max \set {x_n, y_n} - \max \set {l, m} } = \size {x_n - l} < \epsilon$ :if $x_n \le y_n$ then: ::$\size {\max \set {x_n, y_n} - \max \set {l, m} } = \size {y_n - m} < \epsilon$ In either case: : $\size {\max \set {x_n, y_n} - \max \set {l, m} } < \epsilon$ The result follows. {{qed|lemma}} === Case 2: $l > m$ === Let $l > m$. Then: :$l = \max \set {l, m}$ Let $\delta = \dfrac {l - m} 2$. Then: :$\delta > 0$ Let $\epsilon > 0$ be given. Let $\epsilon' = \min \set {\delta, \epsilon}$. Then: :$\epsilon' > 0$ By definition of the limit of a real sequence, we can find $N_1$ such that: :$\forall n > N_1: \size {x_n - l} < \epsilon'$ Similarly we can find $N_2$ such that: :$\forall n > N_2: \size {y_n - m} < \epsilon'$ Let $N = \max \set {N_1, N_2}$. Then if $n > N$, both the above inequalities will be true: :$n > N_1$ :$n > N_2$ From Corollary 3 of Negative of Absolute Value, we have that $\forall n > N$: :$\size {x_n - l} < \epsilon' \implies x_n > l - \epsilon'$ :$\size {y_n - m} < \epsilon' \implies y_n < m + \epsilon'$ As $\epsilon' \le \delta$: {{begin-eqn}} {{eqn | l = l - \epsilon'       | r = l - \delta       | o = \ge }} {{eqn | r = l - \dfrac {l - m} 2 }} {{eqn | r = \dfrac {2 l - l + m} 2 }} {{eqn | r = \dfrac {l + m} 2 }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = m + \epsilon'       | r = m + \delta       | o = \le }} {{eqn | r = m + \dfrac {l - m} 2 }} {{eqn | r = \dfrac {2 m + l - m} 2 }} {{eqn | r = \dfrac {l + m} 2 }} {{end-eqn}} So: :$x_n > l - \epsilon' \ge \dfrac {l + m} 2 \ge  m + \epsilon' > y_n$ Hence: :$x_n = \max \set{x_n, y_n}$ So $\forall n > N$: :$\size {\max \set {x_n, y_n} - \max \set {l, m} }  = \size {x_n - l} < \epsilon' \le \epsilon$ The result follows. {{qed|lemma}} === Case 3: $m > l$ === Similar to Case 2 by interchanging $l$ with $m$ and $x_n$ with $y_n$. {{qed}}
By Sum Less Maximum is Minimum: :$\forall n \in \R: \min \set {x_n, y_n} = x_n + y_n - \max \set {x_n, y_n}$ and :$\min \set {m, l} = m + l - \max \set {m, l}$ By Maximum Rule for Real Sequences: :$\displaystyle \lim_{n \mathop \to \infty} \max \set {x_n, y_n} = \max \set {m, l}$ By the Multiple Rule for Real Sequences: :$\displaystyle \lim_{n \mathop \to \infty} - \max \set {x_n, y_n} = - \max \set {m, l}$ By the Sum Rule for Real Sequences: :$\displaystyle \lim_{n \mathop \to \infty}  x_n + y_n - \max \set {x_n, y_n} = m + l - \max \set {m, l}$ Hence: :$\displaystyle \lim_{n \mathop \to \infty} \min \set {x_n, y_n} = \min \set {l, m}$ {{qed}}
=== (N1) : Positive Definiteness === {{begin-eqn}} {{eqn | lo= \forall x \in S:       | l = \norm x_S       | r = 0 }} {{eqn | ll= \leadstoandfrom       | l = \norm x       | r = 0       | c = Definition of $\norm x_S$ }} {{eqn | ll= \leadstoandfrom       | l = x       | r = 0       | c = Norm axiom (N1) }} {{end-eqn}} {{qed|lemma}} === (N2) : Multiplicativity === {{begin-eqn}} {{eqn | lo= \forall x, y \in S:       | l = \norm {x y}_S       | r = \norm {x y}       | c = Definition of $\norm {\, \cdot \,}_S$ }} {{eqn | r = \norm x \norm y       | c = Norm axiom (N2) }} {{eqn | r = \norm x_S \norm y_S       | c = Definition of $\norm {\, \cdot \,}_S$ }} {{end-eqn}} {{qed|lemma}} === (N3) : Triangle Inequality === {{begin-eqn}} {{eqn | lo= \forall x, y \in S:       | l = \norm {x + y}_S       | r = \norm {x + y}       | c = Definition of $\norm {\, \cdot \,}_S$ }} {{eqn | o = \le       | r = \norm x + \norm y       | c = Norm axiom (N3) }} {{eqn | r = \norm x_S + \norm y_S       | c = Definition of $\norm {\, \cdot \,}_S$ }} {{end-eqn}} {{qed}} Category:Division Rings Category:Norm Theory ky2s7b1nsucwi1zc3srsir3ym7gws58
By the definition of a normed division ring completion then: :$(1): \quad$ there exists a distance-preserving ring monomorphism $\phi: R \to R'$. :$(2): \quad \struct {R', \norm {\, \cdot \,}' }$ is a complete metric space. :$(3): \quad \phi \sqbrk R$ is a dense subspace in $\struct {R', \norm {\, \cdot \,}' }$. By image of a ring homomorphism is a subring then $\phi \sqbrk R$ is a subring of $R'$ and $\phi: R \to \phi \sqbrk R$ is an isomorphism. By Epimorphism from Division Ring to Ring then $\phi \sqbrk R$ is a division subring of $R'$. By Division Subring of Normed Division Ring then $\struct {\phi \sqbrk R, \norm {\, \cdot \,}' }$ is a normed division subring of $\struct {R', \norm {\, \cdot \,}' }$. By Distance-Preserving Surjection is Isometry of Metric Spaces then $\phi: R \to \phi \sqbrk R$ is an isometry. The result follows. {{qed}} Category:Normed Division Rings Category:Complete Metric Spaces Category:Completion of Normed Division Ring jig8wbrux01r8hlv2ykm1x9upahaez2
By Inverse of Algebraic Structure Isomorphism is Isomorphism then: :$\phi: R \to S$ is an ring isomorphism {{iff}} $\phi^{-1}: S \to R$ is also an ring isomorphism. Let $d_R$ and $d_S$ be the metric induced by the norms $\norm {\,\cdot\,}_R$ and $\norm {\,\cdot\,}_S$ respectively. By Inverse of Isometry of Metric Spaces is Isometry then: :$\phi: \struct {R, d_R} \to \struct {S, d_S}$ is an isometry {{iff}} $\phi^{-1}: \struct {S, d_S} \to \struct {R, d_R}$ is also an isometry. The result follows. {{qed}} Category:Normed Division Rings py0czj6bpbe3o9u87vjpmzmaqi3wrxw
Let $d_R$ and $d_S$ be the metric induced by the norms $\norm {\,\cdot\,}_R$ and $\norm {\,\cdot\,}_S$ respectively. === Necessary Condition === Let $\phi: R \to S$ be an isometric isomorphism. Then for $x \in R$: {{begin-eqn}} {{eqn | l = \norm {\map \phi x}_S       | r = \norm {\map \phi x - 0_S}_S       | c = $0_S$ is the zero of $S$ }} {{eqn | r = \norm {\map \phi x - \map \phi {0_R} }_S       | c = Ring Homomorphism Preserves Zero }} {{eqn | r = \map {d_S} {\map \phi x, \map \phi {0_R} }       | c = {{Defof|Metric Induced by Norm on Division Ring}} }} {{eqn | r = \map {d_R} {x, {0_R} }       | c = {{Defof|Isometry (Metric Spaces)}} }} {{eqn | r = \norm {x - {0_R} }_R       | c = {{Defof|Metric Induced by Norm on Division Ring}} }} {{eqn | r = \norm x_R       | c = $0_R$ is the zero of $R$ }} {{end-eqn}} The result follows. {{qed|lemma}} === Sufficient Condition === Let $\phi: R \to S$ satisfy: :$\forall x \in R: \norm {\map \phi x}_S = \norm x_R$ Then for $x, y \in R$: {{begin-eqn}} {{eqn | l = \map {d_S} {\map \phi x, \map \phi y}       | r = \norm {\map \phi x - \map \phi y}_S       | c = {{Defof|Metric Induced by Norm on Division Ring}} }} {{eqn | r = \norm {\map \phi {x - y} }_S       | c = $\phi$ is a Ring Isomorphism }} {{eqn | r = \norm {x - y}_R       | c = by hypothesis }} {{eqn | r = \map {d_R} {x, y}       | c = {{Defof|Metric Induced by Norm on Division Ring}} }} {{end-eqn}} The result follows. {{qed}} Category:Normed Division Rings lry0tga0ndyp1ra3oqf88egnkmdshws
In Holomorphic Function is Analytic, it is shown that:  :$\displaystyle \map f z = \sum_{n \mathop = 0}^\infty \paren {\frac 1 {2 \pi i} \int_{\partial D} \frac {\map f t} {\paren {t - a}^{n + 1} } \rd t} \paren {z - a}^n$  for all $z \in D$. From Cauchy's Integral Formula: General Result, we have:  :$\displaystyle \frac 1 {2 \pi i} \int_{\partial D} \frac {\map f t} {\paren {t - a}^{n + 1} } \rd t = \frac {\map {f^n} a} {n!}$ Hence the result.  {{qed}} Category:Complex Analysis r6d3bo3tpqrs7hrsjjgxuobtbdq2eo7
By definition of $n$th power (for positive $n$): :$x^n = \begin{cases} 1 & : n = 0 \\ x \times x^{n - 1} & : n > 0 \end{cases}$ Thus: {{begin-eqn}} {{eqn | l = x^2       | r = x \times x^1       | c =  }} {{eqn | r = x \times x \times x^0       | c =  }} {{eqn | r = x \times x \times 1       | c =  }} {{eqn | r = x \times x       | c =  }} {{end-eqn}} Hence the result. {{qed}} Category:Square Function 0unisqa74pxlosnkezxy36diu2jkfgx
$\forall x, y \in S$: {{begin-eqn}} {{eqn | l = \norm {x + y}_S       | r = \norm {x + y}       | c = Definition of $\norm {\,\cdot\,}_S$ }} {{eqn | o = \le       | r = \max \set {\norm x, \norm y}       | c = $(\text N 4)$: Ultrametric Inequality }} {{eqn | r = \max \set {\norm x_S, \norm y_S}       | c = Definition of $\norm {\, \cdot \,}_S$ }} {{end-eqn}} {{qed}} Category:Normed Division Rings mnmsno5i821o90pknrspo4twwhbguqt
=== Necessary Condition === Let $\norm {\,\cdot\,}_R$ be a non-archimedean norm. Then for all $x,y \in R$: {{begin-eqn}} {{eqn | l = \norm {x + y}_S       | r = \norm {\map \phi {\map {\phi^{-1} } x} + \map \phi {\map {\phi^{-1} } y} }_S       | c = $\phi$ is a bijection }} {{eqn | r = \norm {\map {\phi^{-1} } x  + \map {\phi^{-1} } y}_R       | c = $\phi$ is an isometry. }} {{eqn | o = \le       | r = \max \set {\norm {\map {\phi^{-1} } x}_R, \norm {\map {\phi^{-1} } y}_R}       | c = $\norm {\,\cdot\,}_R$ is non-archimedean. }} {{eqn | r = \max \set {\norm {\map \phi {\map {\phi^{-1} } x} }_S, \norm {\map \phi {\map {\phi^{-1} } y} }_S}       | c = $\phi$ is an isometry. }} {{eqn | r = \max \set {\norm x_S, \norm y_S}       | c = $\phi$ is a bijection. }} {{end-eqn}} {{qed|lemma}} === Sufficient Condition === Let $\norm {\, \cdot \,}_S$ be a non-archimedean norm. By Inverse of Isometric Isomorphism, $\phi^{-1}: S \to R$ is an isometric isomorphism. By the '''necessary condition''', $\norm {\, \cdot \,}_R$ is non-archimedean. {{qed}} Category:Normed Division Rings m3vzfu8nxtrx8s5lgtieza28ol5q02r
By the definition of a normed division ring completion then: :$(1): \quad$ there exists a distance-preserving ring monomorphism $\phi: R \to R'$. :$(2): \quad \struct {R', \norm {\, \cdot \,}' }$ is a complete metric space. :$(3): \quad \phi \sqbrk R$ is a dense subspace in $\struct {R', \norm {\, \cdot \,}' }$. By Normed Division Ring is Dense Subring of Completion, $\phi \sqbrk R$ is a dense normed division subring of $R'$ and $\phi: R \to \map \phi R$ is an isometric isomorphism. By Isometrically Isomorphic Non-Archimedean Division Rings, $\struct {R, \norm {\, \cdot \,} }$ is non-archimedean {{iff}} $\struct {\phi \sqbrk R, \norm {\, \cdot \,}' }$ is. So it remains to show that $\struct {\phi \sqbrk R, \norm {\, \cdot \,}' }$ non-archimedean {{iff}} $\struct {R', \norm {\, \cdot \,}' }$ is. === Necessary Condition === Let $\struct {\phi \sqbrk R, \norm {\, \cdot \,}' }$ be non-archimedean. Let $x', y' \in R'$. By the definition of a dense subset: :$\map \cl {\phi \sqbrk R} = R'$ By Closure of Subset of Metric Space by Convergent Sequence then: :there exists a sequence $\sequence {x_n'} \subseteq \phi \sqbrk R$ that converges to $x'$, that is, $\displaystyle \lim_{n \mathop \to \infty} x_n' = x'$ :there exists a sequence $\sequence {y_n'} \subseteq \phi \sqbrk R$ that converges to $y'$, that is, $\displaystyle \lim_{n \mathop \to \infty} y_n' = y'$ By Sum Rule for Sequences in Normed Division Ring then: :$\displaystyle \lim_{n \mathop \to \infty} x_n' + y_n' = x' + y'$ By modulus of limit then: :$\displaystyle \lim_{n \mathop \to \infty} \norm {x_n'}' = \norm {x'}'$ :$\displaystyle \lim_{n \mathop \to \infty} \norm {y_n'}' = \norm {y'}'$ :$\displaystyle \lim_{n \mathop \to \infty} \norm {x_n' + y_n'}' = \norm {x' + y'}'$ By Maximum Rule for Real Sequences then: :$\displaystyle \lim_{n \mathop \to \infty} \max \set {\norm {x_n'}', \norm {y_n'}' } = \max \set {\norm {x'}', \norm {y'}'}$ Since $\struct {\phi \sqbrk R, \norm {\, \cdot \,}' }$ is non-archimedean then: :$\forall n: \norm {x_n' + y_n'}' \le \max \set {\norm {x_n'}', \norm {y_n'}' }$ By Inequality Rule for Real Sequences then: :$\displaystyle \norm {x' + y'}' = \lim_{n \mathop \to \infty} \norm {x_n' + y_n'}' \le \lim_{n \mathop \to \infty} \max \set {\norm {x_n'}', \norm {y_n'}' } = \max \set {\norm {x'}', \norm {y'}'}$ The result follows. {{qed|lemma}} === Sufficient Condition === Let $\struct {R', \norm {\, \cdot \,}' }$ be non-archimedean. By Subring of Non-Archimedean Division Ring then $\struct {\phi \sqbrk R, \norm {\, \cdot \,}' }$ is non-archimedean. {{qed}}
The operation of integer multiplication is defined on all integers. Thus: :$\forall x \in \Z: \exists y \in \Z: x^2 = y$ Hence the result by definition of domain. {{qed}} Category:Square Function thaqmabzebfpu0jntuszq9f2o226uxb
Follows directly by definition of the integer square function. {{qed}} Category:Square Function crigqa9nrja5iymme5m6dh111j0ao06
From Order is Preserved on Positive Reals by Squaring, $f$ is strictly increasing on $\R_{\ge 0}$. By definition, a strictly increasing real function is strictly monotone. The result follows from Strictly Monotone Real Function is Bijective. {{qed}}
From Restriction of Real Square Mapping to Positive Reals is Bijection, $f$ is a bijection. By definition of the positive square root: :$y = \sqrt x \iff x = y^2$ for $x, y \in \R_{\ge 0}$. Hence the result. {{qed}}
From Real Square Function is not Injective, $f$ is not an injection. From Real Square Function is not Surjective, $f$ is not a surjection. The result follows by definition of bijection. {{qed}}
We have that Linear Function on Real Numbers is Bijection. Let $y = \map f x$. Then: {{begin-eqn}} {{eqn | l = y       | r = \map f x       | c =  }} {{eqn | r = a x + b       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = \dfrac {y - b} a       | c =  }} {{end-eqn}} and so: :$\forall y \in \R: \map {f^{-1} } y = \dfrac {y - b} a$ {{qed}}
Let $a \ne 0$. Let $y = \map f x$. {{begin-eqn}} {{eqn | l = y       | r = \map f x       | c =  }} {{eqn | r = a x + b       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = \dfrac {y - b} a       | c =  }} {{end-eqn}} and so: :$\forall y \in \R: \exists x \in \R; y = \map f x$ demonstrating that $f$ is surjective. Then: {{begin-eqn}} {{eqn | l = \map f {x_1}       | r = \map f {x_2}       | c =  }} {{eqn | ll= \leadsto       | l = a x_1 + b       | r = a x_2 + b       | c =  }} {{eqn | ll= \leadsto       | l = a x_1       | r = a x_2       | c =  }} {{eqn | ll= \leadsto       | l = x_1       | r = x_2       | c = as $a \ne 0$ }} {{end-eqn}} demonstrating that $f$ is injective. Thus $f$ is a bijection by definition. {{qed|lemma}} Let $a = 0$. Then: :$\forall x \in \R: \map f x = b$ Thus for example: :$\map f 1 = \map f 2$ and $f$ is trivially not injective. Also: :$\forall y \in \R: y \ne b \implies \nexists x \in \R: \map f x = y$ and $f$ is equally trivially not surjective either. Thus $f$ is not a bijection by definition. {{qed}}
{{begin-eqn}} {{eqn | l = \paren {g \circ f} \circ \paren {g \circ f}       | r = g \circ \paren {f \circ g} \circ f       | c = Composition of Mappings is Associative }} {{eqn | r = g \circ I_T \circ f       | c = {{Defof|Right Inverse Mapping}} }} {{eqn | r = g \circ f       | c = {{Defof|Identity Mapping}} }} {{end-eqn}} {{qed}}
To demonstrate set equivalence, it is sufficient to construct a bijection between the two sets. Let $f: \Z \to 2 \Z$ defined as: :$\forall x \in \Z: \map f x = 2 x$ {{begin-eqn}} {{eqn | l = \map f x       | r = \map f y       | c =  }} {{eqn | ll= \leadsto       | l = 2 x       | r = 2 y       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = y       | c =  }} {{end-eqn}} demonstrating injectivity. {{begin-eqn}} {{eqn | l = y       | o = \in       | r = 2 \Z       | c =  }} {{eqn | ll= \leadsto       | lo= \exists x \in \Z:       | l = y       | r = 2 x       | c = {{Defof|Even Integer}} }} {{eqn | ll= \leadsto       | l = y       | o = \in       | r = f \sqbrk \Z       | c =  }} {{end-eqn}} demonstrating surjectivity. Hence by definition $f$ is a bijection. {{qed}}
Proof by Counterexample: Let $S := \set {1, 2, 3}$. Let: {{begin-eqn}} {{eqn | l = \pi       | o = :=       | r = \dbinom {1 \ 2 \ 3} {2 \ 3 \ 1} }} {{eqn | l = \rho       | o = :=       | r = \dbinom {1 \ 2 \ 3} {1 \ 3 \ 2} }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \pi \circ \rho       | r = \dbinom {1 \ 2 \ 3} {2 \ 3 \ 1} \circ \dbinom {1 \ 2 \ 3} {1 \ 3 \ 2}  }} {{eqn | r = \dbinom {1 \ 2 \ 3} {3 \ 2 \ 1} }} {{end-eqn}} while: {{begin-eqn}} {{eqn | l = \rho \circ \pi       | r = \dbinom {1 \ 2 \ 3} {1 \ 3 \ 2} \circ \dbinom {1 \ 2 \ 3} {2 \ 3 \ 1} }} {{eqn | r = \dbinom {1 \ 2 \ 3} {2 \ 1 \ 3} }} {{eqn | o = \ne       | r = \pi \circ \rho }} {{end-eqn}} {{qed}}
Let $d$ be the metric induced by the norm $\norm {\,\cdot\,}$. By Non-Archimedean Norm iff Non-Archimedean Metric then $d$ is a non-Archimedean metric and $\struct {R, d}$ is an ultrametric space. Let $x, y \in R$ and $\norm x \ne \norm y$. By the definition of the non-Archimedean metric $d$ then: :$\norm x = \norm {x - 0} = \map d {x, 0}$ and similarly: :$\norm y = \map d {y, 0}$ By assumption then: :$\map d {x, 0} \ne \map d {y, 0}$ By Three Points in Ultrametric Space have Two Equal Distances then: :$\norm {x - y} = \map d {x, y} = \max \set {\map d {x, 0}, \map d {y, 0} } = \max \set {\norm x, \norm y}$ By Norm of Negative then: :$\norm {y - x} = \norm {x - y} = \max \set {\norm x, \norm y}$ By the definition of the non-Archimedean metric $d$ then: :$\norm y = \norm {0 - \paren {-y} } = \map d {0, -y} = \map d {-y, 0}$ By assumption then: :$\map d {x, 0} \ne \map d {-y, 0}$ By Three Points in Ultrametric Space have Two Equal Distances then: :$\map d {x, -y} = \max \set {\map d {x, 0}, \map d {-y, 0} } = \max \set {\norm x, \norm y}$ By the definition of the non-Archimedean metric $d$ then: :$\map d {x, -y} = \norm {x - \paren {-y} } = \norm {x + y}$ So $\norm {x + y} = \max \set {\norm x, \norm y}$. The result follows. {{qed}}