hcju/proofwikiqa huggingface dataset:
Let $\PP$ be the valuation ideal induced by the non-Archimedean norm $\norm {\,\cdot\,}$, that is: :$\PP = \set{x \in R : \norm{x} \lt 1}$ By Valuation Ideal is Maximal Ideal of Induced Valuation Ring then: :$\PP$ is a maximal left ideal of $\OO$. Let $J \subsetneq \OO$ be any maximal left ideal of $\OO$. Let $x \in \OO \setminus \PP$. {{AimForCont}} $x \in J$. By Norm of Inverse then: :$\norm {x^{-1}} = 1 / \norm x = 1 / 1 = 1$  Hence: :$x^{-1} \in \OO$ Since $J$ is a left ideal then: :$x^{-1} x = 1_R \in J$ Thus: :$\forall y \in \OO: y \cdot 1_R = y \in J$ That is:  :$J = \OO$ This contradicts the assumption that $J \ne \OO$. So: :$x \notin J$ Hence:  :$\paren {\OO \setminus \PP} \cap J = \O$ That is: :$J \subseteq \PP$ Since $J$ and $\PP$ are both maximal left ideals then: :$J = \PP$ The result follows. {{qed}}
Since $p \nmid 1$ then for all $a \in \Z$, $a = \dfrac a 1 \in \Z_{\paren p}$. Hence $\Z \subset \Z_{\paren p}$ is a subring of $\Z_{\paren p}$. Let $i: \Z \to \Z_{\paren p}$ be the inclusion mapping defined by: :$\map i a = a$ By Inclusion Mapping is Monomorphism then $i$ is a ring monomorphism. Let $q: \Z_{\paren p} \to \Z_{\paren p} / p \, \Z_{\paren p}$ be the quotient ring epimorphism from $\Z_{\paren p}$ to $\Z_{\paren p} / p \, \Z_{\paren p}$. By Quotient Ring Epimorphism is Epimorphism, $q$ is a epimorphism. By definition, $\phi = q \circ i$ is the composition of $i$ with $q$. By Composition of Ring Homomorphisms is Ring Homomorphism, $\phi$ is a homomorphism. {{qed}}
Let $\map \ker \phi$ denote the kernel of $\phi$. Then: {{begin-eqn}} {{eqn | l = a       | o = \in       | r = \map \ker \phi       | c = }} {{eqn | ll= \leadstoandfrom       | l = \map \phi a       | r = p \Z_{\ideal p}       | c = {{Defof|Kernel of Ring Homomorphism}} }} {{eqn | ll= \leadstoandfrom       | l = a + p \Z_{\ideal p}       | r = p \Z_{\ideal p}       | c = Definition of $\phi$ }} {{eqn | ll= \leadstoandfrom       | l = a = a - 0        | o = \in       | r = p \Z_{\ideal p}       | c = Element in Right Coset iff Product with Inverse in Subgroup }} {{eqn | ll= \leadstoandfrom       | lo= \exists a' \in \Z:       | l = a = p a'       | r = p a' }} {{eqn | ll= \leadstoandfrom       | l = a       | o = \in       | r = p \Z }} {{end-eqn}} Hence: :$p \Z = \map \ker \phi$ {{qed}}
Let $a / b \in \Z_{\paren p}$, where $a / b$ are in canonical form. Then $p \nmid b$ Let $\F_p$ be the field of integers modulo $p$. By the definition of a field: :$\exists b' \in \Z: b b' \equiv 1 \pmod p$ By the definition of congruence modulo $p$: :$p \divides b b' - 1$ By Divisor Divides Multiple: :$\forall a \in \Z: p \divides a b b' - a$ By Valuation Ideal of P-adic Norm on Rationals then: :$a b' - \dfrac a b = \dfrac {a b b' - a} b \in p \Z_{\paren p}$ By Element in Left Coset iff Product with Inverse in Subgroup: :$\map \phi {a b'} = a b' + p \Z_{\paren p} = a / b + p \Z_{\paren p}$ It follows that: :$\phi : \Z \to \Z_{\paren p} / p \Z_{\paren p}$ is a surjection. {{qed}}
From Equation for Line through Two Points in Complex Plane: Parametric Form 1: :$z = z_1 + t \paren {z_2 - z_1}$ Letting: {{begin-eqn}} {{eqn | l = z       | r = x + i y }} {{eqn | l = z_1       | r = x_1 + i y_1 }} {{eqn | l = z_2       | r = x_2 + i y_2 }} {{end-eqn}} the parametric equations follow by equating real parts and imaginary parts. {{qed}}
Let $z_1$ and $z_2$ be represented by the points $A = \tuple {x_1, y_1}$ and $B = \tuple {x_2, y_2}$ respectively in the complex plane. Let $z$ be an arbitrary point on $L$ represented by the point $P$. :400px We have that $L$ passes through the point: :$\dfrac {z_1 + z_2} 2$ and is perpendicular to the straight line: :$z = z_1 + t \paren {z_2 - z_1}$ {{ProofWanted}} Category:Equation for Perpendicular Bisector of Two Points in Complex Plane azrcyl0u9al6hkpf7oqidu3zb161zkd
A curve on the surface $\mathbf r$ can be specified as $u = \map u t$, $v = \map v t$ The arc length between the points corresponding to $t_0$ and $t_1$ equals: :$\displaystyle J \sqbrk {u, v} = \int_{t_0}^{t_1} \sqrt {E u'^2 + 2 F u'v' + G v'^2} \rd t$ The following derivatives will appear in Euler's Equations: {{begin-eqn}} {{eqn | l = \dfrac {\partial} {\partial u'} \sqrt {E u'^2 + 2 F u'v' + G v'^2}       | r = \frac 1 {2 \sqrt {E u'^2 + 2 F u'v' + G v'^2} } \map {\dfrac {\partial} {\partial u'} } {E u'^2 + 2 F u'v' + G v'^2} }} {{eqn | r = \frac {2 E u' + 2 F_u u'} {2 \sqrt {E u'^2 + 2 F u'v' + G v'^2} } }} {{eqn | r = \frac {E u' + F u'} {\sqrt {E u'^2 + 2 F u'v' + G v'^2} } }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \dfrac {\partial} {\partial u} \sqrt {E u'^2 + 2 F u'v' + G v'^2}       | r = \frac 1 {2 \sqrt {E u'^2 + 2 F u'v' + G v'^2} } \map {\dfrac {\partial} {\partial u} } {E u'^2 + 2 F u'v' + G v'^2} }} {{eqn | r = \frac {E_u u'^2 + 2 F_u u' v' + G_u v'^2} {2 \sqrt {E u'^2 + 2 F u'v' + G v'^2} } }} {{end-eqn}} Analogous relations hold for derivatives {{WRT|Differentiation}} $v$ and $v'$. Euler's Equation together with the results above yield the geodesic equations. {{qed}}
From Newton's First Law of Motion, the total force on $B$ must equal zero in order for $B$ to remain stationary. That is, $\mathbf E$ must be such that: :$\mathbf E + \displaystyle \sum_{k \mathop = 1}^n \mathbf F_k = \bszero$ That is: :$\mathbf E = -\displaystyle \sum_{k \mathop = 1}^n \mathbf F_k$ {{qed}} Category:Force Category:Equilibrants k6ckdj50bq38jsmydrpwo6cbv5ij6sz
{{begin-eqn}} {{eqn | l = \cmod {e^{i z} }       | r = \cmod {\map \exp {i R \, \map \exp {i \theta} } }       | c =  }} {{eqn | r = \cmod {\map \exp {i R \paren {\cos \theta + i \sin \theta} } }       | c =  }} {{eqn | r = \cmod {\map \exp {R \paren {-\sin \theta + i \cos \theta} } }       | c =  }} {{eqn | r = \cmod {\map \exp {- R \sin \theta} \, \map \exp {i \cos \theta} }       | c =  }} {{eqn | r = \map \exp {- R \sin \theta}       | c = Modulus and Argument of Complex Exponential }} {{end-eqn}} {{qed}}
Let: {{begin-eqn}} {{eqn | l = r e^{i \theta}       | r = \displaystyle \sum_{k \mathop = 1}^n z_k       | c = }} {{eqn | r = z_1 + z_2 + \dotsb + z_k       | c = }} {{eqn | r = r_1 \paren {\cos \theta_1 + i \sin \theta_1} + r_2 \paren {\cos \theta_2 + i \sin \theta_2} + \dotsb + r_n \paren {\cos \theta_n + i \sin \theta_n}       | c = {{Defof|Complex Number}} }} {{eqn | r = r_1 \cos \theta_1 + r_2 \cos \theta_2 + \dotsb + r_n \cos \theta_n + i \paren {r_1 \sin \theta_1 + r_2 \sin \theta_2 + \dotsb + r_n \sin \theta_n}       | c = rerranging }} {{end-eqn}} By the definition of the complex modulus, with $z = x + i y$, $r$ is defined as: :$r = \sqrt {\map {\Re^2} z + \map {\Im^2} z}$ Hence {{begin-eqn}} {{eqn | l = r       | r = \sqrt {\map {\Re^2} z + \map {\Im^2} z}       | c = }} {{eqn | l = r       | r = \sqrt {\paren {r_1 \cos \theta_1 + r_2 \cos \theta_2 + \dotsb + r_n \cos \theta_n }^2 + \paren {r_1 \sin \theta_1 + r_2 \sin \theta_2 + \dotsb + r_n \sin \theta_n}^2 }       | c = }} {{end-eqn}} In the above we have two types of pairs of terms: {{begin-eqn}} {{eqn | n = 1       | lo= 1 \le k \le n:       | l = {r_k}^2 \cos^2 {\theta_k}^2 + {r_k}^2 \sin^2 {\theta_k}^2       | r = {r_k}^2 \paren {\cos^2 {\theta_k}^2 + \sin^2 {\theta_k}^2}       | c = }} {{eqn | r = {r_k}^2       | c =Sum of Squares of Sine and Cosine }} {{eqn | n = 2       | lo= 1 \le j < k \le n:       | l = 2 r_j r_k \cos \theta_j \cos \theta_k + 2 {r_j} {r_k} \sin \theta_j \sin \theta_k       | r = 2 r_j r_k \paren {\cos \theta_j \cos \theta_k + \sin \theta_j \sin \theta_k}       | c = }} {{eqn | r = 2 r_j r_k \, \map \cos {\theta_j - \theta_k}       | c = Cosine of Difference }} {{end-eqn}} Hence: :$r = \sqrt {\displaystyle \sum_{k \mathop = 1}^n r_k + \displaystyle \sum_{1 \mathop \le j \mathop < k \mathop \le n} 2 {r_j} {r_k} \, \map \cos {\theta_j - \theta_k} }$ Note that $r > 0$ since $r_k > 0$ for all $k$. Hence we may safely assume that $r > 0$ when determining the argument below. By definition of the argument of a complex number, with $z = x + i y$, $\theta$ is defined as any solution to the pair of equations: :$(1): \quad \dfrac x {\cmod z} = \map \cos \theta$ :$(2): \quad \dfrac y {\cmod z} = \map \sin \theta$ where $\cmod z$ is the modulus of $z$. As $r > 0$ we have that $\cmod z \ne 0$ by definition of modulus. Hence we can divide $(2)$ by $(1)$, to get: {{begin-eqn}} {{eqn | l = \map \tan \theta       | r = \frac y x       | c = }} {{eqn | r = \frac {\map \Im z} {\map \Re z}       | c = }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \theta       | r = \map \arctan {\frac {\map \Im {r e^{i \theta} } } {\map \Re {r e^{i \theta} } } }       | c = }} {{eqn | r = \map \arctan {\dfrac {r_1 \sin \theta_1 + r_2 \sin \theta_2 + \dotsb + r_n \sin \theta_n} {r_1 \cos \theta_1 + r_2 \cos \theta_2 + \dotsb + r_n \cos \theta_n} }       | c = }} {{end-eqn}} {{qed}}
If $\hat {\sigma^2}$ is a biased estimator of $\sigma^2$, then:  :$\displaystyle \expect {\hat {\sigma^2} } \ne \sigma^2$ We have:  {{begin-eqn}} {{eqn	| l = \expect {\hat {\sigma^2} } 	| r = \expect {\frac 1 n \sum_{i \mathop = 1}^n \paren {X_i - \bar X}^2} }} {{eqn	| r = \expect {\frac 1 n \sum_{i \mathop = 1}^n \paren{\paren {X_i - \mu} - \paren {\bar X - \mu} }^2} 	| c = writing $X_i - \bar X = X_i - \bar X - \mu + \mu$ }} {{eqn	| r = \expect {\frac 1 n \sum_{i \mathop = 1}^n \paren{\paren {X_i - \mu}^2 - 2 \paren {\bar X - \mu} \paren {X_i -\mu} + \paren {\bar X - \mu}^2} } 	| c = Square of Difference }} {{eqn	| r = \expect {\frac 1 n  \sum_{i \mathop = 1}^n \paren {X_i - \mu}^2 - \frac 2 n \paren {\bar X - \mu} \sum_{i \mathop = 1}^n \paren {X_i -\mu} + \frac 1 n \paren {\bar X - \mu}^2 \sum_{i \mathop = 1}^n 1} 	| c = Summation is Linear }} {{end-eqn}} We have that:  {{begin-eqn}} {{eqn	| l = \frac 1 n \sum_{i \mathop = 1}^n \paren {X_i - \mu} 	| r = \frac 1 n \sum_{i \mathop = 1}^n X_i - \frac n n \mu 	| c = from $\displaystyle \sum_{i \mathop = 1}^n 1 = n$, noting that $\mu$ is independent of $i$.  }} {{eqn	| r = \bar X - \mu 	| c = by our definition of $\bar X$ }} {{end-eqn}} So: {{begin-eqn}} {{eqn	| l = \expect {\frac 1 n \sum_{i \mathop = 1}^n \paren {X_i - \mu}^2 - \frac 2 n \paren {\bar X - \mu} \sum_{i \mathop = 1}^n \paren {X_i -\mu} + \frac 1 n \paren {\bar X - \mu}^2 \sum_{i \mathop = 1}^n 1} 	| r = \expect {\frac 1 n \sum_{i \mathop = 1}^n \paren {X_i - \mu}^2 - 2 \paren {\bar X - \mu}^2 + \paren {\bar X - \mu}^2} }} {{eqn	| r = \frac 1 n \expect {\sum_{i \mathop = 1}^n \paren {X_i - \mu}^2} - \expect {\paren {\bar X - \mu}^2} 	| c = Linearity of Expectation Function }} {{eqn	| r = \frac 1 n \sum_{i \mathop = 1}^n \expect {\paren {X_i - \mu}^2} - \var {\bar X} 	| c = {{Defof|Variance}}, Linearity of Expectation Function }} {{eqn	| r = \frac 1 n \sum_{i \mathop = 1}^n \var {X_i} - \frac {\sigma^2} n 	| c = {{Defof|Variance}}, Variance of Sample Mean }} {{eqn	| r = \frac n n \sigma^2 - \frac {\sigma^2} n 	| c = $\var {X_i} = \sigma^2$, $\displaystyle \sum_{i \mathop = 1}^n 1 = n$ }} {{eqn	| r = \sigma^2 - \frac {\sigma^2} n }} {{eqn	| o = \ne 	| r = \sigma^2 }} {{end-eqn}} So $\hat {\sigma^2}$ is a biased estimator of $\sigma^2$.  Further, we have:  :$\displaystyle \operatorname{bias} \paren {\hat {\sigma ^2}} = \sigma^2 - \frac {\sigma^2} n - \sigma^2 = -\frac {\sigma^2} n$ {{qed}} Category:Inductive Statistics has08yzmpl19phi5wwcvio4oj4zzyo8
If $\hat{\sigma^2}$ is an unbiased estimator of $\sigma^2$, then:  :$\displaystyle \expect {\hat {\sigma^2}} = \sigma^2$ In Bias of Sample Variance, it is shown that:  :$\displaystyle \expect {\frac 1 n \sum_{i \mathop = 1}^n \paren {X_i - \bar X}^2} = \paren {1 - \frac 1 n} \sigma^2$ By Linearity of Expectation Function:  {{begin-eqn}} {{eqn	| l = n \expect {\frac 1 n \sum_{i \mathop = 1}^n \paren {X_i - \bar X}^2} 	| r = \expect {\sum_{i \mathop = 1}^n \paren {X_i - \bar X}^2} }} {{eqn	| r = \paren {n - 1} \sigma^2 }} {{end-eqn}} So:  {{begin-eqn}} {{eqn	| l = \sigma^2 	| r = \frac 1 {n - 1} \expect {\sum_{i \mathop = 1}^n \paren {X_i - \bar X}^2} }} {{eqn	| r = \expect {\frac 1 {n - 1} \sum_{i \mathop = 1}^n \paren {X_i - \bar X}^2} 	| c = Linearity of Expectation Function }} {{eqn	| r = \expect {\hat {\sigma^2} } }} {{end-eqn}} So $\hat {\sigma^2}$ is an unbiased estimator of $\sigma^2$.  {{qed}} {{Namedfor|Friedrich Wilhelm Bessel|cat = Bessel}} Category:Inductive Statistics hi7tbzjldl7mmosk8kkdcus2epjmpi8
If $\bar X$ is an unbiased estimator of $\mu$, then:  :$\displaystyle \expect {\bar X} = \mu$ We have:  {{begin-eqn}} {{eqn	| l = \expect {\bar X} 	| r = \expect {\frac 1 n \sum_{i \mathop = 1}^n X_i} }} {{eqn	| r = \frac 1 n \sum_{i \mathop = 1}^n \expect {X_i} 	| c = Linearity of Expectation Function }} {{eqn	| r = \frac 1 n \sum_{i = 1}^n \mu 	| c = as $\expect {X_i} = \mu$ }} {{eqn	| r = \frac n n \mu 	| c = as $\displaystyle \sum_{i \mathop = 1}^n 1 = n$ }} {{eqn	| r = \mu }} {{end-eqn}} So $\bar X$ is an unbiased estimator of $\mu$.  {{qed}} Category:Inductive Statistics osbe59f8es82v6b730jl34k4l86lymx
By Characterisation of Non-Archimedean Division Ring Norms then: :$\exists n \in \N$ such that $\norm n > 1$ Let $n_0 = \min \set {n \in \N : \norm n > 1}$ By Norm of Unity then: :$n_0 > 1$ Let $\alpha = \dfrac {\log \norm {n_0} } {\log n_0}$ Since $n_0, \norm n_0 > 1$ then: :$\alpha > 0$ ==== Lemma 1.1 ==== {{:Ostrowski's Theorem/Archimedean Norm/Lemma 1.1}}{{qed|lemma}} ==== Lemma 1.2 ==== {{:Ostrowski's Theorem/Archimedean Norm/Lemma 1.2}}{{qed|lemma}} Hence: :$\forall n \in \N: \norm n = n^\alpha = \size n^\alpha$ By Equivalent Norms on Rational Numbers then $\norm {\, \cdot \,}$ is equivalent to the absolute value $\size {\, \cdot \,}$.
By Characterisation of Non-Archimedean Division Ring Norms then: :$\forall n \in \N: \norm n \le 1$ ==== Lemma 2.1 ==== {{:Ostrowski's Theorem/Non-Archimedean Norm/Lemma 2.1}}{{qed|lemma}} Let $n_0 = \min \set {n \in N : \norm n < 1}$. ==== Lemma 2.2 ==== {{:Ostrowski's Theorem/Non-Archimedean Norm/Lemma 2.2}}{{qed|lemma}} Let $p = n_0$. Let $\alpha = - \dfrac {\log \norm p } {\log p}$ then: :$\norm p = p^{-\alpha} = \paren {p^{-1}}^\alpha = \norm p_p^\alpha$ Let $b \in N$ Let $p \nmid b$. Then $p$ and $b$ are coprime, that is: :$p \perp b$. By Corollary 5 of Three Points in Ultrametric Space have Two Equal Distances: :$\norm b = 1$ By the definition of the $p$-adic norm: :$\norm b_p = 1$  Hence: :$\norm b = 1 = 1^\alpha = \norm b_p^\alpha$ Let $p \divides b$. Let $\nu = \map {\nu_p} b$ where $\nu_p$ is the $p$-adic valuation on $\Z$. Then: :$b = p^\nu c$ where $p \nmid c$ From the above: :$\norm c = 1$ Hence: {{begin-eqn}} {{eqn | l = \norm b       | r = \norm p^\nu \norm {c}       | c = {{Norm Axiom|2}} }} {{eqn | r = \norm p^\nu       | c =  }} {{eqn | r = \norm p_p^{\alpha \nu} }} {{eqn | r = \paren {p^{-1} }^{\alpha \nu}       | c = {{Defof|P-adic Norm|$p$-adic norm}} }} {{eqn | r = p^{-\alpha \nu}       | c =  }} {{eqn | r = \paren {p^{-\nu} }^\alpha       | c =  }} {{eqn | r = \norm b_p^\alpha       | c = {{Defof|P-adic Norm|$p$-adic norm}} }} {{end-eqn}} It has been shown: :$\forall b \in \N: \norm b = \norm b_p^\alpha$ By Equivalent Norms on Rational Numbers: :$\norm {\, \cdot \,}$ is equivalent to the absolute value $\norm {\, \cdot \,}_p$.
By the definition of $\alpha$ then: :$\norm {n_0} = n_0^\alpha$ By the definition of $n_0$ then: :$n_0^\alpha > 1$ Let $n \in \N$. By Basis Representation Theorem then $n$ can be written: :$n = a_0 + a_1 n_0 + a_2 n_0^2 + \cdots + a_s n_0^s$ where $0 \le a_i < n_0$ and $a_s \ne 0$ Since all of the $a_i < n_0$, by choice of $n_0$ then: :$\forall a_i: \norm {a_i} \le 1$ Then: {{begin-eqn}} {{eqn | l = \norm n       | o = \le       | r = \norm {a_0} + \norm {a_1 n_0} + \norm {a_2 n_0^2} + \cdots + \norm {a_s n_0^s}       | c = {{NormAxiom|3}} }} {{eqn | r = \norm {a_0} + \norm {a_1} \norm {n_0} + \norm {a_2} \norm {n_0}^2 + \cdots + \norm {a_s} \norm {n_0}^s       | c = {{NormAxiom|2}} }} {{eqn | r = \norm {a_0} + \norm {a_1} n_0^\alpha + \norm {a_2} n_0^{2 \alpha} + \cdots + \norm {a_s} n_0^{s\alpha}       | c = as $\norm {n_0} = n_0^\alpha$ }} {{eqn | o = \le       | r = 1 + n_0^\alpha + n_0^{2 \alpha} + \cdots + n_0^{s \alpha}       | c = as $\forall a_i: \norm {a_i} \le 1$ }} {{eqn | r = n_0^{s \alpha} \paren {1 + n_0^{-\alpha} + n_0^{-2 \alpha} + \cdots + n_0^{-s \alpha} }       | c =  }} {{eqn | o = \le       | r = n_0^{s \alpha} \paren {\sum_{i \mathop = 0}^\infty \paren{\frac 1 {n_0^\alpha} }^i}        | c =  }} {{eqn | r = n_0^{s \alpha} \paren {\dfrac {n_0^\alpha} {n_0^\alpha - 1} }       | c = Sum of Infinite Geometric Progression (and as $n_0^\alpha > 1$) }} {{eqn | o = \le       | r = n^\alpha \paren {\dfrac {n_0^\alpha} {n_0^\alpha - 1} }       | c = as $n \ge n_0^s$ }} {{end-eqn}} Let $C = \paren {\dfrac {n_0^\alpha} {n_0^\alpha - 1} }$ Hence: :$\norm n \le C n^\alpha$ As $n \in \N$ was arbitrary: :$\forall n \in N: \norm n \le C n^\alpha$ Let $n, N \in N$ Then: :$\norm {n^N} \le C \paren {n^N}^\alpha$ Now: {{begin-eqn}} {{eqn | l = \norm {n^N} \le C \paren {n^N}^\alpha       | o = \leadsto       | r = \norm n^N \le C \paren {n^N}^\alpha       | c = {{NormAxiom|2}} }} {{eqn | o = \leadsto       | r = \norm n^N \le C \paren {n^\alpha}^N       | c =  }} {{eqn | o = \leadsto       | r = \norm n \le \sqrt [N] C n^\alpha       | c = taking $N$th roots }} {{end-eqn}} By Limit of Root of Positive Real Number: :$\sqrt [N] C \to 1$ as $N \to \infty$ By the Multiple Rule for Real Sequences: :$\sqrt [N] C n^\alpha \to n^\alpha$ as $N \to \infty$ By Inequality Rule for Real Sequences, letting $N \to \infty$ for fixed $n$: :$\norm n \le n^\alpha$ The result follows. {{qed}}
By the definition of $\alpha$: :$\norm {n_0} = n_0^\alpha$ By the definition of $n_0$: :$n_0^\alpha > 1$ Let $n \in \N$. By Basis Representation Theorem then $n$ can be written: :$n = a_0 + a_1 n_0 + a_2 n_0^2 + \cdots + a_s n_0^s$ where $0 \le a_i < n_0$ and $a_s \ne 0$ By Bounds for Integer Expressed in Base k: :$n_0^{s + 1} > n \ge n_0^s$ By Lemma 1.1: :$\norm {n_0^{s + 1} - n} \le \paren {n_0^{s + 1} - n}^\alpha$ Hence: {{begin-eqn}} {{eqn | l = \norm n       | o = \ge       | r = \norm {n_0^{s + 1} } - \norm {n_0^{s + 1} - n}       | c = Reverse Triangle Inequality }} {{eqn | r = \norm {n_0}^{s + 1} - \norm {n_0^{s + 1} - n}       | c = {{Norm Axiom|2}} }} {{eqn | r = n_0^{\alpha \paren {s + 1} } - \norm {n_0^{s + 1} - n}       | c = as $\norm {n_0} = n_0^\alpha$ }} {{eqn | o = \ge       | r = n_0^{\alpha \paren {s + 1} } - \paren {n_0^{s + 1} - n}^\alpha       | c = as $\norm {n_0^{s + 1} - n} \le \paren {n_0^{s + 1} - n}^\alpha$ }} {{eqn | o = \ge       | r = n_0^{\alpha \paren {s + 1} } - \paren {n_0^{s + 1} - n_0^s}^\alpha       | c = as $n \ge n_0^s$ }} {{eqn | r = n_0^{\alpha \paren {s + 1} } \paren {1 - \paren {1 - \frac 1 {n_0} }^\alpha}       | c =  }} {{eqn | o = \ge       | r = n^\alpha \paren {1 - \paren {1 - \frac 1 {n_0} }^\alpha}       | c = as $n_0^{s+1}>n$ }} {{end-eqn}} Let $C' = \paren{1 - \paren{1 - \frac 1 {n_0} }^\alpha}$. Then: :$\norm n \ge C' n^\alpha$ As $n \in \N$ was arbitrary: :$\forall n \in N: \norm n \ge C' n^\alpha$ Let $n, N \in N$. Then: :$\norm {n^N} \ge C' \paren {n^N}^\alpha$ Now: {{begin-eqn}} {{eqn | l = \norm {n^N} \ge C' \paren {n^N}^\alpha       | o = \leadsto       | r = \norm n^N \ge C' \paren {n^N}^\alpha       | c = {{NormAxiom|2}} }} {{eqn | o = \leadsto       | r = \norm n^N \ge C' \paren {n^\alpha}^N       | c =  }} {{eqn | o = \leadsto       | r = \norm n \ge \sqrt [N] {C'} n^\alpha       | c = taking $N$th roots }} {{end-eqn}} By Limit of Root of Positive Real Number: :$\sqrt [N] {C'} \to 1$ as $N \to \infty$ By the Multiple Rule for Real Sequences: :$\sqrt [N] {C'} n^\alpha \to n^\alpha$ as $N \to \infty$ By Inequality Rule for Real Sequences, letting $N \to \infty$ for fixed $n$: :$\norm n \ge n^\alpha$ The result follows. {{qed}}
=== Necessary Condition === {{:Equivalent Norms on Rational Numbers/Necessary Condition}}{{qed|lemma}} === Sufficient Condition === {{:Equivalent Norms on Rational Numbers/Sufficient Condition}}{{qed}}
Because $\norm {\, \cdot \,}$ is non-trivial: :$\exists \dfrac a b \in \Q : 0 < \norm {\dfrac a b} \mbox { and } \norm {\dfrac a b} \ne 1$ By Norm of Inverse then: :$\norm {\dfrac a b} > 1 \implies \norm {\dfrac b a} < 1$ Hence either $\norm {\dfrac a b} < 1$ or $\norm {\dfrac b a} < 1$. {{WLOG}} assume $\norm {\dfrac a b} < 1$. By Norm of Quotient then: :$\dfrac {\norm a} {\norm b} < 1$ Hence: :${\norm a} < \norm b$ Let $n = \size a$ and $m = \size b$ where $\size {\,\cdot\,}$ is the absolute value on $\Q$. Then $n, m \in \N$ By Norm of Negative then: :$\norm n = \norm a$ :$\norm m = \norm b$ Hence: :$\norm n < \norm m$ Since $\norm {\, \cdot \,}$ is non-Archimedean then: :$\norm m \le 1$ Hence: :$\norm n < \norm m \le 1$ {{qed}}
Let $z_1, z_2, \dotsc, z_5$ be the complex $5$th roots of unity. From Sum of Powers of Primitive Complex Roots of Unity, setting $s = 1$: :$z_1 + z_2 + z_3 + z_4 + z_5 = 0$ Thus from Complex 5th Roots of Unity: :$1 + \cis 72 \degrees + \cis 144 \degrees + \cis 216 \degrees + \cis 288 \degrees = 0$ Equating the real parts: :$1 + \cos 72 \degrees + \cos 144 \degrees + \cos 216 \degrees + \cos 288 \degrees = 0$ :500px {{qed}}
{{AimForCont}} $n_0$ is a composite number.  Let $n_1, n_2 \in \N$ such that $n_1, n_2 < n_0$ and $n_0 = n_1 n_2$.  By the definition of $n_0$ then: :$\norm {n_1} = 1$  :$\norm {n_2} = 1$  By {{NormAxiom|2}}: :$\norm {n_0} = \norm {n_1 n_2} = \norm {n_1} \norm {n_2} = 1$ This contradicts the assumption that $\norm {n_0} < 1$.  So $n_0$ must be a prime number. {{qed}}
We have: {{begin-eqn}} {{eqn | l = 144 \degrees       | r = 180 \degrees - 36 \degrees       | c =  }} {{eqn | ll= \leadsto       | l = \cos 36 \degrees       | r = -\cos 144 \degrees       | c = Cosine of Supplementary Angle }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = 108 \degrees       | r = 180 \degrees - 72 \degrees       | c =  }} {{eqn | ll= \leadsto       | l = \cos 72 \degrees       | r = -\cos 108 \degrees       | c = Cosine of Supplementary Angle }} {{end-eqn}} Thus: :$\cos 36 \degrees + \cos 72 \degrees + \cos 108 \degrees + \cos 144 \degrees = 0$ :500px {{qed}}
The elements of $U_n = \set {z \in \C: z^n = 1}$ are the solutions to the equation: :$z^n - 1 = 0$ Thus by definition the coefficients of the powers of $z$: :$z^2, z^3, \ldots, z^{n - 1}$ are all zero. The result follows directly from Viète's Formulas. {{qed}}
From Dot Product Operator is Commutative: :$z_1 \circ z_2 = z_2 \circ z_1$ The result follows trivially. {{qed}}
{{begin-eqn}} {{eqn | l = \size {z_2 \times z_1}       | r = \size {-z_1 \times z_2}       | c = Complex Cross Product is Anticommutative }} {{eqn | r = \size {z_1 \times z_2}       | c = {{Defof|Absolute Value}} }} {{end-eqn}} Hence the result. {{qed}}
$\Box ABCD$ can be divided into $2$ triangles: $\triangle ABC$ and $\triangle ADC$. Hence $\mathcal A$ is the sum of the areas of $\triangle ABC$ and $\triangle ADC$. From Area of Triangle in Determinant Form: {{begin-eqn}} {{eqn | l = \map \Area {\triangle ABC}       | r = \dfrac 1 2 \size {\paren {\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \\ \end{vmatrix} } } }} {{eqn | l = \map \Area {\triangle ADC}       | r = \dfrac 1 2 \size {\paren {\begin{vmatrix} x_1 & y_1 & 1 \\ x_4 & y_4 & 1 \\ x_3 & y_3 & 1 \\ \end{vmatrix} } } }} {{end-eqn}} Hence the result. {{qed}}
By Bézout's Identity then: :$\exists n, m \in \Z : m a + n b = 1$   By Norm of Unity then: :$\norm {m a + n b} = 1$ By Corollary 5 of Characterisation of Non-Archimedean Division Ring Norms then: :$\norm a, \norm b, \norm n, \norm m \le 1$ Let $\norm a \lt 1$. By Norm axiom $(\text N 2)$: Multiplicativity: :$\norm {m a} = \norm m \norm a \lt 1$ Hence: :$\norm {m a} < \norm {m a + n b}$ By Corollary 4: :$\norm {n b} = \norm {m a + n b} = 1$ By Norm axiom $(\text N 2)$: Multiplicativity: :$\norm n \norm b = 1$. Hence $\norm b = 1$. The result follows. {{qed}}