hcju/proofwikiqa huggingface dataset:
Let $0_R$ be the zero of $R$, then: :$\sequence {x_n}$ converges to $0_R$ in $\norm {\, \cdot \,}_1 \iff \sequence {x_n}$ converges to $0_R$ in $\norm {\, \cdot \,}_2$ Hence: :$\sequence {x_n}$ is a null sequence in $\norm{\,\cdot\,}_1  \iff \sequence {x_n}$ is a null sequence in $\norm{\,\cdot\,}_2$  {{qed}}
Let $x \in R$. Let $\sequence {x_n}$ be the sequence defined by: $\forall n: x_n = x^n$. {{begin-eqn}} {{eqn | l = \norm x_1 < 1 \quad       | o = \leadstoandfrom       | c = $\sequence {x_n}$ is a null sequence in $\norm {\, \cdot \,}_1$       | cc= Sequence of Powers of Number less than One in Normed Division Ring }} {{eqn | o = \leadstoandfrom       | c = $\sequence {x_n}$ is a null sequence in $\norm {\, \cdot \,}_2$       | cc= Assumption }} {{eqn | o = \leadstoandfrom       | c = $\norm x_2 < 1$       | cc= Sequence of Powers of Number less than One in Normed Division Ring }} {{end-eqn}} {{qed}}
==== Case 1 ==== For all $x \in R: x \ne 0_R$, let $x$ satisfy $\norm x_1 \ge 1$. ===== Lemma 1 ===== {{:Equivalence of Definitions of Equivalent Division Ring Norms/Open Unit Ball Equivalent implies Norm is Power of Other Norm/Lemma 1}} By assumption, for all $x \in R, x \ne 0_R$, then $\norm x_2 \ge 1$. Similarly $\norm {\, \cdot \,}_2$ is the trivial norm. Hence $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ are equal. For $\alpha = 1$ the result follows. {{qed|lemma}} ==== Case 2 ==== Let $x_0 \in R$ such that $x_0 \ne 0_R$ and $\norm {x_0}_1 < 1$. By assumption then $\norm {x_0}_2 < 1$. Let $\alpha = \dfrac {\log \norm {x_0}_1 } {\log \norm {x_0}_2 }$. Then $\norm {x_0}_1 = \norm {x_0}_2^\alpha$. As $\norm {x_0}_1, \norm {x_0}_2 < 1$: :$\log \norm {x_0}_1 < 0$ :$\log \norm {x_0}_2 < 0$ So $\alpha > 0$. ====== Lemma 2 ====== {{:Equivalence of Definitions of Equivalent Division Ring Norms/Open Unit Ball Equivalent implies Norm is Power of Other Norm/Lemma 2}}{{qed}}
From Intersection of Subgroups is Subgroup, $P \cap Q$ is a subgroup of $P$. Thus: :$\order {P \cap Q} = p^c$ for some $c \in \Z$ such that $0 \le c \le a$ where $\order {P \cap Q}$ denotes the order of $P \cap Q$. We have: {{begin-eqn}} {{eqn | l = \order {P Q}       | r = \frac {\order P \order Q} {\order {P \cap Q} }       | c = Order of Subgroup Product }} {{eqn | r = \frac {p^a p^b} {p^c}       | c =  }} {{eqn | r = p^{a + b - c}       | c =  }} {{end-eqn}} {{AimForCont}} $P Q$ is a subgroup of $G$. By Lagrange's Theorem (Group Theory): :$\order {P Q} \divides \order G$ where $\divides$ denotes divisibility. We have that $\order {P Q}$ is a power of $p$. The highest power of $p$ which divides $\order G$ is $p^a$. Thus $P Q$ could have order $p^a$ at most. Thus: :$a + b - c \le a$ That is: :$b \le c$ But $P \cap Q$ is a subgroup of $P$. Hence it must be the case that $b = c$. Thus: :$P \cap Q = Q$ and so $Q$ is a subgroup of $P$. This contradicts the fact that $Q$ is not a subgroup of $P$. It must follow that $P Q$ is not a subgroup of $G$. {{qed}}
By hypothesis there is a basis $B$ of $E$ with $n$ elements. Then $H \cup B$ is a generator for $E$. So by Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set there exists a basis $C$ of $E$ such that $H \subseteq C \subseteq H \cup B$. {{Qed}}
Let $H = \set {\xi_1, \xi_2, \ldots, \xi r}$. Consider the basis $B = \set {\alpha_1, \alpha_2, \ldots, \alpha_n}$ of $E$. Consider the set $G = H \cup B = \set {\xi_1, \xi_2, \ldots, \xi r, \alpha_1, \alpha_2, \ldots, \alpha_n}$. We have that $G$ is a generator of $E$. As $B$ is a basis, it follows that each of $H$ is a linear combination of $B$. Thus $G = H \cup B$ is linearly dependent. Thus one of the elements $\alpha_i$ of $B$ is a linear combination of the preceding elements of $G$. Eliminate this one, and do the same thing with with $G \setminus \set {\alpha_i}$.  Eventually there will exist a set which is a basis of $E$ containing all the elements of $H$. {{Qed}}
By hypothesis, let $H$ be a linearly independent subset of $E$ === Necessary Condition === Let $H$ be a basis for $E$. By definition of dimension of vector space, a basis for $E$ contains exactly $n$ elements. By Bases of Finitely Generated Vector Space have Equal Cardinality, it follows that $H$ also contains exactly $n$ elements. {{qed|lemma}} === Sufficient Condition === Let $H$ contain exactly $n$ elements. By Sufficient Conditions for Basis of Finite Dimensional Vector Space $H$ is itself a basis for $E$. {{Qed}}
=== Necessary Condition === Let $G$ be a basis for $E$. From Cardinality of Basis of Vector Space, $\card G = n$. {{qed}} === Sufficient Condition === Let $\card G = n$. From Sufficient Conditions for Basis of Finite Dimensional Vector Space, $G$ is a basis for $E$. {{qed}}
Let $S$ denote an ISBN-$10$ whose $k$th digit is $d_k$. Let $d_S$ denote the check digit of $S$. Let $S'$ denote the ISBN-$10$ $S$ whose $n$th digit has been transmitted incorrectly, as $d'_n$. Let $d'_S$ denote the check digit calculated on $S'$ according to the algorithm via which calculated $d_S$ on $S$. It will be demonstrated that $d'_S \ne d_S$. By definition of ISBN-$10$: :$d_S = \displaystyle \sum_{k \mathop = 1}^9 k d_k \pmod {11}$ where $d_k$ is assigned the symbol $X$ if it calculates to $10$. Then: :$d'_S = \displaystyle \sum_{k \mathop = 1}^9 k d_k - n d_n + n d'_n \pmod {11}$ and so: :$d'_S - d_S = n \paren {d'_n - d_n} \pmod {11}$ It follows that: :$d'_S - d_S \ne 0$ and so the check digit on $d_S$ is the incorrect check digit for $S'$. {{qed}}
For each term of a sequence in $\map V {n, p}$ there are $p$ possible values. There are $n$ such terms. Hence there are $\underbrace {p \times p \times \cdots \times p}_{n \text { times} } = p^n$ different possible sequences in $\map V {n, p}$. {{qed}}
Recall the vector space axioms: {{:Definition:Vector Space Axioms}} First, the set of sequences $\tuple {x_1, x_2, \ldots, x_n}$, for $x_1, x_2, \ldots, x_n \in \Z_p$, has to be shown to fulfil the abelian group axioms. This follows from: :Integers Modulo m under Addition form Cyclic Group and: :Cyclic Group is Abelian. {{ProofWanted|needs to be finished off, but it's tedious and I lack the patience tonight.}}
From Master Code forms Vector Space, $\map V {n, p}$ is a vector space. By definition, $\tuple {n, k}$ is a subspace of $\map V {n, p}$. The result follows by the fact that a subspace is itself a vector space. {{finish|I lose patience with the fine detail.}}
It is to be demonstrated that $d$ satisfies all the metric space axioms. Let $u, v, w \in \map V {n, p}$ be arbitrary. === Proof of $(\text M 1)$ === By definition of distance: :$\map d {u, u} = 0$ So axiom $(\text M 1)$ holds for $d$. {{qed|lemma}} === Proof of $(\text M 2)$ === Consider $\map d {u, v} + \map d {v, w}$. Let $\map d {u, w} \ne 0$. Then at each term at which $u$ and $w$ are different, those corresponding terms in either $u$ and $v$ or $v$ and $w$ must be different. So every contribution to the value of $\map d {u, w}$ is present in either $\map d {u, v}$ or $\map d {v, w}$. It follows that $\map d {u, v} + \map d {v, w} \ge \map d {u, w}$. So axiom $(\text M 2)$ holds for $d$. {{qed|lemma}} === Proof of $(\text M 3)$ === $\map d {u, v} = \map d {v, u}$ by definition of distance. So axiom $(\text M 3)$ holds for $d$. {{qed|lemma}} === Proof of $(\text M 4)$ === {{begin-eqn}} {{eqn | l = u       | o = \ne       | r = v       | c =  }} {{eqn | ll= \leadsto       | l = \map d {u, v}       | o = >       | r = 0       | c = {{Defof|Distance between Linear Codewords}} }} {{end-eqn}} So axiom $(\text M 4)$ holds for $d$. {{qed|lemma}} Thus $d$ satisfies all the metric space axioms and so is a metric. {{qed}}
Let $f := \displaystyle \min_{u \mathop \in C} \map w u$. Let $\mathbf 0$ denote the codeword in $\map V {n, p}$ consisting of all zeroes. As $C$ is a subspace of $\map V {n, p}$, we have that $\mathbf 0 \in C$. Let $w$ be a codeword with weight $f$. Then: :$\map d {w, \mathbf 0} = f$ so $f \ge \map d C$. Let $u, v \in C$ such that $\map d {u, v} = \map d C$. We have that $C$ is a linear code. Therefore: :$u - v \in C$ where $u - v$ denotes the difference between $u$ and $v$. But $u - v$ has weight $\map d C$. Thus: :$\map d C \le f$ and it follows that $\map d C = f$. {{qed}}
Let $C$ be a linear code whose master code is $V$. Let $c \in C$ be a transmitted codeword. Let $v$ be the received word from $c$. By definition, $v$ is an element of $V$. Let $v$ have a distance $f$ from $c$, where $f \le d - 1$. Thus there have been $f$ transmission errors. As $d$ is the minimum distance it is clear that $v$ cannot be a codeword of $C$. Hence it can be understood that $C$ has detected that $v$ has as many as $d - 1$ transmission errors. {{Qed}}
Let $C$ be a linear code whose master code is $V$. Let $c \in C$ be a transmitted codeword. Let $v$ be the received word from $c$. By definition, $v$ is an element of $V$. Let $v$ have a distance $e$ from $c$, where $2 e + 1 \le d$. Thus there have been $e$ transmission errors. {{AimForCont}} $c_1$ is a codeword of $C$, distinct from $c$, such that $\map d {v, c_1} \le e$. Then: {{begin-eqn}} {{eqn | l = \map d {c, c_1}       | o = \le       | r = \map d {c, v} + \map d {v, c_1}       | c =  }} {{eqn | o = \le       | r = e + e       | c =  }} {{eqn | o = <       | r = d       | c =  }} {{end-eqn}} So $c_1$ has a distance from $c$ less than $d$. But $C$ has a minimum distance $d$. Thus $c_1$ cannot be a codeword of $C$. From this contradiction it follows that there is no codeword of $C$ closer to $v$ than $c$. Hence there is a unique codeword of $C$ which has the smallest distance from $v$. Hence it can be understood that $C$ has corrected the transmission errors of $v$. {{Qed}}
Let $C$ denote the Golay ternary code. By inspection of the standard generator matrix $G$ of $C$: :$G := \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 2 & 2 & 1 \\ 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 2 & 2 \\ 0 & 0 & 1 & 0 & 0 & 0 & 2 & 1 & 0 & 1 & 2 \\ 0 & 0 & 0 & 1 & 0 & 0 & 2 & 2 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 0 & 1 & 2 & 2 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1  \end{pmatrix}$ it is seen that the smallest weight of all the codewords of $C$ that can be found in $G$ is $5$. So it is immediately seen that the minimum distance of $C$ is at least $5$. It remains to be shown that the minimum distance of $C$ is no more than $5$. {{ProofWanted}}
We have that Golay Ternary Code has Minimum Distance 5. The result follows from Error Correction Capability of Linear Code. {{qed}}
We prove the contrapositive. Let $\norm {\, \cdot \,}_1$ be a nontrivial norm. Then: :$\exists y \in R: \norm y_1 \ne 0, \norm y_1 \ne 1$. By Real Numbers form Ordered Field either $\norm y_1 < 1$ or $\norm y_1 > 1$. Suppose $\norm y_1 > 1$. By Norm axiom $(\text N 1)$: Positive Definiteness: :$y \ne 0_R$ By Norm of Inverse in Division Ring: :$\norm {y^{-1} }_1 = \dfrac 1 {\norm y_1} < 1$ So either $\norm y_1 < 1$ or $\norm {y^{-1} }_1 < 1$ So: :$\exists x \in R, x \ne 0_R:\norm x_1 < 1$ The theorem now follows by the Rule of Transposition.  {{qed}} Category:Equivalence of Definitions of Equivalent Division Ring Norms m7xz77776pfypxq1omyk1sb2jg57gwo
Let $G = \paren {\begin{array} {c|c} \mathbf I & \mathbf A \end{array} }$. Let $c \in \map V {n, p}$. Then, by definition of $G$, $c$ is a codeword of $C$ {{iff}} $c$ is of the form $u G$, where $u \in \map V {k, p}$. Thus $c \in C$ {{iff}}: {{begin-eqn}} {{eqn | l = c       | r = u G       | c =  }} {{eqn | r = u \paren {\begin{array} {c {{!}} c} \mathbf I & \mathbf A \end{array} }       | c =  }} {{eqn | r = \paren {\begin{array} {c {{!}} c} u & v \end{array} }       | c =  }} {{end-eqn}} where: :$v = u \mathbf A$ :$\paren {\begin{array} {c|c} u & v \end{array} }$ denotes the $1 \times n$ matrix formed from the $k$ elements of $u$ and the $n - k$ elements of $v$. Let $w \in \map V {n, p}$. $w$ can be expressed in the form: :$w = \paren {\begin{array} {c|c} u_1 & v_1 \end{array} }$ where $u_1 \in \map V {k, p}$. The syndrome of $v$ is then calculated as: {{begin-eqn}} {{eqn | l = \map S v       | r = \paren {\begin{array} {c {{!}} c} -\mathbf A^\intercal & \mathbf I \end{array} } w^\intercal       | c =  }} {{eqn | r = \paren {\begin{array} {c {{!}} c} -\mathbf A^\intercal & \mathbf I \end{array} } \paren {\begin{array} {c {{!}} c} u_1^\intercal & v_1^\intercal \end{array} }       | c =  }} {{eqn | r = -\mathbf A^\intercal u_1^\intercal + v_1^\intercal       | c =  }} {{end-eqn}} It follows that the syndrome of $w$ is zero {{iff}} $w$ is the concatenation of $u_1$ and $v_1$, where: :$v_1^\intercal = \mathbf A^\intercal u_1^\intercal = \paren {u_1 \mathbf A}^\intercal$ Thus the syndrome of $w$ is zero {{iff}} $w$ is a codeword of $C$. {{qed}}
Let $u, v \in \map V {n, p}$. Let $\map S u$ denote the syndrome of $u$. Then: {{begin-eqn}} {{eqn | l = \map S u       | r = \map S v       | c =  }} {{eqn | ll= \leadstoandfrom       | l = P u^\intercal       | r = P v^\intercal       | c = {{Defof|Syndrome}} }} {{eqn | ll= \leadstoandfrom       | l = P \paren {u^\intercal - v^\intercal}       | r = \mathbf 0       | c =  }} {{eqn | ll= \leadstoandfrom       | l = P \paren {u - v}^\intercal       | r = \mathbf 0       | c =  }} {{eqn | ll= \leadstoandfrom       | l = u - v       | o = \in       | r = C       | c = Syndrome is Zero iff Vector is Codeword }} {{end-eqn}} Hence the result from Elements in Same Coset iff Product with Inverse in Subgroup. {{qed}}
Assume that: :$\ds J \sqbrk y = \int_a^b \map F {y, y'} \rd x$ Then: {{begin-eqn}} {{eqn| l = \delta J      | r = 0 }} {{eqn| ll= \leadsto      | l = F_y - \dfrac \d {\d x} F_{y'}      | r = F_y - \paren {\dfrac {\d y} {\d x} \dfrac {\partial F_{y'} } {\partial y} + \dfrac {\d y'} {\d x} \dfrac {\partial F_{y'} } {\partial y'} } }} {{eqn| r = F_y - y' F_{y'y} - y'' F_{y'y'} }} {{end-eqn}} Multiply this differential equation by $y'$. This gives: {{begin-eqn}} {{eqn | l = F_y y' - {y'}^2 F_{y'y} - y'y'' F_{y'y'}       | r = \paren {F_{y} y' + F_{y'} y''} - F_{y'} y'' - y' \paren {y' F_{y'y} + y'' F_{y'y'} } }} {{eqn | r = \dfrac {\d F} {\d x} - y' \paren {\dfrac {\d y} {\d x} \dfrac {\partial F_{y'} } {\partial y} + \dfrac {\d y'} {\d x} \dfrac {\partial F_{y'} } {\partial y'} } - F_{y'} y'' }} {{eqn | r = \dfrac {\d F} {\d x} - y' \dfrac {\d F} {\d x} - \dfrac {\d y'} {\d x} F_{y'} }} {{eqn | r = \dfrac \d {\d x} \paren {F - y' F_{y'} } }} {{eqn | r = 0 }} {{end-eqn}} Integration yields the desired result. {{qed}}
Assume that: :$\ds J \sqbrk y = \int_a^b \map F {x,y'} \rd x$ Euler's equation for $J$ is: :$\dfrac \d {\d x} F_{y'} = 0$ Integration yields: :$F_{y'} = C$  {{qed}}
Assume that: :$\ds J \sqbrk y = \int_a^b \map F {x,y} \rd x$ Then Euler's Equation for $J$ is: :$F_y = 0$ Since $F$ is independent of $y'$, the equation is algebraic. {{qed}}
Substitution of $\rd s$ into $J$ results in the following functional: :$\ds J \sqbrk y = \int_a^b \map f {x, y, y'} \sqrt {1 + y'^2} \rd x$ We can consider this as a functional with the following effective $F$: :$F = \map f {x, y, y'} \sqrt {1 + y'^2}$ Find Euler's Equation: {{begin-eqn}} {{eqn | l = F_y - \dfrac \d {\d x} F_{y'}       | r = f_y \sqrt {1 + y'^2} - \dfrac \d {\d x} \paren {\map f {x, y, y'} \frac {y'} {\sqrt {1 + y'^2} } } }} {{eqn | r = f_y \sqrt {1 + y'^2} - \dfrac \d {\d x} \map f {x, y, y'} \frac {y'} {\sqrt {1 + y'^2} } - \map f {x, y, y'} \dfrac \d {\d x} \frac {y'} {\sqrt {1 + y'^2} } }} {{eqn | r = f_y \sqrt {1 + y'^2} - f_x \frac {y'} {\sqrt {1 + y'^2} } - f_y \frac {y'^2} {\sqrt {1 + y'^2} } - f_{y'} \frac {y' y''} {\sqrt {1 + y'^2} } - f \frac {y'' \sqrt {1 + y'^2} - \frac {y'^2 y''} {\sqrt {1 + y'^2} } } {1 + y'^2} }} {{eqn | r = \frac 1 {\sqrt {1 + y'^2} } \paren {f_y - f_x y' - f_{y'} y' y'' - f \frac {y''} {\paren {1 + y'^2}^{\frac 3 2} } } }} {{end-eqn}} Due to assumptions on $y$, the prefactor does not vanish. By Euler's Equation, the last expression vanishes. {{qed}}
Note that the zero codeword is in $C^+$ as it has a weight of $0$ which is even. Let $c$ and $d$ be of even weight, where $c$ and $d$ agree in $k$ ordinates. Let $\map w c$ denote the weight of $c$. Then: {{begin-eqn}} {{eqn | l = \map w {c + d}       | r = \map w c - k + \map w d - k       | c =  }} {{eqn | r = \map w c + \map w d - 2 k       | c =  }} {{end-eqn}} which is even. Since the negative of a vector $\mathbf v$ in $\Z_2$ equals $\mathbf v$, it follows that the inverse of $c \in C$ is also in $C$. It follows from the Two-Step Subgroup Test that $C^+$ is a subgroup of $C$. Let $C \ne C^+$. Then $C$ contains a codeword $c$ of odd weight. Let $C^-$ denote the subset of $C$ consisting of all the codewords of $C$ which have odd weight. Adding $c$ to each codeword of $C^+$ gives distinct codewords of odd weight, so: :$\order {C^-} \ge \order {C^+}$ Similarly, adding $c$ to each codeword of $C^-$ gives distinct codewords of even weight, so: :$\order {C^-} \le \order {C^+}$ As $C = C^+ \cup C^-$ it follows that: :$\order C = 2 \order {C^+}$ Hence the result. {{qed}}
The area functional of the surface of revolution is: :$\ds A \sqbrk y = 2 \pi \int_{x_0}^{x_1} y \sqrt {1 + y'^2} \rd x$ The integrand does not depend on $x$. By Euler's Equation: :$F - y' F_{y'} = C$ that is: :$y \sqrt {1 + y'^2} - \dfrac {y y'^2} {\sqrt {1 + y'^2} } = C$ which is equivalent to: :$y = C \sqrt {1 + y'^2}$ Solving for $y'$ yields:   :$y' = \sqrt {\dfrac {y^2 - C^2} {C^2} }$ Consider this as a differential equation for $\map x y$: :$\dfrac {\d x} {\d y} = \dfrac C {\sqrt {y^2 - C^2} }$ Integrate this once: :$x + C_1 = C \ln {\dfrac {y + \sqrt {y^2 - C^2} } C}$ Solving for $y$ results in: :$y = C \cosh {\dfrac {x + C_1} C}$ To find the area, substitute this function into the formula for the area: {{begin-eqn}} {{eqn| l = A \sqbrk y      | r = 2 \pi \int_{x_0}^{x_1} y \sqrt {1 + y'^2} \rd x }} {{eqn| r = 2 \pi \int_{x_0}^{x_1} C \, \map {\cosh^2} {\frac {x + C_1} C} \rd x }} {{eqn| r = \paren {x_1 - x_0} C \pi + \frac {\pi C^2} 2 \paren {\map \sinh {\frac {2 \paren {x_1 + C_1} } C} - \map \sinh {\frac {2 \paren {x_0 + C_1} } C} } }} {{end-eqn}} {{Finish|Connect this to the fact that not always this minimizes the area}}
Since $\norm {x_0}_1 < 1$ then $\norm {x_0}_2 < 1$ and: :$\log \norm {x_0}_1 < 0$ :$\log \norm {x_0}_2 < 0$ Hence $\alpha > 0$ Since $\forall x \in R: \norm x_1 < 1 \iff \norm x_2 < 1$:  ===== Lemma 1 ===== {{:Equivalence of Definitions of Equivalent Division Ring Norms/Open Unit Ball Equivalent implies Norm is Power of Other Norm/Lemma 2/Lemma 2.1}} ===== Lemma 2 ===== {{:Equivalence of Definitions of Equivalent Division Ring Norms/Open Unit Ball Equivalent implies Norm is Power of Other Norm/Lemma 2/Lemma 2.2}} Let $x \in R, x \ne 0_R$. ===== Case 1 ===== Let $\norm x_1 = 1$. By Lemma 2 then: :$\norm x_2 = 1$ Hence: :$\norm x_1 = 1 = 1^\alpha = \norm x_2^\alpha$ {{qed|lemma}}  ===== Case 2 ===== Let $\norm x_1 \ne 1$. By Lemma 2 then: :$\norm x_2 \ne 1$ Let $\beta = \dfrac {\log \norm x_1 } {\log \norm x_2 }$. Then $\norm x_1 = \norm x_2^\beta$. ====== Lemma 3 ====== {{:Equivalence of Definitions of Equivalent Division Ring Norms/Open Unit Ball Equivalent implies Norm is Power of Other Norm/Lemma 2/Lemma 2.3}} Hence $\norm x_1 = \norm x_2^\alpha$. </onlyinclude> {{qed}} Category:Equivalence of Definitions of Equivalent Division Ring Norms d424573c8eu49vw4soo5v51psipeox4
For $y \in R$ then: {{begin-eqn}} {{eqn | l = \norm y_1 > 1       | r = \dfrac 1 {\norm y_1 } < 1       | o = \leadstoandfrom       | c =  }} {{eqn | r = \norm {y^{-1} }_1 < 1        | o = \leadstoandfrom       | c = Norm of Inverse in Division Ring }} {{eqn | r = \norm {y^{-1} }_2 < 1        | o = \leadstoandfrom       | c = by assumption }} {{eqn | r = \dfrac 1 {\norm y_2 } < 1       | o = \leadstoandfrom       | c = Norm of Inverse in Division Ring }} {{eqn | r = \norm y_2 > 1       | o = \leadstoandfrom       | c =  }} {{end-eqn}} {{qed}} Category:Equivalence of Definitions of Equivalent Division Ring Norms 7c3kw7jre137tryfpo9a96n979a9bxq
By assumption: :$\forall y \in R:\norm y_1 \ge 1 \iff \norm y_2 \ge 1$ By Lemma 1: :$\forall y \in R:\norm y_1 \le 1 \iff \norm y_2 \le 1$ Hence $\forall y \in R$: {{begin-eqn}} {{eqn | l = \norm y_1 = 1        | o = \leadstoandfrom       | r = \norm y_1 \le 1, \norm y_1 \ge 1 }} {{eqn | o = \leadstoandfrom       | r = \norm y_2 \le 1, \norm y_2 \ge 1       | c = Lemma 1 and by assumption }} {{eqn | o = \leadstoandfrom       | r = \norm y_2 = 1 }} {{end-eqn}}  {{qed}} Category:Equivalence of Definitions of Equivalent Division Ring Norms i3p50dq69v4rnr033jf1x68v04xmtjz
Consider $S_3$ as presented by its Cayley table: {{:Symmetric Group on 3 Letters/Cayley Table}} Consider $D_3$ as presented by its group presentation: {{:Group Presentation of Dihedral Group D3}} and its Cayley table: {{:Dihedral Group D3/Cayley Table}} Let $\phi: S_3 \to D_3$ be specified as: {{begin-eqn}} {{eqn | l = \map \phi {1 2 3}       | r = a       | c =  }} {{eqn | l = \map \phi {2 3}       | r = b       | c =  }} {{end-eqn}} Then by inspection, we see: {{begin-eqn}} {{eqn | l = \map \phi {1 3 2}       | r = a^2       | c =  }} {{eqn | l = \map \phi {1 3}       | r = a b       | c =  }} {{eqn | l = \map \phi {1 2}       | r = a^2 b       | c =  }} {{end-eqn}} and the result follows. {{qed}}
It is noted that :$\paren {\omega^2}^2 = \omega$ and so $\phi$ is the square function. By Roots of Unity under Multiplication form Cyclic Group and Cyclic Group is Abelian, $U_3$ is abelian. Thus for all $a, b \in U_3$: {{begin-eqn}} {{eqn | l = \map \phi a \map \phi b       | r = a^2 b^2 }} {{eqn | r = a b a b       | c = {{Defof|Abelian Group}} }} {{eqn | r = \paren {a b}^2 }} {{eqn | r = \map \phi {a b} }} {{end-eqn}} showing that $\phi$ is a group homomorphism. {{qed}}
Because $x, y \in R \setminus 0_R$: :$\norm x_1 , \norm y_1, \norm x_2 , \norm y_2 > 0$. Because $\norm{x}_1 , \norm {y}_1 \ne 1$, by Lemma 2: :$\norm x_2 , \norm y_2 \ne 1$. Hence: :$\log \norm x_1 , \log \norm y_1, \log \norm x_2, \log \norm y_2 \ne 0$  and $\alpha, \beta$ are well-defined. Let $r = \dfrac n m \in \Q$ be any rational number where $n, m \in \Z$ are integers and $m \ne 0$. Then: {{begin-eqn}} {{eqn | l = \norm y_1^n < \norm x_1^m       | o = \leadstoandfrom       | r = \dfrac {\norm y_1^n} {\norm x_1^m} < 1 }} {{eqn | o = \leadstoandfrom       | r = \norm y_1^n \norm {x^{-1} }_1^m       | c = Norm of Inverse in Division Ring }} {{eqn | o = \leadstoandfrom       | r = \norm {y^n}_1 \norm {\paren {x^{-1} }^m}_1 < 1        | c = Norm Axiom $(\text N 2)$: Multiplicativity }} {{eqn | o = \leadstoandfrom       | r = \norm {y^n \paren {x^{-1} }^m}_1 < 1        | c = Norm Axiom $(\text N 2)$: Multiplicativity }} {{eqn | o = \leadstoandfrom       | r = \norm {y^n \paren {x^{-1} }^m}_2 < 1        | c = By assumption }} {{eqn | o = \leadstoandfrom       | r = \norm {y^n}_2 \norm {\paren {x^{-1} }^m}_2 < 1        | c = Norm Axiom $(\text N 2)$: Multiplicativity }} {{eqn | o = \leadstoandfrom       | r = \norm y_2^n \norm {x^{-1} }_2^m       | c = Norm Axiom $(\text N 2)$: Multiplicativity }} {{eqn | o = \leadstoandfrom       | r = \dfrac {\norm y_2^n} {\norm x_2^m} < 1       | c = Norm of Inverse in Division Ring }} {{eqn | r = \norm y_2^n < \norm x_2^m       | o = \leadstoandfrom }} {{end-eqn}} By Logarithm is Strictly Increasing: :$\log \norm y_1^n < \log \norm x_1^m \iff \log \norm y_2^n < \log \norm x_2^m$ By Sum of Logarithms: :$n \log \norm y_1 < m \log \norm x_1 \iff n \log \norm y_2 < m \log \norm x_2$ Because $m, \log \norm x_1, \log \norm x_2 \ne 0$: :$r = \dfrac n m < \dfrac {\log \norm x_1} {\log \norm y_1} \iff r = \dfrac n m < \dfrac {\log \norm x_2} {\log \norm y_2}$ By Between two Real Numbers exists Rational Number: :$\dfrac {\log \norm x_1} {\log \norm y_1} = \dfrac {\log \norm x_2} {\log \norm y_2}$ Because $\log \norm x_2, \log \norm y_2 \ne 0$: :$\dfrac {\log \norm x_1} {\log \norm x_2} = \dfrac {\log \norm y_1} {\log \norm y_2}$ That is: :$\alpha = \beta$ {{qed}} Category:Equivalence of Definitions of Equivalent Division Ring Norms 0w9fh9fzzjpsmeh7kgmw6d8fc585lm2