hcju/proofwikiqa huggingface dataset:
Let $\BB$ be the set of open $\epsilon$-balls in $M$. Let $\BB_H$ be the set of open $\epsilon$-balls in $\struct {H, d_H}$. Let $U \in \tau_{d_H}$. By the definition of the topology induced by the metric $d_H$: :$\exists \AA_H \subseteq \BB_H: U = \bigcup \AA_H$ Let $\AA = \set {B': B' \in \BB, B' \cap H \in \AA_H}$. Let $V = \bigcup \set {B' : B' \in \AA}$. By the definition of the topology induced by the metric $d$: :$V \in \tau$ By the definition of the subspace metric: :$\forall B \in \BB_H: \exists B' \in \BB: B = B' \cap H$ Hence: {{begin-eqn}} {{eqn | l = U       | r = \bigcup \set {B' \cap H : B' \in \AA} }} {{eqn | r = \paren {\bigcup \set {B' : B' \in \AA} } \cap H       | c = Intersection Distributes over Union }} {{eqn|r= V \cap H }} {{end-eqn}} By the definition of the subspace topology induced by $\tau$ on $H$: :$U \in \tau_H$ Hence: :$\tau_{d_H} \subseteq \tau_H$. {{qed|lemma}} Let $U \in \tau_H$. By the definition of the subspace topology induced by $\tau$ on $H$: :$\exists V \in \tau : U = V \cap H$ By the definition of the topology induced by the metric $d$: :$\exists \AA \subseteq \BB: V = \bigcup \set {B' : B' \in \AA}$ Hence: {{begin-eqn}} {{eqn | l = U       | r = \paren {\bigcup \set {B' : B' \in \AA} } \cap H }} {{eqn | r = \bigcup \set {B' \cap H : B' \in \AA}       | c = Intersection Distributes over Union }} {{end-eqn}} By the definition of the subspace metric: :$\forall B' \in \BB: B' \cap H \in \BB_H$ By the Definition of the topology induced by the metric $d_H$: :$U \in \tau_{d_H}$ Hence: :$\tau_H \subseteq \tau_{d_H}$ By Equivalence of Definitions of Set Equality: :$\tau_H = \tau_{d_H}$ {{qed}} Category:Metric Spaces Category:Topology mmcdd43jmur3u3f18rwm2yx8tqqj6lg
Note that $H \cap K$ is a subgroup of $H$. From Index of Intersection of Subgroups, we have: :$\index G {H \cap K} \le \index G H \index G K$ Setting $G = H$, we have: :$\index H {H \cap K} \le \index H H \index H K$ {{finish|This does not get us where we want}} {{CircularStructure|This is Index in Subgroup, which is used to prove the main theorem}}
Let $\sequence {x_n}$ and $\sequence {y_n}$ be Cauchy sequences in $\mathcal C$. Then: {{begin-eqn}} {{eqn | l = \sequence {x_n} + \mathcal N = \sequence {y_n} + \mathcal N       | o = \leadstoandfrom       | r = \sequence {x_n} - \sequence {y_n} \in \mathcal N       | c = Cosets are Equal iff Product with Inverse in Subgroup }} {{eqn | r = \lim_{n \mathop \to \infty} x_n - y_n = 0_R       | o = \leadstoandfrom       | c = Definition of $\mathcal N$ }} {{eqn | r = \lim_{n \mathop \to \infty} \norm {x_n - y_n} = 0       | o = \leadstoandfrom       | c = {{Defof|Convergent Sequence in Normed Division Ring}} }} {{eqn | r = \lim_{n \mathop \to \infty} \map d {x_n - y_n} = 0       | o = \leadstoandfrom       | c = {{Defof|Metric Induced by Norm on Division Ring}} }} {{eqn | r = \sequence {x_n} \sim \sequence {y_n}       | o = \leadstoandfrom       | c = Definition of equivalence relation $\sim$ }} {{end-eqn}} Hence: :$\sequence {x_n}$ and $\sequence {y_n}$ belong to the same equivalence class in $\mathcal C \,\big / \mathcal N$ {{iff}} $\sequence {x_n}$ and $\sequence {y_n}$ belong to the same equivalence class in $\tilde {\mathcal C}$. The result follows. {{qed}}
By Lemma 1 of Quotient of Cauchy Sequences is Metric Completion we have that: :$\mathcal C \,\big / \mathcal N = \tilde {\mathcal C}$ Let $\eqclass {x_n} {}$ and $\eqclass {x_n} {}$ be equivalence classes in $\mathcal C \,\big / \mathcal N = \tilde {\mathcal C}$. Then: {{begin-eqn}} {{eqn | l = \map {d'} {\eqclass {x_n}{}, \eqclass {x_n}{} }       | r = \norm {\eqclass {x_n}{} - \eqclass {x_n}{} }       | c = Definition of $d'$ }} {{eqn | r = \norm {\eqclass {x_n - y_n}{} }       | c = {{Defof|Quotient Ring}} Addition }} {{eqn | r = \lim_{n \mathop \to \infty} \norm {x_n - y_n}       | c = {{Defof|Norm on Division Ring}}: $\norm {\,\cdot\,}$ on $\mathcal C \,\big / \mathcal N$ }} {{eqn | r = \lim_{n \mathop \to \infty} \map d {x_n, y_n}       | c = {{Defof|Metric Induced by Norm on Division Ring}}: by the norm $\norm {\,\cdot\,}$ on $R$ }} {{eqn | r = \map {\tilde d} {\eqclass {x_n}{}, \eqclass {x_n}{} }       | c = Definition of $\tilde d$ }} {{end-eqn}} The result follows. {{qed}}
Note that: :$\mathbf I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ is the identity for (conventional) matrix multiplication of order $2$. We have: {{begin-eqn}} {{eqn | l = \mathbf a^2       | r = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}       | c =  }} {{eqn | r = \begin{bmatrix} 0 \times 0 + 1 \times \paren {-1} & 0 \times 1 + 1 \times 0 \\ \paren {-1} \times 0 + 0 \times \paren {-1} & \paren {-1} \times 1 + 0 \times 0 \end{bmatrix}       | c = {{Defof|Matrix Product (Conventional)}} }} {{eqn | r = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}       | c =  }} {{end-eqn}} and so: {{begin-eqn}} {{eqn | l = \mathbf a^4       | r = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}       | c =  }} {{eqn | r = \begin{bmatrix} \paren {-1} \times \paren {-1} + 0 \times 0 & \paren {-1} \times 0 + 0 \times \paren {-1} \\ 0 \times \paren {-1} + \paren {-1} \times 0 & 0 \times 0 + \paren {-1} \times \paren {-1} \end{bmatrix}       | c = {{Defof|Matrix Product (Conventional)}} }} {{eqn | r = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}       | c =  }} {{eqn | r = \mathbf I       | c =  }} {{end-eqn}} Next we have: {{begin-eqn}} {{eqn | l = \mathbf b^2       | r = \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix} \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix}       | c =  }} {{eqn | r = \begin{bmatrix} 0 \times 0 + i \times i & 0 \times i + i \times 0 \\ \paren i \times 0 + 0 \times i & i \times i + 0 \times 0 \end{bmatrix}       | c = {{Defof|Matrix Product (Conventional)}} }} {{eqn | r = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}       | c =  }} {{end-eqn}} and so: :$\mathbf b^2 = \mathbf a^2$ Then we have: {{begin-eqn}} {{eqn | l = \mathbf a \mathbf b \mathbf a       | r = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}       | c =  }} {{eqn | r = \begin{bmatrix} \paren {-1} \times 0 + 0 \times i & \paren {-1} \times i + 0 \times 0 \\ \paren 0 \times 0 + \paren {-1} \times i & 0 \times i + \paren {-1} \times 0 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}       | c = {{Defof|Matrix Product (Conventional)}} }} {{eqn | r = \begin{bmatrix} 0 & -i \\ -i & 0 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}       | c =  }} {{eqn | r = \begin{bmatrix} 0 \times \paren {-1} + \paren {-i} \times 0 & 0 \times 0 + \paren {-i} \times \paren {-1} \\ \paren {-i} \times \paren {-1} + 0 \times 0 & \paren {-i} \times 0 + 0 \times \paren {-1} \end{bmatrix}       | c = {{Defof|Matrix Product (Conventional)}} }} {{eqn | r = \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix}       | c =  }} {{eqn | r = \mathbf b       | c =  }} {{end-eqn}} Thus $\gen {\mathbf a = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}, \mathbf b = \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix} }$ fulfils the conditions for the group presentation of $\Dic 2$: {{:Quaternion Group/Group Presentation}} Hence the result. {{qed}} Category:Quaternion Group pt2e4tktnyza63mtq1yg1ju5syuu1r7
We have that $15 = 3 \times 5$. Thus: :$15$ is square-free :$5 \equiv 2 \pmod 3$ :$3 \equiv 3 \pmod 5$ The conditions are fulfilled for Condition for Nu Function to be 1. Thus $\map \nu {15} = 1$ and so all groups of order $15$ are cyclic. {{Qed}}
From Number of Sylow p-Subgroups in Group of Order 15: :the number of Sylow $3$-subgroups is in the set $\set {1, 4, 7, \ldots}$ :the number of Sylow $5$-subgroups is in the set $\set {1, 6, 11, \ldots}$. From the Fifth Sylow Theorem :the number of Sylow $3$-subgroups is a divisor of $15$ :the number of Sylow $5$-subgroups is a divisor of $15$. Combining the above: :there is a unique Sylow $3$-subgroup in $G$, which we will denote $P$ :there is a unique Sylow $5$-subgroup in $G$, which we will denote $Q$. Let $x \in G$ be of order $3$. Then by Non-Trivial Group has Non-Trivial Cyclic Subgroup, $\gen x$ has $3$ elements. Thus $\gen x$ is a Sylow $3$-subgroup and so: :$\gen x = P$ and so $x$ is one of the $2$ non-identity elements of $P$. Similarly, let $y \in G$ be of order $5$. Then by a similar argument: :$\gen y = Q$ and so $y$ is one of the $4$ non-identity elements of $Q$. So $G$ has: :$1$ element of order $1$ (that is, the identity) :$2$ elements of order $3$ :$4$ elements of order $5$ which leaves $8$ elements whose order is still to be determined. From Order of Element Divides Order of Finite Group, they are all of order $1$, $3$, $5$ or $15$. As the elements of order $1$, $3$ and $5$ have been accounted for, they must all be of order $15$. So $G$ has $8$ distinct elements of order $15$. Hence $G$ must be cyclic. {{Qed}}
{{AimForCont}} $G$ is non-abelian. Let $n_3$ denote the number of elements of $G$ of order $3$. From Number of Elements of Order p in Group of Order pq is Multiple of q, $n_3$ is a multiple of $5$. From Number of Order p Elements in Group with m Order p Subgroups, $n_3$ is a multiple of $2$. Therefore $n_3$ is a multiple of $10$. Let $n_5$ denote the number of elements of $G$ of order $5$. From Number of Elements of Order p in Group of Order pq is Multiple of q, $n_5$ is a multiple of $3$. From Number of Order p Elements in Group with m Order p Subgroups, $n_5$ is a multiple of $4$. Therefore $n_5$ is a multiple of $12$. Together with the identity element which has order $1$, that makes $1 + 12 a + 10 b = 15$ for some positive integers $a$ and $b$. This is impossible. Hence by Proof by Contradiction it follows that $G$ must be abelian. Since $15$ is a product of $2$ distinct primes, by Abelian Group of Semiprime Order is Cyclic, $G$ is cyclic. {{qed}}
From Lagrange's Theorem: :$\dfrac {\order G} {\order N} = \index G N$ where $\index G N$ is the index of $N$ in $G$. By definition of index: :$\index G N = \order {G / N}$ {{qed}}
=== $(1)$ implies $(2)$ === Let $f: A \to A$ be an involutive mapping by definition 1. Then by definition: :$(1): \quad \forall x \in A: \map f {\map f x} = x$ Let $\map f x = y$. Then by substituting $y$ for $\map f x$ into $(1)$: :$\forall x \in A: \map f y = x$ Thus $f: A \to A$ is an involutive mapping by definition 2. {{qed|lemma}} === $(2)$ implies $(1)$ === Let $f: A \to A$ be an involutive mapping by definition 2. Then by definition: :$\forall x, y \in A: \map f x = y \implies \map f y = x$ and so substituting $\map f x$ for $y$ we have: :$\map f {\map f x} = x$ Thus $f: A \to A$ is an involutive mapping by definition 1. {{qed|lemma}} === $(1)$ iff $(3)$ === This is demonstrated in Mapping is Involution iff Bijective and Symmetric. {{qed}} Category:Involutions b0vpa1nflm5aor9yhynpvugkyrma69i
Let $G$ be a non-abelian finite simple group. Let $t \in G$ which is not the identity. By definition of a simple group and Center of Group is Normal Subgroup: :either $\map Z G = G$ or $\map Z G$ is the trivial group. By definition of an abelian group: :$\map Z G = G$ {{iff}} $G$ is abelian Hence we must have $\map Z G$ is the trivial group. Thus $t \notin \map Z G$. From definition of center: :$\exists x \in G: t x \ne x t$ For this $x$, $x \notin \map {C_G} t$. Hence $\map {C_G} t \ne G$. {{Questionable|So trivial it's scary: something stronger was proved </br> Remark: the identity is a self-inverse element with $\map {C_G} e {{=}} G$ trivially}}
By Norm Sequence of Cauchy Sequence has Limit then: :for each $\eqclass {x_n}{}$ the $\displaystyle \lim_{n \mathop \to \infty} \norm{x_n}$ exists. Suppose $\eqclass {x_n}{} = \eqclass {y_n}{}$. By Left Cosets are Equal iff Difference in Subgroup then: :$\sequence {x_n} - \sequence {y_n} =  \sequence {x_n - y_n} \in \mathcal N$ By Equivalent Cauchy Sequences have Equal Limits of Norm Sequences then: :$\displaystyle \lim_{n \mathop \to \infty} \norm{x_n} = \lim_{n \mathop \to \infty} \norm{y_n}$ Hence: :$\displaystyle \norm { \eqclass {x_n}{} }_1 = \lim_{n \mathop \to \infty} \norm{x_n} = \lim_{n \mathop \to \infty} \norm{y_n} =  \norm { \eqclass {x_n}{} }_1$ The result follows. {{qed}}
By Quotient Ring of Cauchy Sequences is Division Ring the zero of $\CC \,\big / \NN$ is $\eqclass {0_R} {}$. {{begin-eqn}} {{eqn | l = \norm {\eqclass {0_R} {} }_1 = 0       | o = \leadstoandfrom       | r = \lim_{n \mathop \to \infty} \norm {x_n} = 0       | c = Definition of $\norm {\,\cdot\,}_1$ }} {{eqn | o = \leadstoandfrom       | r = \sequence {x_n} \in \NN       | c = Definition of $\NN$ }} {{eqn | o = \leadstoandfrom       | r = \eqclass {x_n} {} = \eqclass {0_R} {}       | c = Left Cosets are Equal iff Product with Inverse in Subgroup }} {{end-eqn}} The result follows. {{qed}}
Let $\eqclass {x_n} {}, \eqclass {y_n} {} \in \CC \,\big / \NN$ {{begin-eqn}} {{eqn | l = \norm {\eqclass {x_n} {} \eqclass {y_n} {} } _1       | r = \norm {\eqclass {x_n y_n} {} }_1       | c = Multiplication on quotient ring }} {{eqn | r = \lim_{n \mathop \to \infty} \norm {x_n y_n}       | c = Definition of $\norm {\,\cdot\,}_1$ }} {{eqn | r = \lim_{n \mathop \to \infty} \norm {x_n} \norm {y_n}       | c = {{NormAxiom|2}} }} {{eqn | r = \lim_{n \mathop \to \infty} \norm {x_n} \times \lim_{n \to \infty} \norm {y_n }       | c = Product Rule for Real Sequences }} {{eqn | r = \norm {\eqclass {x_n} {} }_1 \times \norm {\eqclass {y_n} {} } _1       | c = Definition of $\norm {\,\cdot\,}_1$ }} {{end-eqn}} {{qed}}
Let $\eqclass {x_n} {}, \eqclass {y_n} {} \in \CC \,\big / \NN$ {{begin-eqn}} {{eqn | l = \norm {\eqclass {x_n} {} + \eqclass {y_n} {} } _1       | r = \norm {\eqclass {x_n + y_n} {} }_1       | c = Addition on quotient ring }} {{eqn | r = \lim_{n \mathop \to \infty} \norm {x_n + y_n}       | c = Definition of $\norm {\,\cdot\,}_1$ }} {{end-eqn}} By {{NormAxiom|3}}: :$\forall n: \norm {x_n + y_n} \le \norm {x_n} + \norm {y_n}$ So: {{begin-eqn}} {{eqn | l = \lim_{n \mathop \to \infty} \norm {x_n + y_n}       | o = \le       | r = \lim_{n \mathop \to \infty} \norm {x_n} + \norm {y_n}       | c = Inequality Rule for Real Sequences  }} {{eqn | r = \lim_{n \mathop \to \infty} \norm {x_n} + \lim_{n \mathop \to \infty} \norm {y_n}       | c = Sum Rule for Real Sequences }} {{eqn | r = \norm {\eqclass {x_n} {} }_1 + \norm {\eqclass {y_n} {} } _1       | c = Definition of $\norm {\,\cdot\,}_1$ }} {{end-eqn}} {{qed}}
We have from Inverse of Identity Element is Itself that $e$ is invertible. Hence $e \in U$ and so $U \ne \O$. Let $x, y \in U$. As $x$ and $y$ are invertible, it follows that $x^{-1}$ and $y^{-1}$ both exist in $S$. Both $x^{-1}$ and $y^{-1}$ also have inverses $x$ and $y$ respectively, and so themselves are invertible. Hence both $x^{-1} \in U$ and $y^{-1} \in U$. Then we have: {{begin-eqn}} {{eqn | l = \paren {x \circ y} \circ \paren {y^{-1} \circ x^{-1} }       | r = x \circ \paren {y \circ y^{-1} } \circ x^{-1}       | c = $\circ$ is associative in $S$ }} {{eqn | r = x \circ e \circ x^{-1}       | c =  }} {{eqn | r = x \circ x^{-1}       | c =  }} {{eqn | r = e       | c =  }} {{end-eqn}} and similarly for $\paren {y^{-1} \circ x^{-1} } \circ \paren {x \circ y}$ Hence $x \circ y$ is invertible. The result follows by the Two-Step Subgroup Test. {{qed}}
By Limit of Subsequence equals Limit of Sequence then $\sequence {y_n}$ is convergent with: :$\displaystyle \lim_{n \mathop \to \infty} y_n = l$ Let $\epsilon > 0$ be given. Let $\epsilon' = \dfrac {\epsilon  {\norm l}^2 } {2}$. Then: :$ \epsilon' > 0$ As $\sequence {y_n} \to l$, as $n \to \infty$, we can find $N_1$ such that: :$\forall n > N_1: \norm {y_n - l} < \epsilon'$ As $\sequence {y_n}$ converges to $l \ne 0$, by Sequence Converges to Within Half Limit: :$\exists N_2 \in \N: \forall n > \N_2: \dfrac {\norm l} 2 < \norm {y_n}$ or equivalently: :$\exists N_2 \in \N: \forall n > \N_2: 1 < \dfrac {2 \norm {y_n} } {\norm l}$ Let $N = \max \set {N_1, N_2}$.  Then $\forall n > N$: :$(1): \quad \norm {y_n - l} < \epsilon'$ :$(2): \quad 1 < \dfrac {2 \norm {y_n} } {\norm l}$ Hence: {{begin-eqn}} {{eqn | l = \norm { {y_n }^{-1} - l^{-1} }       | o = <       | r = \dfrac {2 \norm {y_n} } {\norm l} \norm { {y_n}^{-1} - l^{-1} }       | c = from $(1)$ }} {{eqn | r = \dfrac 2 {\norm l^2} \paren {\norm { {y_n } } \norm { {y_n}^{-1} - l^{-1} } \norm l}       | c = multiplying and dividing by $\norm l$ }} {{eqn | r = \dfrac 2 {\norm l^2} \norm {y_n \paren { {y_n}^{-1} - l^{-1} } l}       | c = {{NormAxiom|2}} }} {{eqn | r = \dfrac 2 {\norm l^2} \norm {y_n y_n^{-1} l - y_n l^{-1} l}       | c = {{Ring-axiom|D}} }} {{eqn | r = \dfrac 2 {\norm l^2} \norm {l - y_n}       | c = Inverse Property of Division Ring }} {{eqn | o = <       | r = \dfrac 2 {\norm l^2} \epsilon'       | c = from $(2)$ }} {{eqn | r = \dfrac {2} {\norm l^2} \paren { \dfrac {\epsilon \norm l^2 } 2}       | c = Definition of $\epsilon'$ }} {{eqn | r = \epsilon       | c = cancelling terms }} {{end-eqn}} Hence: :$\sequence { {y_n}^{-1} }$ is convergent with $\displaystyle \lim_{n \mathop \to \infty} {y_n}^{-1} = l^{-1}$. {{qed}} Category:Combination Theorem for Sequences in Normed Division Rings s9m3b9wd0jhdnb1n3cmd1fpdiz6vbae
=== $(1)$ implies $(2)$ === Let $n$ be an odd-times odd integer by definition 1. Then by definition: :$n > 1$ :$n$ is not prime. Thus: :$n = r s$ where $1 < r < p$ and $1 < s < p$. As $n$ is odd: :$2 \nmid n$ and so: :$2 \nmid r$ and: :$2 \nmid s$ So both $r$ and $s$ are odd integers greater than $1$ such that $n = r s$. Thus $n$ is an odd-times odd integer by definition 2. {{qed|lemma}} === $(2)$ implies $(1)$ === Let $n$ be an odd-times odd integer by definition 2. Then by definition: :$n = r s$ where $r > 1, s > 1$ are odd integers. Thus $n$ is not prime by definition. As $r$ and $s$ are odd: :$2 \nmid r$ and :$2 \nmid s$ and so: :$2 \nmid r s$ So $n = r s$ is odd. As $r > 1$ and $s > 1$ it follows from Divisor Relation on Positive Integers is Partial Ordering that $n > 1$. Thus $n$ is an odd-times odd integer by definition 1. {{qed}} Category:Odd-Times Odd Integers 3mxvcvamzd7h3dvgol5s08njoh1mgu8
Let $x_0 \in R$. Let $\epsilon \in \R_{>0}$.  Let $x \in R: \norm {x - x_0} < \epsilon$. Then: {{begin-eqn}} {{eqn | l = \size {\norm x - \norm {x_0} }       | o = \le       | r = \norm {x - x_0}       | c = Reverse triangle inequality }} {{eqn | o = <       | r = \epsilon }} {{end-eqn}} By the definition of metric induced by a norm and the definition of a continuous mapping, $\norm {\,\cdot\,}$ is continuous. {{qed}} Category:Normed Division Rings Category:Norm Theory 81dpy5f0xjivpcxu1h0pvr5llskobwv
Proof by induction: For all $n \in \N_{>0}$, let $\map P n$ be the proposition: :$\displaystyle \sum_{j \mathop = 1}^n {F_j}^2 = F_n F_{n + 1}$ === Basis for the Induction === $\map P 1$ is the case ${F_1}^2 = 1 = F_3 - 1$, which holds from the definition of Fibonacci numbers. {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 1}^1 {F_j}^2       | r = {F_1}^2       | c =  }} {{eqn | r = 1 \times 1       | c =  }} {{eqn | r = F_1 \times F_2       | c =  }} {{end-eqn}} demonstrating that $\map P 1$ holds. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\displaystyle \sum_{j \mathop = 1}^k {F_j}^2 = F_k F_{k + 1}$ Then we need to show: :$\displaystyle \sum_{j \mathop = 1}^{k + 1} {F_j}^2 = F_{k + 1} F_{k + 2}$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 1}^{k + 1} {F_j}^2       | r = \sum_{j \mathop = 1}^k {F_j}^2 + {F_{k + 1} }^2       | c =  }} {{eqn | r = F_k F_{k + 1} + {F_{k + 1} }^2       | c = Induction Hypothesis }} {{eqn | r = \paren {F_k + F_{k + 1} } F_{k + 1}       | c =  }} {{eqn | r = F_{k + 2} F_{k + 1}       | c = {{Defof|Fibonacci Number}} }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \ge 1: \displaystyle \sum_{j \mathop = 1}^n {F_j}^2 = F_n F_{n + 1}$ {{qed}}
By Norm of Negative of Unity then: :$\norm{-1_R} = 1$. Then: {{begin-eqn}} {{eqn | l = \norm{-x}       | r = \norm{-1_R \circ x}       | c = Product with Ring Negative }} {{eqn | r = \norm{-1_R} \norm{x}       | c = Norm axiom (N2) (Multiplicativity) }} {{eqn | r = \norm{x}       | c = Norm of Negative of Unity }} {{end-eqn}} as desired. {{qed}}
By the norm axiom (N2) (Multiplicativity) then: :$\forall x, y \in R: \norm {x \circ y} = \norm{x} \norm{y}$ In particular: :$\norm{1_R} = \norm{1_R \circ 1_R} = \norm{1_R} \norm{1_R}$ By the norm axiom (N1) (Positive defintiteness) then: :$\norm{1_R} \ne 0$ So $\norm{1_R}$ has an inverse in $\R$. By multiplying by this inverse, then: :$ \norm{1_R} \norm{1_R} =\norm{1_R} \iff \norm{1_R} = 1$ as desired. {{qed}}
By Product of Ring Negatives then: :$-1_R \circ -1_R = 1_R \circ 1_R=1_R$. So: {{begin-eqn}} {{eqn | l = \norm{-1_R}^{2}       | r = \norm{-1_R} \norm{-1_R} }} {{eqn | r = \norm{-1_R \circ -1_R}       | c = Norm axiom (N2) (Multiplicativity) }} {{eqn | r = \norm{1_R}       | c = Product of Ring Negatives }} {{eqn | r = 1       | c = Norm of Unity of Division Ring }} {{end-eqn}} Thus: :$\norm{-1_R} = \pm 1$ By the norm axiom (N1) (Positive defintiteness) then:  :$\norm{-1_R} \ge 0$ Hence: :$\norm{-1_R} = 1$ {{qed}}
The digits of a repunit are by definition all $1$. Thus the product of the digits of a repunit is $1$. By One Divides all Integers, $1$ is a divisor of $n$. Hence the result, by definition of Zuckerman number.
Then: {{begin-eqn}} {{eqn | l = \norm {x - y}       | r = \norm {x + \paren {-y} } }} {{eqn | o = \le       | r = \norm x + \norm {-y}       | c = Norm axiom $(\text N 3)$: Triangle Inequality }} {{eqn | r = \norm x + \norm y       | c = Norm of Negative }} {{end-eqn}} as desired. {{qed}}
Let $x \ne 0_R$. By Norm axiom $(\text N 1)$: Positive Definiteness: :$\norm x \ne 0$ So: {{begin-eqn}} {{eqn| l = \norm x \norm {x^{-1} }      | r = \norm {x \circ x^{-1} }      | c = Norm axiom $(\text N 2)$: Multiplicativity }} {{eqn| r = \norm {1_R}      | c = {{Defof|Product Inverse}} }} {{eqn| r = 1      | c = Norm of Unity of Division Ring }} {{eqn| ll= \leadsto      | l = \norm {x^{-1} }      | r = \dfrac 1 {\norm x}      | c =  }} {{end-eqn}} {{qed}}
Let $y \ne 0_R$. By Norm axiom (N1) (Positive Definiteness) then: :$\norm {y} \ne 0$ So: {{begin-eqn}} {{eqn| l = \norm{x \circ y^{-1} }      | r = \norm{x} \norm{y^{-1} }      | c = Norm axiom (N2) (Multiplicativity) }} {{eqn| r = \dfrac {\norm{x} } {\norm{y} }      | c = Norm of Inverse }} {{end-eqn}} Similarly: :$\norm{y^{-1}x} = \dfrac {\norm{x}}{\norm{y}}$ {{qed}}
Let $n \in \N_{\gt 0}$. Let $\norm {x^n} = 1$. By Norm axiom (N2) (Multiplicativity) then: :$\norm x^n = 1$ Since $\norm x \ge 0$, by Positive Real Complex Root of Unity then: :$\norm x = 1$ as desired. {{qed}}
Let $n \in \N_{\gt 0}$. Then: {{begin-eqn}} {{eqn| l = \norm {n \cdot 1_R}      | r = = \norm {1_R + 1_R + \dots + 1_R}      | o = }} {{eqn| o = }} {{eqn| o =       | r = \le \underbrace {\norm {1_R} + \norm {1_R} + \dots + \norm {1_R} }_{n \, times}      | c = Norm axiom (N3) (Triangle Inequality) }} {{eqn| o = }} {{eqn| o =      | r = = \underbrace {1 + 1 + \dots + 1 }_{n \, times}      | c = Norm of Unity }} {{eqn| o = }} {{eqn| o =      | r = = n }} {{end-eqn}} {{qed}}