hcju/proofwikiqa huggingface dataset:
We are given that: :$\phi \circ q_\RR = f$ is well-defined. Note that from Quotient Mapping is Surjection, $q_\RR$ is a surjection. === Sufficient Condition === Let $\phi$ be a surjection. Then from Composite of Surjections is Surjection: :$\phi \circ q_\RR$ is a surjection. {{qed|lemma}} === Necessary Condition === Let $f = \phi \circ q_\RR$ be a surjection. Then from Surjection if Composite is Surjection: :$\phi$ is a surjection. {{qed|lemma}} Hence the result. {{qed}}
From Condition for Mapping from Quotient Set to be Well-Defined, $\phi$ is well-defined {{iff}}: :$\forall x, y \in S: \tuple {x, y} \in \RR \implies \map f x = \map f y$ By definition of injection, $\phi$ is injective {{iff}}: :$\map \phi {\eqclass x \RR} = \map \phi {\eqclass y \RR} \implies \eqclass x \RR = \eqclass y \RR$ that is: :$\map \phi {\eqclass x \RR} = \map \phi {\eqclass y \RR} \implies \tuple {x, y} \in \RR$ By definition of $\phi$: :$\map \phi {\eqclass x \RR} = \map f x$ and so: :$\map f x = \map f y \implies \tuple {x, y} \in \RR$ Hence the result. {{qed}}
From Condition for Mapping from Quotient Set to be Well-Defined, $\phi$ is well-defined {{iff}}: :$\forall x, y \in S: \tuple {x, y} \in \RR \implies \map f x = \map f y$ From Quotient Mapping is Surjection, $q_\RR$ is a surjection. Suppose $\psi: S / \RR \to T$ is another well-defined mapping defined as: :$\psi \circ q_\RR = f$ Then we have: {{begin-eqn}} {{eqn | l = \phi \circ q_\RR       | r = \psi \circ q_\RR       | c =  }} {{eqn | ll= \leadsto       | l = \phi       | r = \psi       | c = Surjection iff Right Cancellable }} {{end-eqn}} Hence the result. {{qed}}
Consider the commutative diagram: ::$\begin {xy} \xymatrix@L + 2mu@ + 1em {  S \ar[rr]^*{f}    \ar[dd]_*{q_S}    \ar[ddrr]^*{q_T \circ f} & & T \ar[dd]^*{q_T} \\ & & \\ S / \RR_S \ar@{-->}[rr]_*{g} & & T / \RR_T } \end {xy}$ We consider the mapping $q_T \circ f: S \to T / \RR_T$. From Condition for Mapping from Quotient Set to be Well-Defined: :there exists a mapping $g: S / \RR_S \to T / \RR_T$ such that $g \circ q_S = q_T \circ f$ {{iff}}: :$\forall x, y \in S: \tuple {x, y} \in \RR_S \implies \map {\paren {q_T \circ f} } x = \map {\paren {q_T \circ f} } y$ Hence: {{begin-eqn}} {{eqn | l = \map {\paren {q_T \circ f} } x       | r = \map {\paren {q_T \circ f} } y       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \map {q_t} {\map f x}       | r = \map {q_t} {\map f y}       | c = {{Defof|Composition of Mappings}} }} {{eqn | ll= \leadstoandfrom       | l = \map f x       | o = \RR_T       | r = \map f y       | c = {{Defof|Quotient Mapping}} }} {{end-eqn}} {{Qed}}
=== Closedness === Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $K$. Suppose, $\sequence {x_n}_{n \mathop \in \N}$ converges to $L \in K$.  Then there is a subsequence $\sequence {x_{n_k}}_{k \mathop \in \N}$ convergent to $L' \in K$.  But $\sequence {x_{n_k}}_{k \mathop \in \N}$ is a subsequence $\sequence {x_n}_{n \mathop \in \N}$. Hence, $\sequence {x_{n_k}}_{k \mathop \in \N}$ converges to $L$. By uniqueness of limits, $L = L'$. By definition, $K$ is closed. === Boundedness === Suppose, $K$ is not bounded. Then: :$\forall n \in \N : \exists x_n \in K: \norm {x_n} > n$ Therefore, no subsequence of $\sequence {x_n}_{n \mathop \in \N}$ is convergent. Hence, $K$ is not compact. This is in contradiction with the assumption. Thus, $K$ is bounded.
Let $z = x + i y$. By definition of the complex plane, it can be represented by the point $\tuple {x, y}$. By the Distance Formula, the distance $d$ of $z$ from the origin is: {{begin-eqn}} {{eqn | l = d       | r = \sqrt {\paren {x - 0}^2 + \paren {y - 0}^2}       | c =  }} {{eqn | r = \sqrt {x^2 + y^2}       | c =  }} {{end-eqn}} which is precisely the modulus of $z$. {{qed}}
An example of the use of Third Isomorphism Theorem/Groups/Corollary 2. {{ProofWanted|details}}
A subgroup is by definition a subset which is a group. But neither $G$ nor $H$ are actually subsets of their cartesian product $G \times H$. Hence the result. {{qed}} Category:Group Direct Products Category:Subgroups ninzsg4wxsaaotzi11ntmic8a0nz9sn
Consider the mapping $g: \C \to \R$ defined as: :$\forall z \in \C: \map f z = \map \Re z$ where $\map \Re z$ denotes the real part of $z$. From Real Part as Mapping is Endomorphism for Complex Addition, this is a projection from $\C$ to $\R$. The result follows from Canonical Injection is Right Inverse of Projection. {{Qed}}
We need to show that $\phi: \tuple {x, y} \mapsto x + y i$ is a group isomorphism. === $\phi$ is a group homomorphism === {{begin-eqn}} {{eqn | ll = \forall a, b, c, d \in \R:       | l = \map \phi {a, b} + \map \phi {c, d}       | r = \paren {a + b i} + \paren {c + d i} }} {{eqn | r = \paren {a + c} + \paren {b + d} i       | c = Complex Addition is Commutative }} {{eqn | r = \map \phi {a + c, b + d} }} {{end-eqn}} {{qed|lemma}} === $\phi$ is bijective === We show that $\phi^{-1}: z \mapsto \tuple {\map \Re z, \map \Im z}$ is the inverse of $\phi$. We have: :$\map \phi {\map {\phi^{-1} } z} = \map \phi {\map \Re z, \map \Im z} = \map \Re z + i \map \Im z = z$ :$\map {\phi^{-1} } {\map \phi {a, b} } = \map {\phi^{-1} } {a + b i} = \tuple {a, b}$ hence $\phi^{-1}$ is the inverse of $\phi$. {{qed|lemma}} Hence $\phi$ is a group isomorphism, and thus the direct product $\struct {\R, +} \times \struct {\R, +}$ is isomorphic with $\struct {\C, +}$. {{qed}}
Let $\tuple {a, b}$ and $\tuple {c, d}$ be pairs of non-zero real numbers: :$\tuple {a, b} \in \R_{\ne 0} \times \R_{\ne 0}$ :$\tuple {c, d} \in \R_{\ne 0} \times \R_{\ne 0}$ Then by definition of group direct product: :$\tuple {a, b} \times \tuple {c, d} = \tuple {a \times c, b \times d}$ However, by interpreting $\tuple {a, b}$ and $\tuple {c, d}$ as complex numbers: :$\tuple {a, b} \times \tuple {c, d} = \tuple {a \times c - b \times d, b \times c + a \times d}$ by definition of complex multiplication. Hence the result. {{Qed}}
We have that $\II$ is a non-empty subset of the complex numbers $\C$. Indeed, for example: :$0 + 0 i \in \II$ Now, let $0 + i x, 0 + i y \in \II$. Then we have: {{begin-eqn}} {{eqn | l = \paren {0 + i x} + \paren {-\paren {0 + i y} }       | r = \paren {0 + i x} - \paren {0 + i y}       | c = Inverse for Complex Addition }} {{eqn | r = i \paren {x - y}       | c = {{Defof|Complex Subtraction}} }} {{eqn | o = \in       | r = \II       | c =  }} {{end-eqn}} Hence the result by the One-Step Subgroup Test. {{qed}}
Let $0 + i x \in \II$. We have: {{begin-eqn}} {{eqn | l = \paren {0 + i x} \times \paren {0 + i x}       | r = \paren {0 - x^2} + i \paren {0 \times x + 0 \times x}       | c = {{Defof|Complex Multiplication}} }} {{eqn | r = -x^2       | c =  }} {{eqn | o = \notin       | r = \II       | c =  }} {{end-eqn}} So $\struct {\II, \times}$ is not closed. Hence $\struct {\II, \times}$ is not a group. {{qed}}
We note that $S$ is a subset of the set of all real functions on $\R$. From Set of All Self-Maps is Semigroup, we have that $\circ$ is associative. Consider the real function $I: \R \to \R$ defined as: :$\forall x \in \R: \map I x = 1 \times x + 0$ We have that: :$I \in S$ :$I$ is the identity mapping. So $S$ is not empty. Then we note: Let $f, g \in S$ such that: :$\map f x = r_1 x + s_1$ :$\map g x = r_2 x + s_2$ For all $x \in \R$, we have: {{begin-eqn}} {{eqn | l = \map {g \circ f} x       | r = \map g {\map f x}       | c = {{Defof|Composition of Mappings}} }} {{eqn | r = \map g {r_1 x + s_1}       | c = Definition of $f$ }} {{eqn | r = r_2 {r_1 x + s_1} + s_2       | c = Definition of $f$ }} {{eqn | r = \paren {r_2 r_1} x + \paren {r_2 s_1 + s_2}       | c =  }} {{eqn | r = r x + s       | c = where $r = r_2 r_2$ and $s = r_2 s_1 + s_2$ }} {{end-eqn}} This demonstrates that $\struct {S, \circ}$ is closed. Let $\phi: \R \to \R$ be the real function defined as: :$\forall x \in \R: \map \phi x = p x + q$ where $p \in \R_{\ne 0}$ and $q \in \R$. Then: {{begin-eqn}} {{eqn | l = y       | r = p x + q       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = \dfrac {y - q} p       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = \dfrac 1 p y + \paren {-\dfrac q p}       | c =  }} {{end-eqn}} Thus we define $\psi: \R \to \R$ be the real function defined as: :$\forall y \in \R: \map \psi y = \dfrac 1 p y + \paren {-\dfrac q p}$ where $\dfrac 1 p \in \R_{\ne 0}$ and $-\dfrac q p \in \R$. It is seen that $\psi \in S$. For all $x \in \R$, we have: {{begin-eqn}} {{eqn | l = \map {\phi \circ \psi} x       | r = \map \phi {\map \psi x}       | c = {{Defof|Composition of Mappings}} }} {{eqn | r = \map \phi {\dfrac 1 p x - \dfrac q p}       | c = Definition of $\psi$ }} {{eqn | r = p {\dfrac 1 p x -\dfrac q p} + q       | c = Definition of $\phi$ }} {{eqn | r = \dfrac p p x - \dfrac {p q} p + q       | c =  }} {{eqn | r = x       | c =  }} {{end-eqn}} Hence: :$\phi \circ \psi = I$ and so $\psi$ is the inverse of $\phi$. We have then that $\struct {S, \circ}$ is closed under inversion. The result follows by the Two-Step Subgroup Test. {{qed}}
Let $S' = \set {0, 1, 2, 3}$. Let $\phi: S \to S'$ be the bijection: {{begin-eqn}} {{eqn | l = \map \phi 2       | r = 0 }} {{eqn | l = \map \phi 3       | r = 1 }} {{eqn | l = \map \phi 4       | r = 2 }} {{eqn | l = \map \phi 1       | r = 3 }} {{end-eqn}} By inspection, the Cayley table presented above is in the same form as the Cayley table for the cyclic group of order $4$: :$\begin{array}{r|rrrr} +_4 & 0 & 1 & 2 & 3 \\ \hline 0 & 0 & 1 & 2 & 3  \\ 1 & 1 & 2 & 3 & 0 \\ 2 & 2 & 3 & 0 & 1 \\ 3 & 3 & 0 & 1 & 2 \\ \end{array}$ {{qed}}
From the Division Laws for Groups: :$g h = e \implies g = e h^{-1} = h^{-1}$ Also by the Division Laws for Groups: :$g h = e \implies h = g^{-1} e = g^{-1}$ {{qed}}
From the Division Laws for Groups: :$h g = e \implies g = e h^{-1} = h^{-1}$ Also by the Division Laws for Groups: :$h g = e \implies h = g^{-1} e = g^{-1}$ {{qed}}
=== Case $1$: $m + 1$ is composite === Suppose $m + 1$ is composite. Then: :$\exists p, q > 1: m + 1 = p q$ By Division Theorem, for each $k$ with $0 \le z \le m$: :$\exists i, j \in \N: 0 \le i \le q - 1, \, 0 \le j \le p - 1: k = i + q j$ Thus: {{begin-eqn}} {{eqn | l = n       | r = \sum_{k \mathop = 0}^m 10^{r k} }} {{eqn | r = \sum_{i \mathop = 0}^{q - 1} \sum_{j \mathop = 0}^{p - 1} 10^{r \paren {i + q j} } }} {{eqn | r = \paren {\sum_{i \mathop = 0}^{q - 1} 10^{r i} } \paren {\sum_{j \mathop = 0}^{p - 1} 10^{r q j} } }} {{end-eqn}} Both sums are greater than $1$, so $n$ is composite. {{qed|lemma}} === Case $2$: $m + 1$ is an odd prime and $r \ne m + 1$ === Notice that:  :$\displaystyle \sum_{k \mathop = 0}^m 10^{r k} \times R_r = R_{r \paren {m + 1} }$ where $R_i$ is the $i$th repunit. Suppose $m + 1$ and $r$ are coprime. By Condition for Repunits to be Coprime, $R_{m + 1}$ and $R_r$ are coprime. By Euclid's Lemma: :$R_{m + 1} \divides \dfrac {R_{r \paren {m + 1} } } {R_r} = n$ Suppose $r^2 \divides m + 1$. By Divisors of Repunit with Composite Index: :$R_r \divides R_{r^2}$ and: :$R_{r^2} \divides R_{r \paren {m + 1} }$ So we have: :$\dfrac {R_{r^2}} {R_r} \divides \dfrac {R_{r \paren {m + 1} } } {R_r} = n$ {{qed|lemma}} === Case $3$: $m = 1$, $r$ is not a power of $2$ === We write $n = 1 + 10^r$. Since $r$ is not a power of $2$, $r$ has an odd factor greater than $1$. Write $r = x \paren {2 y + 1}$.  Then: {{begin-eqn}} {{eqn | l = n       | r = 1 + 10^r }} {{eqn | r = 1 + 10^{x \paren {2 y + 1} } }} {{eqn | r = 1 + \paren {10^x}^{2 y + 1} }} {{end-eqn}} Since $1 + \paren {-1}^{2 y + 1} = 0$, by Polynomial Factor Theorem, $1 + z^{2 y + 1} = \paren {1 + z} \map Q z$ for some polynomial $Q$. Therefore $1 + 10^x$ is a factor of $1 + \paren {10^x}^{2 y + 1} = n$. Thus all cases are covered. {{qed}} mzz55jfgl0cdv9bmnodhdjp299yv1ga
Consider the odd integers $1$ and $3$. We have that $1 + 3 = 4$. But $4$ is not odd. Thus addition on $\struct {S, +}$ is not closed. Hence $\struct {S, +}$ is not a group, let alone a subgroup of $\struct {\Z, +}$ {{qed}}
First it is noted that $H \subseteq G$, and that the identity mapping is an element of $H$. Hence to demonstrate that $H$ is a subgroup of $G$, one may use the Two-Step Subgroup Test. Let $\phi \in H$ be a transposition. Then from Transposition is Self-Inverse, $\phi^{-1} \in H$. So $H$ is closed under inversions. Let $a, b, c \in S$. Let $\pi, \phi \in H$ such that: :$\pi = \tuple {a \ b}$ :$\phi = \tuple {b \ c}$ Then by inspection: :$\phi \circ \pi = \tuple {a \ c \ b}$ Thus $\phi \circ \pi$ is a $3$-cycle and so not a transposition. Hence $H$ is not closed under composition. So, by the Two-Step Subgroup Test, $H$ is not a subgroup of $G$. {{Qed}}
We have that $S$ is the same thing as $\R_{>0}$, the set of strictly positive real numbers: :$\R_{>0} = \set {x \in \R: x > 0}$ From Strictly Positive Real Numbers under Multiplication form Uncountable Abelian Group, $\struct {S, \times}$ is a group. Hence as $S$ is a group which is a subset of $\struct {\C_{\ne 0}, \times}$, it follows that $\struct {S, \times}$ is a subgroup of $\struct {\C_{\ne 0}, \times}$. {{qed}}
By Euler's Identity, $\phi$ can also be expressed as: :$\forall x \in \R: \map \phi x = e^{i x}$ From Homomorphism from Reals to Circle Group, $\phi$ is a homomorphism from $\struct {\R, +}$ to the circle group $\struct {K, \times}$. From Circle Group is Infinite Abelian Group, we note that $\struct {K, \times}$ is a subgroup of the multiplicative group of complex numbers. {{qed}}
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $F$ with a limit point $x$ in $X$. === Definition 1 implies Definition 2 === Let $F$ be closed in $V$ according to Definition 1: :'''$F$ is closed in $V$''' {{iff}} its complement $X \setminus F$ is open in $V$. {{AimForCont}} $x \notin F$. Since $F$ is closed, $X \setminus F$ is open. Let $\map {B_\epsilon} x := \set {y \in X : \norm {y - x} < \epsilon}$ be an open ball. Then there exists $\map {B_\epsilon} x$, which belongs to $X \setminus F$: :$\forall x \in X \setminus F: \exists \epsilon \in \R_{>0}: \map {B_\epsilon} x \subset X \setminus F$ $\sequence {x_n}_{n \mathop \in \N}$ converges to $x$. Therefore: :$\displaystyle \forall \epsilon \in \R_{> 0} : \exists N \in \N : \forall n \in N : n > N \implies \norm {x_n - x} < \epsilon$ Hence: :$\norm {x_{N + 1} - x} < \epsilon$. Then we have: :$x_{N + 1} \in \map {B_\epsilon} x \subset X \setminus F$ But at the same time: :$x_{N + 1} \in F$ and we reached a contradiction. Hence there is no such $x$ which is not in $F$. It follows that every limit point of $F$ is also a point of $F$. That is, $F$ is closed in $V$ according to Definition 2. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $F$ be closed in $V$ according to Definition 2: :'''$F$ is closed (in $V$)''' {{iff}} every limit point of $F$ is also a point of $F$. We are given that $x$ is a limit point of $F$. Let $x \in F$. {{AimForCont}} $X \setminus F$ is not open. Then: :$\neg \paren {\forall x \in X \setminus F : \exists \epsilon \in \R_{> 0} : \map {B_\epsilon} x \subseteq X \setminus F}$ By Assertion of Existence, this is equivalent to $\paren \star$: :$\exists x \in X \setminus F : \forall \epsilon \in \R_{> 0} : \map {B_\epsilon} x \cap F \ne \empty$ Set $\displaystyle \epsilon = \frac 1 n$ with $n \in \N$. Then: :$\exists n \in \N : x_n \in \map {B_\epsilon} x \cap F$ So there is a sequence $\sequence {x_n}_{n \mathop \in \N}$ in $F$ such that: :$\displaystyle \forall n \in \N : \norm {x - x_n} < \frac 1 n$ and $x_n$ converges to $x$ in $F$. By $\paren \star$: :$x \in X \setminus F$ That is: :$x \notin F$ This contradicts the condition in the definition that $x \in F$. Hence, $X \setminus F$ has to be open. That is, $F$ is closed in $V$ according to Definition 1. {{qed}}
Let $z_0 \in \C_{\ne 0}$. Write $z_0 = r_0 e^{i \theta}$, where $r_0 > 0$ and $\theta \in \hointr 0 {2 \pi}$. We will show that: :$z_0 S = \set {z \in \C: \exists r \in \R_{>0}: z = r e^{i \theta}}$ Pick any $w \in z_0 S$. Then there exists some $x \in S$ such that $w = z_0 x$. Note that $x \in \R_{>0}$ and $r_0 x \in \R_{>0}$. Thus: :$w = z_0 x = r_0 x e^{i \theta} \in \set {z \in \C: \exists r \in \R_{>0}: z = r e^{i \theta}}$ This shows that: :$z_0 S \subseteq \set {z \in \C: \exists r \in \R_{>0}: z = r e^{i \theta}}$ Pick any $w \in \set {z \in \C: \exists r \in \R_{>0}: z = r e^{i \theta}}$. Write $w = r e^{i \theta}$, where $r \in \R_{>0}$. Note that $\dfrac r {r_0} \in \R_{>0}$. Then: :$w = \paren {\dfrac r {r_0}} r_0 e^{i \theta} = \paren {\dfrac r {r_0}} z_0 \in z_0 S$ This shows that: :$z_0 S \supseteq \set {z \in \C: \exists r \in \R_{>0}: z = r e^{i \theta}}$ By definition of set equality: :$z_0 S = \set {z \in \C: \exists r \in \R_{>0}: z = r e^{i \theta}}$ {{qed}}
Let $\textbf p$ denote the equivalence class that is identified with the prime number $p$.  By Rational Numbers are Dense Subfield of P-adic Numbers:  :the constant sequence $\tuple {p, p, p, \dots}$ represents $\mathbf  p \in \Q_p$. From P-adic Number times P-adic Norm is P-adic Unit there exists $n \in \Z$: :$\mathbf p^n \mathbf a \in \Z^\times_p$ :$\norm {\mathbf a}_p =  p^n$  where $\Z^\times_p$ is the set of $p$-adic units. Let $l = -n$. From P-adic Norm of P-adic Expansion is determined by First Nonzero Coefficient: :$l = \min \set {i: i \ge m \land d_i \ne 0}$ and :$l = \min \set {i: i \ge k \land e_i \ne 0}$ Now: {{begin-eqn}} {{eqn | l = m       | o = <       | r = 0 }} {{eqn | ll = \leadsto       | l = d_m       | o = \ne       | r = 0       | c = {{Defof|P-adic Expansion|$p$-adic Expansion}} }} {{eqn | ll = \leadsto       | l = m       | r = l       | c = Definition of $l$ }} {{eqn | ll = \leadsto       | l = l       | o = <       | r = 0 }} {{end-eqn}} Similarly: :$k < 0 \leadsto k = l$ === Proof of statement $(1)$ === ==== Case $l < 0$ ==== Let $l < 0$. Then: {{begin-eqn}} {{eqn | l = l       | o = <       | r = 0 }} {{eqn | ll = \leadsto       | l = m       | o = <       | r = 0       | c = as $m \le l$ }} {{eqn | ll = \leadsto       | l = m       | r = l       | c = from above }} {{end-eqn}} Similarly: :$l < 0 \leadsto k = l$ So:  :$m = l = k$. {{qed|lemma}} ==== Case $l \ge 0$ ==== Let $l \ge 0$. Then: {{begin-eqn}} {{eqn | l = l       | o = \ge       | r = 0 }} {{eqn | ll = \leadsto       | l = m       | o = \not <       | r = 0       | c = as $m < 0 \leadsto l < 0$ }} {{eqn | ll = \leadsto       | l = m       | r = 0       | c = {{Defof|P-adic Expansion|$p$-adic Expansion}} }} {{end-eqn}} Similarly: :$l \ge  0 \leadsto k = 0$ So: :$m = 0 = k$. {{qed|lemma}} === Proof of statement $(2)$ === By definition of $l$, for all $i$ such that $m \le i < l$: :$d_i = 0$ :$e_i = 0$ So for all $m \le i < l$: :$d_i = e_i$. Consider the series: :$\displaystyle \sum_{i \mathop = l}^\infty d_i p^i$ From P-adic Expansion Less Intial Zero Terms Represents Same P-adic Number :$\displaystyle \sum_{i \mathop = l}^\infty d_i p^i$ is a representative of $\mathbf a$ From Multiple Rule for Cauchy Sequences in Normed Division Ring: :$\displaystyle p^{-l} \sum_{i \mathop = l}^\infty d_i p^i = \sum_{i \mathop = l}^\infty d_i p^{i - l} = \sum_{i \mathop = 0}^\infty d_{i + l} p^i$ is a representative of $\mathbf p^{-l} \mathbf a \in \Z^\times_p$. By definition of a $p$-adic expansion: :$\forall i \in N : 0 \le d_{i + 1} < p - 1$ Thus the series: :$\displaystyle \sum_{i \mathop = 0}^\infty d_{i + l} p^i$ is a $p$-adic expansion that represents $\mathbf p^{-l} \mathbf a \in \Z^\times_p$. Similarly the series: :$\displaystyle \sum_{i \mathop = 0}^\infty e_{i + l} p^i$ is a $p$-adic expansion that represents $\mathbf p^{-l} \mathbf a \in \Z^\times_p$. From Equivalence Class in P-adic Integers Contains Unique P-adic Expansion it follows that: :$\forall i \in N : d_{i + 1} = e_{i + 1}$ That is: :$\forall i \ge l : d_i = e_i$ The result follows. {{qed}} Category:P-adic Number Theory mwdiermydodd3oeh7n1i816o0pj40x5
By Completion of Normed Division Ring and Normed Division Ring is Field iff Completion is Field, the valued field $\struct {\Q_p, \norm {\, \cdot \,}_p}$ is the completion of $\struct {\Q, \norm {\,\cdot\,}_p}$. By Embedding Division Ring into Quotient Ring of Cauchy Sequences, the mapping $\phi: \Q \to \Q_p$ is a distance-preserving monomorphism.  By Normed Division Ring is Dense Subring of Completion, $\Q$ is isometrically isomorphic to $\map \phi \Q$ which is a dense subfield of $\Q_p$. {{qed}} Category:P-adic Number Theory kx44qwig7mg8nrmfsh6dj99n0tsgbet
=== Sufficient Condition === Let $G$ be an edgeless graph. By definition of edgeless graph, $G$ has no edges. That means every vertex of $G$ is incident with no edges. That is, every vertex of $G$ is of degree $0$. Hence the result, by definition of $0$-regular graph. {{qed|lemma}} === Necessary Condition === Let $G$ be a $0$-regular graph. Then every vertex is incident with no edges. So as there are no vertices with incident edges, there can be no edges. Hence $G$ is an edgeless graph. {{qed}} Category:Edgeless Graphs Category:Regular Graphs dbyiy7ppzz1uxdszsl35wv9mc9vjlij
Because $N_n$ has no edges, it has no cycles longer than $0$. Thus in particular, it has no odd cycles. The result follows from Graph is Bipartite iff No Odd Cycles. {{qed}} Category:Edgeless Graphs Category:Bipartite Graphs onj0q7fygjn9c7y8vyoibcwtir8973h
Because there are $n$ vertices in $N_n$, it cannot have more than $n$ components. {{AimForCont}} $N_n$ has fewer than $n$ components. That would mean that at least $2$ vertices of $N_n$ are connected. But to be connected, they need to be joined by at least one edge. But that contradicts the fact that $N_n$ has no edges. Hence by Proof by Contradiction it is not possible for $N_n$ to have fewer than $n$ components. Thus it must be the case that $N_n$ has exactly $n$ components. {{qed}} Category:Edgeless Graphs i1btb0ilnm3j7habcqpc9vlv4f7gamr
By definition, $K_1$ has $1$ vertex. From Complete Graph is Regular, $K_1$ is $0$-regular. Hence the result from Graph is 0-Regular iff Edgeless. {{qed}} Category:Edgeless Graphs Category:Complete Graphs oyi3m2zbnw1gysum6opypbdjzd67alx
By definition of complete graph, each vertex of $K_n$ is adjacent to every other vertex of $K_n$. Let $\overline {K_n}$ denote the complement of $K_n$. Let $v$ be a vertex of $\overline {K_n}$. By definition of complement, $v$ is adjacent to all the vertices of $\overline {K_n}$ that it is not in $K_n$. But that means $v$ is adjacent to no vertices of $\overline {K_n}$. That means $v$ is a $0$-regular graph. The result follows from Graph is 0-Regular iff Edgeless. {{qed}} Category:Complete Graphs Category:Edgeless Graphs mknrfnhayej228c135eavzllqu3tfrp
By definition, a tree is a simple connected graph with no circuits. $N_1$ is trivially connected graph. As $N_1$ is edgeless, it has no edges. But a circuit is a closed trail with at least one edge. Hence the result. {{qed}} Category:Edgeless Graphs Category:Tree Theory 6phu3rw7kr82fneduj3igc2jjqvflem
By definition, a '''forest''' is a simple graph whose components are all trees. From Edgeless Graph of Order n has n Components $N_n$ has as many components as it has vertices. Each of those vertices is an instance of $N_1$, the edgeless graph with $1$ vertex. The result follows from Edgeless Graph of Order 1 is Tree. {{qed}} Category:Edgeless Graphs Category:Forests r75yjqg1zlbbdptu1zc8s9gp9crist9