hcju/proofwikiqa huggingface dataset:
The proof proceeds by induction. For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition: :$\size {K_n} = \dfrac {n \paren {n - 1} } 2$ First we explore the degenerate case $\map P 0$: {{begin-eqn}} {{eqn | l = \size {K_0}       | r = 0       | c = as $K_0$ is the null graph }} {{eqn | r = \dfrac {0 \paren {0 - 1} } 2       | c =  }} {{end-eqn}} Thus $\map P 0$ is seen to hold. === Basis for the Induction === $\map P 1$ is the case: {{begin-eqn}} {{eqn | l = \size {K_1}       | r = 0       | c = Complete Graph of Order 1 is Edgeless }} {{eqn | r = \dfrac {1 \paren {1 - 1} } 2       | c =  }} {{end-eqn}} Thus $\map P 1$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown 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 the induction hypothesis: :$\size {K_k} = \dfrac {k \paren {k - 1} } 2$ from which it is to be shown that: :$\size {K_{k + 1} } = \dfrac {\paren {k + 1} k} 2$ === Induction Step === This is the induction step: Let $K_{k + 1}$ be constructed by adding a new vertex $v_{k + 1}$ to the complete graph $K_k$. To do so, it is necessary to add an edge to join $v_{k + 1}$ to every vertex of $K_k$. Thus there are a total of $k$ edges more in $K_{k + 1}$ than there are in $K_k$. So: {{begin-eqn}} {{eqn | l = \size {K_{k + 1} }       | r = \size {K_k} + k       | c = from the above analysis }} {{eqn | r = \dfrac {k \paren {k - 1} } 2 + k       | c = Induction Hypothesis }} {{eqn | r = \dfrac {k^2 - k + 2 k} 2       | c =  }} {{eqn | r = \dfrac {k^2 + k} 2       | c =  }} {{eqn | r = \dfrac {\paren {k + 1} k} 2       | c =  }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{\ge 0}: \size {K_n} = \dfrac {n \paren {n - 1} } 2$ {{qed}} Category:Complete Graphs Category:Proofs by Induction 4tr8m4mditkjcw3gxkmd1g28qg49b7o
By Corollary to Cauchy Sequences form Ring with Unity, $\CC$ is a commutative ring of Cauchy sequences with the ring operations defined by: :$+ : \quad \forall \sequence {x_n}, \sequence {y_n} \in \CC$: :::$\quad \sequence {x_n} + \sequence {y_n} = \sequence {x_n + y_n}$ :$\circ : \quad \forall \sequence {x_n}, \sequence {y_n} \in \CC$: :::$\quad \sequence {x_n} \circ \sequence {y_n} = \sequence {x_n y_n}$ By Corollary to Null Sequences form Maximal Left and Right Ideal, $\NN$ is a maximal ideal of $\CC$.  By Corollary to Quotient Ring of Cauchy Sequences is Division Ring, the quotient ring $\Q_p = \CC \big / \NN$ is a field with field operations defined by: :$+ :\quad \forall \sequence {x_n} + \NN, \sequence {y_n} + \NN \in \CC \, \big / \NN$: :::$\quad \paren {\sequence {x_n} + \NN} + \paren {\sequence {y_n} + \NN} = \paren {\sequence {x_n} + \sequence {y_n} } + \NN$ :$\circ :\quad \forall \sequence {x_n} + \NN, \sequence {y_n} + \NN \in \CC \, \big / \NN$:  :::$\quad \paren {\sequence {x_n} + \NN} \paren {\sequence {y_n} + \NN} = \paren {\sequence {x_n} \circ \sequence {y_n} } + \NN$ Putting the operations on $\CC$ with those on $\CC \, \big / \NN$ gives: :$+ :\quad \forall \sequence {x_n} + \NN, \sequence{y_n} + \NN \in \CC \, \big / \NN$: :::$\quad \paren {\sequence {x_n} + \NN} + \paren {\sequence {y_n} + \NN} = \sequence {x_n + y_n} + \NN$ :$\circ :\quad \forall \sequence {x_n} + \NN, \sequence {y_n} + \NN \in \CC \, \big / \NN$:  :::$\quad \paren {\sequence {x_n} + \NN} \paren {\sequence {y_n} + \NN} = \sequence {x_n y_n} + \NN$ {{qed}} Category:P-adic Number Theory f5milhfzya6uhar91iybfwf5106kj9d
By definition, $K_n$ is the simple graph of order $n$ such that every vertex of $K_n$ is adjacent to all other vertices. So, let $G$ have the largest size of all simple graphs of order $n$. Then by definition of largest size, it is not possible for another edge to be added to $G$. The only way that could be is if all the vertices of $G$ are already adjacent to all other vertices. Hence $G = K_n$ by definition. The size of $G$ then follows from Size of Complete Graph. {{qed}}
It is seen that examples exist of simple graphs which fulfil the criteria where the order of $G$ is no greater than $3$: :400px The cases where the order of $G$ is $1$ or $2$ are trivial. When the order of $G$ is $3$, the criteria can be verified by inspection. Let the order of $G = \struct {V, E}$ be $4$ or more. Let $v_1, v_2, v_3, v_4 \in V$. Suppose every vertex is adjacent to every other vertex. As $v_1$ is adjacent to $v_2$, there exists the edge $v_1 v_2$. As $v_3$ is adjacent to $v_4$, there exists the edge $v_3 v_4$. But $v_1 v_2$ and $v_3 v_4$ both join two distinct pairs of vertices. Thus $v_1 v_2$ and $v_3 v_4$ are not adjacent, by definition. So when there are $4$ or more vertices in $G$, it cannot fulfil the criteria. {{qed}}
Suppose there exists a set of $3$ vertices that are connected. Then at least one of these vertices has at least $2$ edges. That would mean that at least $2$ edges were incident with the same vertex. That is, that at least $2$ edges were adjacent to each other. So, for a simple graph to fulfil the conditions, vertices can exist only in $2$-vertex components. So such a simple graph must have an even number of vertices. {{qed}}
By definition of star graph, one vertex is adjacent to all the $n - 1$ other vertices. So the size of $S_n$ is at least $n - 1$. All the other vertices are of degree $1$, and the edges they are incident with are the ones joining them to the distinguished vertex. Those edges have already been counted. So the size of $S_n$ is exactly $n - 1$. {{qed}} Category:Star Graphs Category:Star Graphs ot5nmixbbq2jsccss1d7ef75lpdn6gq
In order for the one vertex in question to be adjacent to all the others, it needs to be incident to $n - 1$ edges. This is the smallest number of edges required. Hence $G$ is a star graph. From Size of Star Graph, the size of $G$ is thus $n - 1$. {{qed}}
If $G$ has no edges, then all the vertices are incident to all the edges vacuously. So any of the edgeless graphs $N_n$ for order $n \in \Z_{\ge 0}$ fulfils the criterion. Suppose $G$ has more than $2$ vertices $v_1, v_2, v_3$ and at least one edge. {{WLOG}}, let one edge be $v_1 v_2$. But $v_3$ cannot be incident to edge $v_1 v_2$. So $G$ can have no more than $2$ vertices. Furthermore, there can be only one edge joining those two vertices. The result follows from inspection of $K_2$ and $S_2$. {{qed}}
=== Necessary Condition === {{:Heine-Borel Theorem/Necessary Condition}}{{qed|lemma}} === Sufficient Condition === {{:Heine-Borel Theorem/Sufficient Condition}}{{qed}}
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $K$. $\sequence {x_n}_{n \mathop \in \N}$ is bounded. Hence, $\sequence {x_n}_{n \mathop \in \N}$ has a convergent subsequence $\sequence {x_{n_k}}_{k \mathop \in \N}$. Denote the limit $\displaystyle \lim_{k \mathop \to \infty} \sequence {x_{n_k}} = L$. By assumption, $K$ is closed. By equivalence of definitions, $L \in K$. By definition, $K$ is compact.
A cycle graph is defined as a (simple) graph which consists of a single cycle. So a cycle graph consists of just one component, and hence is connected. A cycle, at base, is a closed walk with the properties: :all its edges are distinct :all its vertices (except for the start and end) are distinct {{finish|I hate trying to prove obvious things}}
The proof proceeds by induction. For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition: :$\size {D_n} = n \paren {n - 1}$ === Basis for the Induction === $\map P 1$ is the case: {{begin-eqn}} {{eqn | l = \size {D_n}       | r = 0       | c = as $D_n$ has no arcs }} {{eqn | r = 1 \times \paren {1 - 1}       | c =  }} {{end-eqn}} Thus $\map P 1$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown 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 the induction hypothesis: :$\size {D_k} = k \paren {k - 1}$ from which it is to be shown that: :$\size {D_{k + 1} } = \paren {k + 1} k$ === Induction Step === This is the induction step: Let $D_{k + 1}$ be constructed by adding a new vertex $v_{k + 1}$ to $D_k$. To do so, it is necessary to add $2$ arcs to join $v_{k + 1}$ to every vertex of $D_k$. Thus there are a total of $2 k$ arcs more in $D_{k + 1}$ than there are in $D_k$. So: {{begin-eqn}} {{eqn | l = \size {D_{k + 1} }       | r = \size {D_k} + 2 k       | c = from the above analysis }} {{eqn | r = k \paren {k - 1} + 2 k       | c = Induction Hypothesis }} {{eqn | r = k^2 - k + 2 k       | c =  }} {{eqn | r = k^2 + k       | c =  }} {{eqn | r = \paren {k + 1} k       | c =  }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{\ge 0}: \size {D_n} = n \paren {n - 1}$ {{qed}}
From P-adic Norm is Non-Archimedean Norm, $p$-adic norm is a non-Archimedean norm on the rationals $\Q$. Since $\sequence {x_n}$ does not converge to $0$, from Non-Null Cauchy Sequence in Non-Archimedean Norm is Eventually Stationary: :$\exists N \in \N: \forall n, m > N: \norm {x_n}_p = \norm {x_m}_p$ By definition of the $p$-adic norm on $\Q$: :$\exists v \in \Z: \norm {x_{N + 1} }_p = p^{-v}$ Therefore: :$\forall n > N: \norm {x_n}_p = \norm {x_{N + 1} }_p = p^{-v}$ Let $\sequence {y_n}$ be the subsequence of $\sequence {\norm {x_n}}$ defined by: :$\forall n: y_n = \norm {x_{N + n} }_p$ Then $\sequence {y_n}$ is the constant sequence $\tuple {p^{-v}, p^{-v}, p^{-v}, \dotsc}$. Therefore: {{begin-eqn}} {{eqn | l = \displaystyle \lim_{n \mathop \to \infty} \norm {x_n}_p       | r = \displaystyle \lim_{n \mathop \to \infty} y_n       | c = Limit of Subsequence equals Limit of Real Sequence }} {{eqn | r = p^{-v}       | c = Constant Sequence Converges to Constant in Normed Division Ring }} {{end-eqn}} {{qed}} Category:P-adic Norm of p-adic Number is Power of p mv1t19x0fjiu86lqchnqfz3jrxmpzha
By definition of the $p$-adic numbers $\Q_p$, the rational numbers $\Q$ are dense in $\Q_p$. By the definition of a dense subset then $\map \cl \Q = \Q_p$. By Closure of Subset of Metric Space by Convergent Sequence then: :there exists a sequence $\sequence {x_n} \subseteq \Q$ that converges to $x$. That is: :$\displaystyle \lim_{n \mathop \to \infty} x_n = x$ From Modulus of Limit: :$\displaystyle \lim_{n \mathop \to \infty} \norm{x_n}_p = \norm x_p$ By Convergent Sequence in Normed Division Ring is Cauchy Sequence, $\sequence {x_n}$ is a Cauchy sequence in $\struct {\Q_p, \norm {\,\cdot\,}_p}$. From Sequence is Cauchy in P-adic Norm iff Cauchy in P-adic Numbers, $\sequence {x_n}$ is a Cauchy sequence in $\struct {\Q, \norm {\,\cdot\,}_p}$. From Lemma: :$\exists v \in \Z: \norm x_p = \lim_{n \mathop \to \infty} \norm{x_n}_p = p^{-v}$
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers as a quotient of Cauchy sequences. That is, $\Q_p$ is the quotient ring $\CC \, \big / \NN$ where: :$\CC$ denotes the commutative ring of Cauchy sequences over $\struct {\Q, \norm {\,\cdot\,}_p}$ :$\NN$ denotes the set of null sequences in $\struct {\Q, \norm {\,\cdot\,}_p}$. Then $x$ is a left coset in $\CC \, \big / \NN$. Let $\sequence{x_n}$ be any Cauchy sequence in $x$. From Lemma: :$\exists v \in \Z: \displaystyle \lim_{n \mathop \to \infty} \norm{x_n}_p = p^{-v}$ By definition of the $p$-adic norm on the $p$-adic numbers as a quotient of Cauchy sequences: :$\norm x_p = \displaystyle \lim_{n \mathop \to \infty} \norm{x_n}_p = p^{-v}$
A loop-digraph is a graph such that: :it is not necessarily antireflexive :it is not necessarily symmetric. Hence a loop-digraph is an ordered pair: :$G = \struct {V, E}$ such that: :$V$ is a set of objects :$E$ is a relation on $V$. This defines a relational structure. {{qed}}
First we note that when $n = 2$ there is one edge in $K_n$. So if you start at one vertex $u$ and travel along that edge to the other vertex $v$, you cannot return to $u$ except by using that same edge. Consequently $K_2$ is not Hamiltonian. {{qed|lemma}} Let $n = 4$. Let us take two vertices $u, v$ of $K_n$: From Complete Graph is Regular, the degrees of $u$ and $v$ are given by: :$\map \deg u = \map \deg v = n - 1$ Let us remove the edge $u v$ which joins them. Let the resulting graph be denoted $G$. We now have that: :$\map \deg u = \map \deg v = n - 2$ From Ore's Theorem, $G$ is Hamiltonian if for each pair of non-adjacent vertices $u, v \in V$: :$\deg u + \deg v \ge n$ In $G$, we have that: Hence: {{begin-eqn}} {{eqn | l = \deg u + \deg v       | r = 2 \paren {n - 2}       | c =  }} {{eqn | r = n + \paren {n - 4}       | c =  }} {{eqn | o = \ge       | r = n       | c = as $n \ge 4$ }} {{end-eqn}} Thus $G$ is Hamiltonian. A Hamiltonian graph with another edge added is still Hamiltonian. Thus restoring the edge $u v$ to $G$ to turn it back into $K_n$ means that $K_n$ is Hamiltonian. {{qed|lemma}} When $n = 3$ we cannot use this theorem, as you then find for the resulting $G$: :$\deg u + \deg v = 2$ which is less than $3$. Instead we inspect the complete graph $K_3$ and see it is is the cycle graph $C_3$ :200px The result follows from Cycle Graph is Hamiltonian. {{qed}} Category:Complete Graphs Category:Hamiltonian Graphs cqxn2bajtfwoms7u8vcqv9cvu9gkvym
A undirected graph that contains a cycle that contains every vertex of the graph (but not necessarily every edge). A cycle graph is a graph which consists of a single cycle. So all the vertices of $C_n$ are on the same cycle. The result follows. {{Qed}} Category:Cycle Graphs Category:Hamiltonian Graphs 4nys7723i3gtakbnu2v9e1oz3ij68cc
By definition, a simple graph has no loops or multiple edges. So a vertex can be incident to only as many edges that will join it to all the other vertices once each. There are $\card V - 1$ other vertices. Hence the result. {{qed}}
Suppose we are given the degrees of each of the vertices of $G$. By definition, $\card V$ is simply the count of the vertices. So if we have been given the degrees of each of the vertices, we must know how many vertices there are. Then from the Handshake Lemma: :$\displaystyle \sum_{v \mathop \in V} \map \deg v = 2 \card E$ That is, the sum of the degrees of all the vertices is equal to twice $\card E$. So to find the size of $G$, add up the degrees of all the vertices and divide by $2$. {{qed}}
The following $2$ simple graphs both have order $4$ and size $4$: :400px But: :the graph on the left has vertices with degrees $1, 2, 2, 3$ :the graph on the right has vertices with degrees $2, 2, 2, 2$. {{qed}}
Let $G$ be of order $p$, size $q$ and $r$-regular. Then by definition $G$ has: :$p$ vertices each of which is of degree $r$. :$q$ edges. Thus: {{begin-eqn}} {{eqn | l = q       | r = \dfrac 1 2 \sum_{v \mathop \in V} \map \deg v       | c = Handshake Lemma }} {{eqn | r = \dfrac 1 2 \sum_{v \mathop \in V} r       | c = {{Defof|Regular Graph}} }} {{eqn | r = \dfrac 1 2 p r       | c = {{Defof|Order of Graph}} }} {{end-eqn}} If $p r$ is odd, that means $p$ is odd and $r$ is odd. That means there is an odd number of odd vertices. From the corollary to the Handshake Lemma, this is not possible Thus for $G$ to exist, it is necessary for $p r$ to be even. The result follows. {{Qed}}
Proof by Counterexample: :400px Consider a bijection $\phi: \map V {G_1} \to \map V {G_2}$, where $G_1$ is the graph on the left and $G_2$ is the graph on the right. The vertices $v_1$, $v_2$ and $v_5$ of $G_2$ are each adjacent to both of the others. Because $\phi$ is a bijection, it must map $3$ vertices of $G_1$ to $v_1$, $v_2$ and $v_5$. For $\phi$ to be an isomorphism, two of the vertices of $G_1$ are adjacent {{iff}} the two image vertices in $G_2$ udner $\phi$ are also adjacent. So the $3$ vertices of $G_1$ which map to $v_1$, $v_2$ and $v_5$ of $G_2$ must also each be adjacent to both of the others. But $G_1$ does not contain $3$ such vertices. It follows that there is no such isomorphism from $\map V {G_1}$ to $\map V {G_2}$. That is, $G_1$ and $G_2$ are not isomorphic. {{qed}}
There is only one bijection from $\map V {G_1}$ to $\map V {G_2}$. There are no vertices adjacent to the sole vertex in $\map V {G_1}$ There are no vertices adjacent to the sole vertex in $\map V {G_2}$. Hence the result, vacuously. {{qed}}
Let $\phi$ be a bijection from $\map V {G_1}$ to $\map V {G_2}$. This is possible because $\card {\map V {G_1} } = \card {\map V {G_2} } = n$. Let $v_a \in \map V {G_1}$. By definition, $v_a$ is adjacent to no other vertices of $G_1$. Similarly by definition, $\map \phi {v_a}$ is adjacent to no other vertices of $G_2$. Thus, vacuously, $v_a$ and $v_b$ are adjacent in $G_1$ {{iff}} $\map \phi {v_a}$ and $\map \phi {v_b}$ are adjacent in $G_2$ Hence the result {{qed}} Category:Graph Isomorphisms q141xpa8ckxti7c24gq03oxnujxpixq
The fact that the $3$ graphs given are not isomorphic follows from Vertex Condition for Isomorphic Graphs. The vertices have degrees as follows: :Graph $1$: $2, 2, 1, 1$ :Graph $2$: $3, 1, 1, 1$ :Graph $3$: $2, 2, 2, 0$ {{finish|Now to prove there are no more isomorphism groups.}}
From the definition of a first-order reaction, the rate of change of the quantity of the substance is proportional to the quantity of the substance present at any time. As the rate of change is a decrease, this rate will be negative. Thus the differential equation governing this reaction is given by: :$-\dfrac {\d x} {\d t} = k x$ for some $k \in \R_{>0}$. {{qed}}
Let $x_0$ denote the ratio of carbon-14 to carbon-12 in $Q$ at the time of its death. Let $x$ denote the ratio of carbon-14 to carbon-12 in $Q$ after time $t$. Thus: :$r = \dfrac x {x_0}$ It is assumed that the rate of decay of carbon-14 is a first-order reaction. Hence we use: {{begin-eqn}} {{eqn | l = x       | r = x_0 e^{-k t}       | c = First-Order Reaction }} {{eqn | ll= \leadsto       | l = \dfrac x {x_0}       | r = e^{-k t}       | c =  }} {{eqn | ll= \leadsto       | l = -k t       | r = \map \ln {\dfrac x {x_0} }       | c =  }} {{eqn | ll= \leadsto       | l = -k t       | r = \map \ln r       | c =  }} {{eqn | ll= \leadsto       | l = t       | r = -\dfrac {\ln r} k       | c =  }} {{end-eqn}} Chemical analysis tells us that after $10$ years, $99.876 \%$ of the carbon-14 that was present in the organic matter still remains. Thus at time $t = 10$, we have: {{begin-eqn}} {{eqn | l = 0.99876 x_0       | r = x_0 e^{-10 k}       | c = substituting for $t$ and $x$ in $(1)$ }} {{eqn | ll= \leadsto       | l = \ln 0.99876       | r = -10 k       | c =  }} {{eqn | ll= \leadsto       | l = k       | o = \approx       | r = 0.000124       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 1 k       | o = \approx       | r = 8060       | c =  }} {{end-eqn}} Hence the result. {{qed}}
Let $d$ be the metric induced by the norm $\norm {\, \cdot \,}$. By definition of a complete normed division ring, the metric space $\struct{R, d}$ is complete. From Leigh.Samphier/Sandbox/Inclusion Mapping on Normed Division Subring is Distance Preserving Monomorphism, $i : S \to R$ is a distance-preserving ring monomorphism. From Image under Inclusion Mapping: :$\map {i^\to} S = S$ Thus $\map {i^\to} S$ is dense in $\struct {R, \norm {\, \cdot \,} }$. It follows that $\struct {R, \norm {\, \cdot \,} }$ is a completion of $\struct {S, \norm {\, \cdot \,}}$ by definition. {{qed}} Category:Completion of Normed Division Ring aat9tvf411affe5v08pi40bw0ltscgi
{{begin-eqn}} {{eqn | l = \norm x_1       | r = \norm {x - 0}_1       | c =  }} {{eqn | r = \norm {\map \phi x - \map \phi 0}_2       | c = As $\phi$ is distance-preserving }} {{eqn | r = \norm {\map \phi {x - 0} }_2       | c = As $\phi$ is a ring homomorphism }} {{eqn | r = \norm {\map \phi x}_2       | c = }} {{end-eqn}} {{qed}} Category:Completion of Normed Division Ring jwy6taffeba678zhowexgahutv7yko8
Let $x \in R_2$. By definition of the completion $\struct {R_2, \norm {\, \cdot \,}_2 }$: :$\map {\phi^\to} {R_1}$ is a dense subset of $\struct {R_2, \norm {\, \cdot \,}_2 }$. By the definition of a dense subset: :$\map \cl {\map {\phi^\to} {R_1}} = R_2$ By Closure of Subset of Metric Space by Convergent Sequence: :there exists a sequence $\sequence {y_n} \subseteq \map {\phi^\to} {R_1}$ that converges to $x$ That is: :$\displaystyle \lim_{n \mathop \to \infty} y_n = x$ From Injection to Image is Bijection, we can define: :$\forall n \in \N : x_n = \map {\phi^{-1}} {y_n}$ Then $\sequence{x_n}$ is a sequence in $R_1$: :$\displaystyle \lim_{n \mathop \to \infty} \map \phi {x_n} = \lim_{n \mathop \to \infty} y_n = x$ {{qed}} Category:Completion of Normed Division Ring 618jsapz7vqhlo8rv16ux84u37xkck8
From Leigh.Samphier/Sandbox/Complete Normed Division Ring is Completion of Dense Subring: :$\struct {R, \norm {\, \cdot \,} }$ is a completion of $\struct {S, \norm {\, \cdot \,}}$  where the inclusion mapping $i : S \to R$ is the required distance-preserving ring monomorphism. From Leigh.Samphier/Sandbox/Element of Completion is Limit of Sequence in Normed Division Ring: :for all $x \in R$, there exists a sequence $\sequence{x_n}$ in $S$: ::$\quad x = \displaystyle \lim_{n \mathop \to \infty} \map i {x_n}$ That is: :for all $x \in R$, there exists a sequence $\sequence{x_n}$ in $S$: ::$\quad x = \displaystyle \lim_{n \mathop \to \infty} x_n$ {{qed}} Category:Completion of Normed Division Ring pcx12a1d0qgl2952fx99gp4wd4wj91j
From Inclusion Mapping on Subring is Monomorphism, $i$ is a ring monomorphism. Let $d$ be the metric induced by the norm $\norm {\, \cdot \,}$. Let $d_S$ be the metric induced by the norm $\norm {\, \cdot \,}_S$. Thus: {{begin-eqn}} {{eqn | lo = \forall x, y \in S :        | l = \map d {\map i x, \map i y}       | r = \map d {x, y}       | c = {{Defof|Inclusion Mapping}} }} {{eqn | r = \norm {x - y}       | c = {{Defof|Metric Induced by Norm on Division Ring}} }} {{eqn | r = \norm {x - y}_S       | c = {{Defof|Normed Division Subring|Norm on Subring}} }} {{eqn | r = \map {d_S} {x, y}       | c = {{Defof|Metric Induced by Norm on Division Ring}} }} {{end-eqn}} It follows that $i$ is distance-preserving by definition. {{qed}} Category:Completion of Normed Division Ring ijjb8zqa2nqgvagkiab34103jzk9rrw
=== Case $1$: $p \divides a$ === Let $q = \dfrac a p$. Then $\dfrac 1 p = \dfrac q a$. So the decimal expansion of $\dfrac 1 p$ in base $a$ is $0.q$ and terminates. {{qed|lemma}} === Case $2$: $p \nmid a$ === Let $d$ be the order of $a$ modulo $p$. By definition, $d$ is the smallest integer such that: :$a^d \equiv 1 \pmod p$ By definition of modulo arithmetic: :$\exists x \in \Z: a^d - 1 = xp$ We can rearrange the terms to achieve the following expression: {{begin-eqn}} {{eqn | l = \frac 1 p       | r = \frac x {a^d - 1}       | c =  }} {{eqn | r = \frac x {a^d} \frac 1 {1 - \frac 1 {a^d} }       | c = factorizing }} {{eqn | r = \frac x {a^d} \paren {1 + \paren {\frac 1 {a^d} } + \paren {\frac 1 {a^d} }^2 + \paren {\frac 1 {a^d} }^3 + \cdots}       | c = From Power Series Expansion of Reciprocal of 1 + x, noting that $-1 < -\dfrac 1 {a^d} < 1$ }} {{eqn | r = \paren {\frac 1 {a^d} } x + \paren {\frac 1 {a^d} }^2 x + \paren {\frac 1 {a^d} }^3 x + \cdots       | c = rerranging }} {{end-eqn}} so the decimal expansion of $\dfrac 1 p$ in base $a$ is $0.xxx...$ and is periodic of length $d$, which is the order of $a$ modulo $p$. {{qed}} Category:Reciprocals Category:Prime Numbers fnwzgo3j164ndm49huzfp9r4ja857n0
When $p \divides 10$, $\dfrac 1 p$ expressed in decimal notation is a terminating fraction: {{begin-eqn}} {{eqn | l = \dfrac 1 2       | r = 0 \cdotp 5 }} {{eqn | l = \dfrac 1 5       | r = 0 \cdotp 2 }} {{end-eqn}} So let $p$ be such that $p \nmid 10$. From Period of Reciprocal of Prime, the period of recurrence is the order of $10$ modulo $p$. That is, it is the smallest integer $d$ such that: :$10^d \equiv 1 \pmod p$ From Fermat's Little Theorem: :$10^{p - 1} \equiv 1 \pmod p$ Hence the maximum period of recurrence occurs when the order of $10$ modulo $p$ is $p - 1$. To demonstrate that the maximum period of recurrence occurs for at least one prime number, we take $7$ as an example. $7$ is a prime number which is not a divisor of $10$. From Period of Reciprocal of 7 is of Maximal Length: :$\dfrac 1 7 = 0 \cdotp \dot 14285 \dot 7$ recurring with period $7 - 1 = 6$. {{qed}} Category:Reciprocals Category:Prime Numbers g5np210vmwpxc9l37wsqtyugnsgbj3f