hcju/proofwikiqa huggingface dataset:
=== Integral Form implies Continued Fraction Form === Let $G$ be the integral form of the Euler-Gompertz constant. Then by definition: :$G = \displaystyle \int_0^\infty \dfrac {e^{-u} } {1 + u} \rd u$ {{ProofWanted}} Thus $G$ equals the continued fraction form of the Euler-Gompertz constant. {{qed|lemma}} === Continued Fraction Form implies Integral Form === Let $G$ be the continued fraction form of the Euler-Gompertz constant. Then by definition: :$G = \cfrac 1 {2 - \cfrac {1^2} {4 - \cfrac {2^2} {6 - \cfrac {3^2} {8 - \cfrac {4^2} {10 - \dotsb} } } } }$ {{ProofWanted}} Thus $G$ equals the integral form of the Euler-Gompertz constant. {{qed}}
By Sum to Infinity of 2x^2n over n by 2n Choose n: {{begin-eqn}} {{eqn | l = \frac {2 x \arcsin x} {\sqrt {1 - x^2} }       | r = \sum_{n \mathop = 1}^\infty \frac {\paren {2 x}^{2 n} } {n \dbinom {2 n} n} }} {{eqn | l = \frac {2 \paren {\frac 1 2} \arcsin \frac 1 2} {\sqrt {1 - \paren {\frac 1 2}^2} }       | r = \sum_{n \mathop = 1}^\infty \frac {\paren {2 \paren {\frac 1 2} }^{2 n} } {n \dbinom {2 n} n}       | c = substituting $x = \dfrac 1 2$ }} {{eqn | l = \frac {\frac \pi 6} {\sqrt {\frac 3 4} }       | r = \sum_{n \mathop = 1}^\infty \frac 1 {n \dbinom {2 n} n}       | c = Sine of $30^\circ$ }} {{eqn | l = \sum_{n \mathop = 1}^\infty \frac 1 {n \dbinom {2 n} n}       | r = \frac {\pi \sqrt 3} 9 }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \phi       | r = \dfrac 1 {\phi - 1}       | c = {{Defof|Golden Mean|index = 3}} }} {{eqn | ll= \leadsto       | l = \dfrac 1 \phi       | r = \phi - 1       | c = taking reciprocals of both sides }} {{end-eqn}} The numeric approximation follows from the decimal expansion of the Golden Mean: :{{Definition:Golden Mean/Decimal Expansion}}{{qed}}
That no rational number such that $x \ge \dfrac {\pi^2} 6 - 1$ can be so expressed follows from Riemann Zeta Function of 2: :$\displaystyle \sum_{n \mathop = 1}^n \dfrac 1 {n^2} = 1 + \dfrac 1 {2^2} + \dfrac 1 {3^2} + \dotsb = \dfrac {\pi^2} 6$ That is, using ''all'' the  reciprocals of distinct squares, you can never get as high as $\dfrac {\pi^2} 6 - 1$. It remains to be shown that for all rational numbers $x$ less than $\dfrac {\pi^2} 6 - 1$, you can make $x$ with a subset of them. {{ProofWanted}}
It needs to be shown that $\mathscr I$ satisfies the matroid axioms $(I1)$, $(I2)$ and $(I3)$. === Matroid Axiom $(I1)$ === Now: {{begin-eqn}} {{eqn | l = \size \O       | r = 0       | c = Cardinality of Empty Set }} {{eqn | o = \le       | r = k       | c = Assumption }} {{end-eqn}} By definition of the uniform matroid of rank $k$: :$\O \in \mathscr I$ It follows that $\mathscr I$ satisfies matroid axiom $(I1)$. {{qed|lemma}} === Matroid Axiom $(I2)$ === Let $X \in \mathscr I$. Let $Y \subseteq X$. Then: {{begin-eqn}} {{eqn | l = \size Y       | o = \le       | r = \size X       | c = Cardinality of Subset of Finite Set }} {{eqn | o = \le       | r = k       | c = {{Defof|Uniform Matroid|uniform matroid of rank $k$}} }} {{end-eqn}} By definition of the uniform matroid of rank $k$: :$Y \in \mathscr I$ It follows that $\mathscr I$ satisfies matroid axiom $(I2)$. {{qed|lemma}} === Matroid Axiom $(I3)$ === Let $U, V \in \mathscr I$. Let $\size V < \size U$. From Intersection is Subset: :$U \cap V \subseteq V$ :$U \cap V \subseteq U$ Then: {{begin-eqn}} {{eqn | l = \size {U \cap V}       | o = \le       | r = \size V       | c = Cardinality of Subset of Finite Set }} {{eqn | o = <       | r = \size U       | c = Assumption }} {{end-eqn}} Then: :$U \cap V \neq U$ It follows that: :$U \cap V \subsetneqq U$ Then: {{begin-eqn}} {{eqn | l = \exists x       | o = \in       | r = U \setminus {U \cap V}       | c = Set Difference with Proper Subset }} {{eqn | r = U \setminus V       | c = Set Difference with Intersection is Difference }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \size {V \cup \set x}       | r = \size V + \size {\set x}       | c = Corollary to Cardinality of Set Union }} {{eqn | r = \size V + 1       | c = Cardinality of Singleton }} {{eqn | o = \le       | r = \size U       | c = Assumption }} {{eqn | o = \le       | r = k       | c = {{Defof|Uniform Matroid|uniform matroid of rank $k$}} }} {{end-eqn}} By definition of the uniform matroid of rank $k$: :$V \cup \set x \in \mathscr I$ It follows that $\mathscr I$ satisfies matroid axiom $(I3)$. {{qed}}
Let $S$ have cardinality $n$. Let $\struct{S, \mathscr I_{n,n}}$ be the uniform matroid of rank $n$. From Cardinality of Proper Subset of Finite Set, every subset of $S$ has cardinality less than or equal to $n$. It follows that $\mathscr I_{n,n} = \powerset S$. From Uniform Matroid is Matroid, then $\struct{S, \powerset S}$ Is a matroid. {{qed}}
It needs to be shown that $\mathscr I$ satisfies the matroid axioms $(I1)$, $(I2)$ and $(I3)$. === Matroid Axiom $(I1)$ === {{ProofWanted}} === Matroid Axiom $(I2)$ === {{ProofWanted}} === Matroid Axiom $(I3)$ === {{ProofWanted}}
Let $\mathscr Z = \set {Z \subseteq Y \setminus X : X \cup Z \in \mathscr I}$ Note that $\O \in \mathscr Z $ So $\mathscr Z \ne \O$ Let $Z_0 \in \mathscr Z : \size {Z_0} = \max \set {\size Z : Z \in \mathscr Z}$ {{AimForCont}}: :$\size {X \cup Z_0} < \size Y$ By matroid axiom $(\text I 3)$: :$\exists y \in Y \setminus \paren{X \cup Z_0} : X \cup Z_0 \cup \set y \in \mathscr I$ By choice of $Z_0$ and $y$: :$Z_0 \cup \set y \subseteq Y \setminus X$ So: :$Z_0 \cup \set y \in \mathscr Z$ Then: {{begin-eqn}} {{eqn | l = \size {Z_0 \cup \set y}       | r = \size {Z_0} + \size {\set y}       | c = Corollary to Cardinality of Set Union }} {{eqn | r = \size {Z_0} + 1       | c = Cardinality of Singleton }} {{eqn | o = >       | r = \size {Z_0} }} {{end-eqn}} This contradicts the choice of $Z_0$. It follows that: {{begin-eqn}} {{eqn | l = \size Y       | o = \le       | r = \size {X \cup Z_0} }} {{eqn | r = \size X + \size {Z_0}       | c = Corollary to Cardinality of Set Union }} {{eqn | ll= \leadsto       | l = \size {Z_0}       | o = \ge       | r = \size Y - \size X }} {{end-eqn}} From Finite Set Contains Subset of Smaller Cardinality: :$\exists Z \subseteq Z_0 : \size Z = \size Y - \size X$ Then: {{begin-eqn}} {{eqn | l = \size {X \cup Z}       | r = \size X + \size Z       | c = Corollary to Cardinality of Set Union }} {{eqn | r = \size Y }} {{end-eqn}} We have: {{begin-eqn}} {{eqn | l = \card Z       | r = \card Y - \card X }} {{eqn | o = >       | r = 0       | c = As $\card X < \card Y$ }} {{eqn | ll = \leadsto       | l = Z       | o = \neq       | r = \O       | c = Cardinality of Empty Set }} {{end-eqn}} {{qed}}
By definition of the rank function: :$\map \rho S = \max \set{\size X : X \subseteq S, X \in \mathscr I}$ Let $B_1$ be an independent subset such that: :$\size {B_1} = \map \rho S$ It is shown that $B_1$ is a base. Let $X$ be an independent superset of $B_1$. From Cardinality of Subset of Finite Set: :$\size {B_1} \le \size X$ By choice of $B_1$: :$\size X \le \size {B_1}$ It follows that: :$\size X = \size {B_1}$ From Cardinality of Proper Subset of Finite Set: :$X = B_1$ It follows that $B_1$ is a maximal independent subset. That is, $B_1$ is a base. It is now shown that any other base has the same cardinality as $B_1$. Let $B_2$ be any other base. By choice of $B_1$: :$\size {B_2} \le \size {B_1}$ Aiming for a contradiction, suppose $\size {B_2} < \size {B_1}$. From Independent Set can be Augmented by Larger Independent Set there exists $Z \subseteq B_1 \setminus B_2$ such that: :$B_2 \cup Z \in \mathscr I$ :$\size{B_2 \cup Z} = \size{B_1}$ This contradicts the maximality of $B_2$ in $\mathscr I$. Then: :$\size {B_2} = \size {B_1} = \map \rho S$  {{qed}}
=== Necessary Condition === Let $\mathscr B$ be the set of bases of the matroid on $M = \struct{S, \mathscr I}$ {{:Leigh.Samphier/Sandbox/Matroid Satisfies Base Axiom/Necessary Condition}}{{qed|lemma}} === Sufficient Condition === Let $\mathscr B$ satisfies the base axiom: {{:Definition:Base Axiom (Matroid)}} {{:Leigh.Samphier/Sandbox/Matroid Satisfies Base Axiom/Sufficient Condition}}{{qed}}
Let $\mathbb P$ denote the set of prime numbers. Every set of consecutive primes whose sum is less than $x$ will contribute $1$ to the sum: :$\map f 1 + \map f 2 + \dotsb + \map f x$ The number of such sets of $r$ primes is clearly at most: :$\map \pi {x / r}$ and at least: :$\map \pi {x / r} - r$ where $\pi$ is the prime-counting function. Hence: :$\displaystyle \sum_{r \mathop = 1}^k \paren {\map \pi {x / r} - r} \le \map f 1 + \map f 2 + \dotsb + \map f x \le \sum_{r \mathop = 1}^k \map \pi {x / r}$ where $k$ is determined by: :$(1): \quad p_1 + p_2 + \dotsb + p_k < x < p_1 + p_2 + . . . + p_{k + 1}$ From $(1)$ and the well known $p_r \asymp r \ln r$ we find: :$(2): \quad k \asymp \sqrt {\dfrac x {\ln x} }$ Here $f \asymp g$ denotes that $\dfrac f g$ is bounded above and below by positive numbers for large arguments. This implies: :$\displaystyle \sum_{r \mathop = 1}^k r = \map o x$ where $\map o x$ denotes little-o notation. Hence, if we can show that: :$\displaystyle \sum_{r \mathop = 1}^k \map \pi {x / r} \approx \ln 2$ it will follow that: :$\map f 1 + \map f 2 + \dotsb + \map f x \approx x \ln 2$ Using $(2)$ and the Prime Number Theorem, the following estimates are justified: {{begin-eqn}} {{eqn | l = \sum_{r \mathop = 1}^k \map \pi {x / r}       | o = \approx       | r = \sum_{r \mathop = 1}^k \dfrac x {r \, \map \ln {x / r} }       | c =  }} {{eqn | o = \approx       | r = \int_1^k \dfrac {x \rd r} {r \, \map \ln {x / r} }       | c =  }} {{eqn | r = \bigintlimits {-x \map \ln {\map \ln {x / r} } } 1 k       | c =  }} {{eqn | r = x \paren {\ln \ln x - \map {\ln \ln} {x / k} }       | c =  }} {{eqn | o = \approx       | r = x \paren {\ln \ln x - \ln \ln \sqrt {x \ln x} }       | c =  }} {{eqn | o = \approx       | r = \ln 2       | c =  }} {{end-eqn}} {{qed}}
By Sum to Infinity of 2x^2n over n by 2n Choose n: {{begin-eqn}} {{eqn | l = \sum_{n \mathop = 1}^\infty \frac {\paren {2 x}^{2 n} } {n \dbinom {2 n} n}       | r = \frac {2 x \arcsin x} {\sqrt {1 - x^2} } }} {{eqn | ll= \leadsto       | l = \sum_{n \mathop = 1}^\infty \frac {2 n \paren {2^{2 n} x^{2 n - 1} } } {n \dbinom {2 n} n}       | r = \frac {\paren {2 \arcsin x + \frac {2 x} {\sqrt {1 - x^2} } } \sqrt {1 - x^2} - 2 x \arcsin x \paren {\frac 1 2} \paren {-2 x} \paren {1 - x^2}^{-\frac 1 2} } {1 - x^2}       | c = differentiating both sides {{WRT|Differentiation}} $x$ }} {{eqn | r = \frac {2 x + 2 \arcsin x \sqrt {1 - x^2} + 2 x^2 \arcsin x \paren {1 - x^2}^{-\frac 1 2} } {1 - x^2} }} {{eqn | r = \frac {2 x} {1 - x^2} + \frac {2 \arcsin x \paren {\paren {1 - x^2} + x^2} } {\paren {1 - x^2}^{\frac 3 2} } }} {{eqn | r = \frac {2 x} {1 - x^2} + \frac {2 \arcsin x} {\paren {1 - x^2}^{\frac 3 2} } }} {{eqn | ll= \leadsto       | l = \sum_{n \mathop = 1}^\infty \frac 4 {\dbinom {2 n} n}       | r = \frac {2 \paren {\frac 1 2} } {1 - \paren {\frac 1 2}^2} + \frac {2 \arcsin \paren {\frac 1 2} } {\paren {1 - \paren {\frac 1 2}^2}^{\frac 3 2} }       | c = substituting $x = \dfrac 1 2$ }} {{eqn | r = \frac 1 {\frac 3 4} + \frac {2 \paren {\frac \pi 6} } {\paren {\frac 3 4}^{\frac 3 2} }       | c = Sine of $30 \degrees$ }} {{eqn | r = \frac 4 3 + \frac \pi 3 \times \frac 8 {3 \sqrt 3} }} {{eqn | r = \frac {36 + 8 \pi \sqrt 3} {27} }} {{eqn | ll= \leadsto       | l = \sum_{n \mathop = 1}^\infty \frac 1 {\dbinom {2 n} n}       | r = \frac {9 + 2 \pi \sqrt 3} {27} }} {{end-eqn}} {{qed}}
From Coherent Sequence is Partial Sum of P-adic Expansion there exists a unique $p$-adic expansion of the form: :$\displaystyle \sum_{n \mathop = 0}^\infty d_n p^n$ such that: :$\forall n \in \N: \alpha_n = \displaystyle \sum_{i \mathop = 0}^n d_i p^i$ From P-adic Expansion Converges to P-adic Number: :$\exists x \in \Q_p : \displaystyle \lim_{n \mathop \to \infty} \alpha_n = x$ By definition of a coherent sequence: :$\forall n \in \N: \alpha_n \in \Z \subseteq \Z_p$ By definition, the $p$-adic integers $\Z_p$ is the closed ball $\map {B^-_1} 0$. From Closed Ball of Non-Archimedean Division Ring is Clopen, $\Z_p$ is closed in $\norm{\,\cdot\,}_p$. From Subset of Metric Space contains Limits of Sequences iff Closed: :$x \in \Z_p$ The result follows. {{qed}}
From P-adic Integer is Limit of Unique Coherent Sequence of Integers, there exists a unique coherent sequence $\sequence{\alpha_n}$ such that: :$\displaystyle \lim_{n \mathop \to \infty} \alpha_n = x$ From Coherent Sequence is Partial Sum of P-adic Expansion, there exists a unique $p$-adic expansion of the form: :$\displaystyle \sum_{n \mathop = 0}^\infty d_n p^n$ such that: :$\forall n \in \N: \alpha_n = \displaystyle \sum_{i \mathop = 0}^n d_i p^i$ That is: :$\displaystyle \lim_{n \mathop \to \infty} \sum_{i \mathop = 0}^n d_i p^i = x$ {{qed}}
Let $\Z_p$ be the $p$-adic integers. From P-adic Number times Integer Power of p is P-adic Integer: :$\exists m \in \Z_{\ge 0} : x p^m \in \Z_p$ From P-adic Integer is Limit of Unique P-adic Expansion, there exists a unique $p$-adic expansion of the form: :$\displaystyle \sum_{n \mathop = 0}^\infty d_n p^n$ such that: :$\displaystyle \lim_{n \mathop \to \infty} \sum_{i \mathop = 0}^n d_i p^i = x p^m$ From Multiple Rule for Sequences in Normed Division Ring: :$\displaystyle \lim_{n \mathop \to \infty} \sum_{i \mathop = 0}^n d_i p^{i-m} = x$ Re-indexing from $-m$: :$\displaystyle \lim_{n \mathop \to \infty} \sum_{i \mathop = -m}^n d_i’ p^i = x$ where: :$\forall n \in \Z_{\ge -m} : d_i’ = d_{i+m}$ Then the $p$-adic expansion $\displaystyle \sum_{n \mathop = -m}^\infty d_n’ p^n$ converges to $x$. {{qed}}
=== Case 1 === Let $m = n$.  Then $X = S$ is a subset $X \subseteq S$ such that: :$\size X = m$ {{qed|lemma}} === Case 2 === Let $m = 0$.  Then $X = \O$ is a subset $X \subseteq S$ such that: :$\size X = m$ {{qed|lemma}} === Case 3 === Let $0 < m < n$. By definition of the cardinality of a finite set: :$S \sim \N_{< n}$ where: :$\sim$ denotes set equivalence :$\N_{<n}$ is the set of all natural numbers less than $n$ Consider the set of all natural numbers less than $m$, $\N_{< m}$. {{begin-eqn}} {{eqn | l = k       | o = \in       | r = \N_{< m} }} {{eqn | ll = \leadstoandfrom       | l = k       | o = <       | r = m       | c = {{Defof| Initial Segment of Natural Numbers|set of all natural numbers less than $m$}} }} {{eqn | ll = \leadsto       | l = k       | o = <       | r = n       | c = As $m < n$ and the ordering on the natural numbers is transitive }} {{eqn | ll = \leadstoandfrom       | l = k       | o = \in       | r = \N_{< n}       | c = {{Defof| Initial Segment of Natural Numbers|set of all natural numbers less than $n$}} }} {{end-eqn}} By the definition of a subset: :$\N_{< m} \subseteq \N_{< n}$ Let $i : \N_{< m} \to \N_{< n}$ be the inclusion mapping. Let $f : \N_{< n} \to S$ be a bijection between $\N_{< n}$ and $S$. From Composite of Injections is Injection, the composite $f \circ i : \N_{< m} \to S$ is an injection. Let $X = \Img {f \circ i}$ be the image of $f \circ i$. By definition of the image of a mapping: :$X \subseteq S$ From Injection to Image is Bijection: :$\N_{< m} \sim X$ By definition of the cardinality of $X$: :$\size X = m$ {{qed}}
From Trivial Group is Group, we have that there does exist a group of cardinality $1$. From Group is not Empty, there can be no group of smaller order. {{qed}}
From the definition of a coherent sequence, it needs to be shown that $\sequence{\alpha_n}$ is a sequence of integers such that: :$(1): \quad \forall n \in \N: 0 \le \alpha_n < p^{n + 1}$ :$(2): \quad \forall n \in \N: \alpha_{n + 1} \equiv \alpha_n \pmod {p^{n + 1}}$ That the sequence $\sequence{\alpha_n}$ is a sequence of integers follows immediately from the assumption that the series begins at $n = 0$ and so the terms of each summation are integers.  First note: {{begin-eqn}} {{eqn | lo = \forall n \in \N: \quad       | l = \alpha_{n + 1}       | r = \displaystyle \sum_{i \mathop = 0}^{n + 1} d_i p^i }} {{eqn | r = \displaystyle \sum_{i \mathop = 0}^n d_i p^i + d_{n + 1} p^{n + 1} }} {{eqn | r = \alpha_n + d_{n + 1} p^{n + 1} }} {{end-eqn}} === Partial Sums satisfy (1) === This is proved by induction on $n$: ==== Basis for the Induction ==== $n = 0$  By definition: $\alpha_0 = d_0$ By definition of a $p$-adic expansion: :$0 \le d_0 < p$ So shown for basis for the induction. ==== Induction Hypothesis ==== This is our induction hypothesis: :$0 \le \alpha_k < p^{k + 1}$ Now we need to show true for $n=k+1$: :$0 \le \alpha_{k + 1} < p^{k + 2}$ ==== Induction Step ==== This is our induction step: {{begin-eqn}} {{eqn | l = 0       | o = \le       | r = \alpha_k + d_{k + 1} p^{k + 1}       | c = as $\alpha_k, d_{k + 1}, p \ge 0$ }} {{eqn | o = <        | r = p^{k + 1} + d_{k + 1} p^{k + 1}       | c = induction hypothesis  }} {{eqn | o = \le        | r = p^{k + 1} + \paren{p - 1} p^{k + 1}       | c = {{Defof|P-adic Expansion|$p$-adic expansion}} means $0 \le d_{k + 1} < p$ }} {{eqn | r = p \cdot p^{k + 1} }} {{eqn | r = p^{k + 2} }} {{end-eqn}} Since: :$\alpha_{k + 1} = \alpha_k + d_{k + 1} p^{k + 1}$ then: :$0 \le \alpha_{k + 1} < p^{k + 2}$ By induction: :$\forall n \in \N: 0 \le \alpha_n < p^{n + 1}$ {{qed|lemma}} === Partial Sums satisfy (2) === Let $n \in \N$.   Since: :$\alpha_{n + 1} - \alpha_n = d_{n + 1}p^{n + 1}$ Then:  :$p^{n + 1} \divides \alpha_{n+1} - \alpha_n$ So: :$\alpha_{n + 1} \equiv \alpha_n \pmod {p^{n + 1} }$ The result follows. {{qed}}
From P-adic Expansion is a Cauchy Sequence in P-adic Norm, the sequence of partial sums of the series: :$\displaystyle \sum_{n \mathop = m}^\infty d_n p^n$ is a Cauchy Sequence. By definition, the sequence of partial sums of the series: :$\displaystyle \sum_{n \mathop = m}^\infty d_n p^n$ is a representative of the equivalence class containing the sequence of partial sums. {{qed}}
Let $\map f x = e^x \paren {x - 1} - e^{-x} \paren {x + 1}$. Then if $\map f c = 0$, $c$ is a root of $e^x \paren {x - 1} = e^{-x} \paren {x + 1}$. Notice that: :$\map f 1 = e^1 \paren {1 - 1} - e^{-1} \paren {1 + 1} = -\dfrac 2 e < 0$ :$\map f 2 = e^2 \paren {2 - 1} - e^{-2} \paren {2 + 1} = e^2 - \dfrac 3 {e^2} > 0$ By Intermediate Value Theorem: :$\exists c \in \openint 1 2: \map f c = 0$. This shows that our equation has a root between $1$ and $2$. The exact value of this root can be found using any numerical method, e.g. Newton's Method. {{ProofWanted|Analysis needed of the Kepler Equation}}
Let $i \in \N$. Let $x = \dfrac a b: a, b \in \Z \text{ and } b \neq 0$. {{WLOG}} we can assume that $\dfrac a b$ is in canonical form. By Valuation Ring of P-adic Norm on Rationals: :$\dfrac a b \in \Z_{\paren p} = \set {\dfrac c d \in \Q : p \nmid d}$ So $p \nmid b$. Since $p \nmid b$, by Prime not Divisor implies Coprime then $p^i \perp b$. By Integer Combination of Coprime Integers, there exists $m, l \in \Z: mb + lp^i = 1$. Then: {{begin-eqn}} {{eqn | l = \norm {\dfrac a b - am}_p       | r = \norm {\dfrac a b}_p \norm {1 - a m b}_p       | c = Norm axiom $(\text N 2)$ (Multiplicativity) }} {{eqn | o = \le       | r = \norm {1 - m b}_p       | c = As $\norm {\dfrac a b}_p \le 1$ }} {{eqn | r = \norm {l p^i}_p       | c = As $1 - mb = lp^i$ }} {{eqn | o = \le       | r = p^{-i}       | c = {{Defof|P-adic Norm|$p$-adic norm on integers}} }} {{end-eqn}} As $am \in \Z$, the result follows with $\alpha = am$. {{qed}}
Let $\sequence{\beta_n}$ be a sequence representing $\mathbf a$. That is, $\sequence{\beta_n}$ is a Cauchy sequence in the $p$-adic norm $\norm{\,\cdot\,}_p$ on the rational numbers $\Q$. By definition of a Cauchy sequence: :$\forall j \in \N : \exists \mathop {\map M j} \in \N: \forall m, n \in \N: m, n \ge \map M j : \norm {x_n - x_m} < p^{-j}$ For all $j \in \N$, let: :$\map N j = \max \set{j, \map M j}$ From P-adic Norm of p-adic Number is Power of p, :$\forall j \in \N : \exists \mathop {\map N j} \ge j : \forall m, n \in \N: m, n \ge \map N j : \norm {\beta_n - \beta_m} \le p^{-\paren{j + 1}}$ === Lemma 1 === {{:Equivalence Class in P-adic Integers Contains Unique Coherent Sequence/Lemma 1}} {{qed|lemma}} From Unique Integer Close to Rational in Valuation Ring of P-adic Norm: :for all $j \in \N$ there exists $\alpha_j \in \Z$: ::$(1): \quad \norm{\beta_{\map N j} - \alpha_j}_p \le p^{-\paren{j + 1}}$ ::$(2): \quad  0 \le \alpha_j \le p^{j + 1} - 1$ === Lemma 2 === {{:Equivalence Class in P-adic Integers Contains Unique Coherent Sequence/Lemma 2}} {{qed|lemma}} By definition of the $p$-adic norm, :$\forall j \in \N : \alpha_{j + 1} \equiv \alpha_j \pmod {p^{j + 1}}$ Then $\sequence{\alpha_j}$ is a coherent sequence by definition.  From Corollary of Characterisation of Cauchy Sequence in Non-Archimedean Norm, $\sequence{\alpha_j}$ is a Cauchy sequence. === Lemma 3 === {{:Equivalence Class in P-adic Integers Contains Unique Coherent Sequence/Lemma 3}} {{qed|lemma}} Then $\sequence{\alpha_j}$ is a representative of $\mathbf a$. === Lemma 4 === {{:Equivalence Class in P-adic Integers Contains Unique Coherent Sequence/Lemma 4}} {{qed}}
From Equivalence Class in P-adic Integers Contains Unique Coherent Sequence, $a$ has exactly one representative coherent sequence. Let $\sequence{\alpha_n}$ be the coherent sequence such that: :$\sequence{\alpha_n}$ is a representative of $\mathbf a$. From Coherent Sequence is Partial Sum of P-adic Expansion, there exists a $p$-adic expansion of the form: :$\displaystyle \sum_{n \mathop = 0}^\infty d_n p^n$ such that: :$\forall n \in \N: \alpha_n = \displaystyle \sum_{i \mathop = 0}^n d_i p^i$ Then $\displaystyle \sum_{n \mathop = 0}^\infty d_n p^n$ is a representative of $\mathbf a$. Let $\displaystyle \sum_{n \mathop = 0}^\infty d'_n p^n$ be a $p$-adic expansion that is a representative of $\mathbf  a$. From Partial Sums of P-adic Expansion forms Coherent Sequence, the partial sums: :$\forall n \in \N: \alpha'_n = \displaystyle \sum_{i \mathop = 0}^n d'_i p^i$ are a coherent sequence that is a representative of $\mathbf a$. Since $\sequence{\alpha_n}$ is the only coherent sequence that is a  representative of $a$, then: :$\sequence{\alpha'_n} = \sequence{\alpha_n}$ It follows that: :$\forall n \in \N : d'_n = d_n$ Then $\displaystyle \sum_{n \mathop = 0}^\infty d_n p^n$ is the only $p$-adic expansion that is a representative of $\mathbf a$. {{qed}}