nyuuzyou/wikis huggingface dataset:
P-adic Closed Ball is Instance of Closed Ball of a Norm

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $a \in \Q_p$.
Let $\epsilon \in \R_{>0}$ be a strictly positive real number.
Let $B \subseteq \Q_p$.
Then: 
:$B$ is a closed ball in $p$-adic numbers with radius $\epsilon$ and centre $a$
:
:$B$ is a closed ball of the normed division ring $\struct {\Q_p, \norm {\,\cdot\,}_p}$ with radius $\epsilon$ and centre $a$ 
That is, the definition of a closed ball in $p$-adic numbers is a specific instance of the general definition of a closed ball in a normed division ring.
== Proof ==
From P-adic Numbers form Non-Archimedean Valued Field:
:the $p$-adic numbers $\struct {\Q_p, \norm {\,\cdot\,}_p}$ form a non-Archimedean valued field.
The definition of a closed ball in $p$-adic numbers is identical to the definition of a closed ball of a normed division ring with respect to the norm $\norm {\,\cdot\,}_p$.
Category:P-adic Number Theory
P-adic Expansion Less Intial Zero Terms Represents Same P-adic Number

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $a$ be a $p$-adic number, that is left coset, in $\Q_p$.
Let $\ds \sum_{i \mathop = m}^\infty d_i p^i$ be a $p$-adic expansion that represents $a$.
Let $l$ be the first index $i \ge m$ such that $d_i \ne 0$
Then the series:
:$\ds \sum_{i \mathop = l}^\infty d_i p^i$
also represents $a$.
== Proof ==
Let $\sequence {\alpha_n}$ be the sequence of partial sums:
:$\ds \forall n \in \N: \alpha _n = \sum_{i \mathop = 0}^n d_{n + m} p^{n + m}$
Let $\sequence {\beta_n}$ be the sequence of partial sums:
:$\ds \forall n \in \N: \beta _n = \sum_{i \mathop = 0}^n d_{n + l} p^{n + l}$
Then:
By definition of $l$:
:$m \le l$
So:
:$\forall n \in \N : n + l - m \ge n$
Thus $\sequence {\beta_n}$ is a subsequence of $\sequence {\alpha_n}$ by definition.
From Subsequence of Cauchy Sequence in Normed Division Ring is Cauchy Sequence:
:$\ds \sum_{i \mathop = l}^\infty d_i p^i$ is a Cauchy Sequence in $\Q$.
From Subsequence is Equivalent to Cauchy Sequence:
:$\ds \lim_{n \mathop \to \infty} {\alpha_n - \beta_n} = 0$
That is, the sequence $\sequence {\alpha_n - \beta_n}$ is a null sequence.
By definition of $p$-adic number:
:$\sequence {\alpha_n}$ and $\sequence {\beta_n}$ represent the same $p$-adic number
Since $\sequence {\alpha_n}$ represents $a$, it follows that $\sequence {\beta_n}$ represents $a$.
Category:P-adic Number Theory
P-adic Expansion Representative of P-adic Number is Unique

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $a$ be a $p$-adic number, that is left coset, in $\Q_p$.
Let $\ds \sum_{i \mathop = m}^\infty d_i p^i$ and $\ds \sum_{i \mathop = k}^\infty e_i p^i$ be $p$-adic expansions that represent $a$.
Then:
:$(1) \quad m = k$
:$(2) \quad \forall i \ge m : d_i = e_i$
That is, the $p$-adic expansions $\ds \sum_{i \mathop = m}^\infty d_i p^i$ and $\ds \sum_{i \mathop = k}^\infty e_i p^i$ are identical.
== Proof ==
From P-adic Number times P-adic Norm is P-adic Unit there exists $n \in \Z$:
:$p^n a \in \Z^\times_p$
:$\norm a_p = p^n$ 
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:
Similarly:
:$k < 0 \leadsto k = l$
=== Proof of statement $(1)$ ===
==== Case $l < 0$ ====
Let $l < 0$.
Then:
Similarly:
:$l < 0 \leadsto k = l$
So: 
:$m = l = k$.
==== Case $l \ge 0$ ====
Let $l \ge 0$.
Then:
Similarly:
:$l \ge 0 \leadsto k = 0$
So:
:$m = 0 = k$.
=== 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:
:$\ds \sum_{i \mathop = l}^\infty d_i p^i$
From P-adic Expansion Less Intial Zero Terms Represents Same P-adic Number
:$\ds \sum_{i \mathop = l}^\infty d_i p^i$
is a representative of $a$
From Multiple Rule for Cauchy Sequences in Normed Division Ring:
:$\ds p^{-l} \sum_{i \mathop = l}^\infty d_i p^i = \sum_{i \mathop = l}^\infty d_i p^{i - l} = \sum_{i \mathop = 0}^\infty d_{i + l} p^i$
is a representative of $p^{-l} a \in \Z^\times_p$.
By definition of a $p$-adic expansion:
:$\forall i \in N : 0 \le d_{i + 1} < p - 1$
Thus the series:
:$\ds \sum_{i \mathop = 0}^\infty d_{i + l} p^i$
is a $p$-adic expansion that represents $p^{-l} a \in \Z^\times_p$.
Similarly the series:
:$\ds \sum_{i \mathop = 0}^\infty e_{i + l} p^i$
is a $p$-adic expansion that represents $p^{-l} a \in \Z^\times_p$.
From P-adic Integer has Unique P-adic Expansion Representative 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.
Category:P-adic Number Theory
P-adic Expansion is a Cauchy Sequence in P-adic Norm

== Theorem ==
Let $p$ be a prime number.
Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rationals numbers $\Q$.
Let $\ds \sum_{n \mathop = m}^\infty d_n p^n$ be a $p$-adic expansion.
Then the sequence of partial sums of the series:
:$\ds \sum_{n \mathop = m}^\infty d_n p^n$
is a Cauchy sequence in the valued field $\struct{\Q, \norm{\,\cdot\,}_p}$.
=== Corollary 1 ===
=== Corollary 2 ===
== Proof ==
Let $\sequence {s_N}$ be the sequence of partial sums defined by:
:$\forall N \in \Z_{\ge m}: s_N = \ds \sum_{n \mathop = m}^N d_n p^n$
From Sequence of Consecutive Integers Modulo Power of p is Cauchy in P-adic Norm:
:the sequence $\sequence {s_N}$ is a Cauchy sequence if:
::$\forall N \in \Z_{\ge m}: s_{N + 1} \equiv s_N \pmod {p^n}$
Now for all $N \in \Z_{\ge m}$:
The result follows.
== Sources ==
* : $\S 1.4$ The field of $p$-adic numbers $\Q_p$
Category:P-adic Number Theory
Category:P-adic Expansion is a Cauchy Sequence in P-adic Norm
P-adic Expansion is a Cauchy Sequence in P-adic Norm/Converges to P-adic Number

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $\ds \sum_{n \mathop = m}^\infty d_n p^n$ be a $p$-adic expansion.
Then the sequence of partial sums of the series:
:$\ds \sum_{n \mathop = m}^\infty d_n p^n$
converges to a $p$-adic number in $\struct{\Q_p, \norm{\,\cdot\,}_p}$.
== Proof ==
From P-adic Expansion is a Cauchy Sequence in P-adic Norm, the sequence of partial sums of the series:
:$\ds \sum_{n \mathop = m}^\infty d_n p^n$
is a Cauchy sequence in the rationals $\Q$ with the $p$-adic norm.
From Sequence is Cauchy in P-adic Norm iff Cauchy in P-adic Numbers, the sequence of partial sums of the series:
:$\ds \sum_{n \mathop = m}^\infty d_n p^n$
is a Cauchy sequence in the $p$-adic numbers $\Q_p$ with the $p$-adic norm.
From P-adic Numbers form Completion of Rational Numbers with P-adic Norm:
:the $p$-adic numbers $\Q_p$ form a complete normed division ring.
By definition of a complete normed division ring, every Cauchy sequence is convergent in $\struct{\Q_p, \norm{\,\cdot\,}_p}$.
== Sources ==
* : $\S 1.4$ The field of $p$-adic numbers $\Q_p$
Category:P-adic Expansion is a Cauchy Sequence in P-adic Norm
P-adic Expansion is a Cauchy Sequence in P-adic Norm/Represents a P-adic Number

== Theorem ==
Let $p$ be a prime number.
Let $\ds \sum_{n \mathop = m}^\infty d_n p^n$ be a $p$-adic expansion.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic Numbers.
Then the sequence of partial sums of the series:
:$\ds \sum_{n \mathop = m}^\infty d_n p^n$
represents a $p$-adic number of $\struct {\Q_p,\norm {\,\cdot\,}_p}$.
== Proof ==
From P-adic Expansion is a Cauchy Sequence in P-adic Norm, the sequence of partial sums of the series:
:$\ds \sum_{n \mathop = m}^\infty d_n p^n$
is a Cauchy Sequence.
Then the sequence of partial sums of the series:
:$\ds \sum_{n \mathop = m}^\infty d_n p^n$
is a representative of a $p$-adic number by definition.
== Sources ==
* : $\S 1.4$ The field of $p$-adic numbers $\Q_p$
Category:P-adic Expansion is a Cauchy Sequence in P-adic Norm
P-adic Expansion of P-adic Unit

== Theorem ==
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers for some prime $p$.
Let $\Z_p$ be the $p$-adic integers.
Let $a \in \Z_p$.
Let $\ldots a_n \ldots a_3 a_2 a_1 a_0$ be the canonical expansion of $a$.
Then:
:$a$ is a $p$-adic unit $a_0 \ne 0$
== Proof ==
From P-adic Unit has Norm Equal to One:
:a is a $p$-adic unit $\norm a_p = 1 = p^0$
By definition of the canonical expansion:
:$a$ is the limit of the $p$-adic expansion $\ds \sum_{n \mathop = 0}^\infty a_n p^n$
From P-adic Norm of P-adic Expansion is determined by First Nonzero Coefficient:
:$\norm a_p = p^0$ $a_0 \ne 0$
The result follows.
== Sources ==
* : $\S 3.3$ Exploring $\Q_p$: Problem $105$
* : $\S 1.5$ Arithmetical operations in $\Q_p$: Proposition $1.36$
Category:P-adic Units
P-adic Integer has Unique Coherent Sequence Representative

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $a$ be a $p$-adic number, that is left coset, in $\Q_p$ such that $\norm a _p \le 1$.
Then $a$ has exactly one representative that is a coherent sequence.
=== Corollary ===
== Proof ==
Let $\sequence{\beta_n}$ be a sequence representing $a$.
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 ===
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 ===
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 Sequence of Consecutive Integers Modulo Power of p is Cauchy in P-adic Norm:
:$\sequence{\alpha_j}$ is a Cauchy sequence.
=== Lemma 3 ===
Then $\sequence{\alpha_j}$ is a representative of $a$.
=== Lemma 4 ===
== Also see ==
* Coherent Sequence Converges to P-adic Integer
* P-adic Integer has Unique P-adic Expansion Representative
* P-adic Number has Unique P-adic Expansion Representative
== Sources ==
* : $\S 1.4$ The field of $p$-adic numbers $\Q_p$: Lemma $1.30$
Category:P-adic Number Theory
Category:P-adic Integer has Unique Coherent Sequence Representative
P-adic Integer has Unique Coherent Sequence Representative/Lemma 1

== Theorem ==
Let $p$ be a prime number.
Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rational numbers $\Q$.
Let $\sequence {\beta_n}$ be a Cauchy sequence in $\struct {\Q, \norm {\,\cdot\,}_p}$ such that:
:$\forall j \in \N : \exists \mathop {\map N j} \ge j : \forall m, n \in \N: m, n \ge \map N j: \norm {\beta_n - \beta_m} \le p^{-\paren {j + 1} }$
Then:
:$\forall j \in \N: \norm {\beta_{\map N {j + 1} } - \beta_{\map N j} }_p \le p^{-\paren {j + 1} }$
== Proof ==
Let $j \in N$
Suppose $\map N {j + 1} \ge \map N j$
By definition:
:$\norm{\beta_{\map N {j + 1} } - \beta_{\map N j} }_p \le p^{-\paren {j + 1} }$
Now suppose $\map N j \ge \map N {j + 1}$
Then:
In either case:
:$\norm {\beta_{\map N {j + 1} } - \beta_{\map N j} }_p \le p^{-\paren{j + 1} }$
Since $j$ was arbitrary, then:
:$\forall j \in \N: \norm {\beta_{\map N {j + 1} } - \beta_{\map N j} }_p \le p^{-\paren {j + 1} }$
Category:P-adic Integer has Unique Coherent Sequence Representative
P-adic Integer has Unique Coherent Sequence Representative/Lemma 2

== Theorem ==
Let $p$ be a prime number.
Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rational numbers $\Q$.
Let $\sequence {\gamma_n}$ be a Cauchy sequence in $\struct {\Q, \norm {\,\cdot\,}_p}$ such that:
:$\forall j \in \N: \norm {\gamma_{j + 1} - \gamma_j }_p \le p^{-\paren {j + 1} }$
Let $\sequence {\alpha_n}$ be a sequence in $\Q$ such that:
:$\forall j \in \N: \norm {\alpha_j - \gamma_j }_p \le p^{-\paren {j + 1} }$
Then:
:$\forall j \in \N: \norm {\alpha_{j + 1} - \alpha_j }_p \le p^{-\paren {j + 1} }$
== Proof ==
For all $j \in \N$:
Category:P-adic Integer has Unique Coherent Sequence Representative
P-adic Integer has Unique Coherent Sequence Representative/Lemma 3

== Theorem ==
Let $p$ be a prime number.
Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rational numbers $\Q$.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $\sequence{\beta_n}$ be a Cauchy sequence in $\struct {\Q, \norm {\,\cdot\,}_p}$ such that:
:$\forall j \in \N : \exists \mathop {\map N j} \ge j : \forall m, n \in \N: m, n \ge \map N j : \norm {\beta_n - \beta_m} \le p^{-\paren {j + 1} }$
Let $\sequence{\alpha_n}$ be a Cauchy sequence in $\struct {\Q, \norm {\,\cdot\,}_p}$ such that:
:$\forall j \in \N: \norm {\alpha_j - \beta_{\map N j} }_p \le p^{-\paren {j + 1} }$
Then: 
:$\sequence {\alpha_n}$ and $\sequence {\beta_n}$ are representatives of the same $p$-adic number in $\Q_p$.
== Proof ==
From Representatives of same P-adic Number iff Difference is Null Sequence, it needs only to be shown that $\sequence {\alpha_j - \beta_j}$ is a null sequence.
Let $\epsilon \in \R_{> 0}$.
From Sequence of Powers of Number less than One, the sequence $\sequence {p^n}$ is a null sequence.
Then there exists $j \in \N$ such that $p^{-j} < \epsilon$.
Then for all $i \ge \map N j$:
It follows that $\sequence {\alpha_n - \beta_n}$ is a null sequence by definition.
Category:P-adic Integer has Unique Coherent Sequence Representative
P-adic Integer has Unique Coherent Sequence Representative/Lemma 4

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $a$ be a $p$-adic number, that is left coset, in $\Q_p$ such that $\norm a _p \le 1$.
Let $\sequence {\alpha_j}$ be a coherent sequence that represents $a$.
Then:
:$\sequence {\alpha_j}$ is the only coherent sequence that represents $a$.
== Proof ==
Let $\sequence {\alpha'_j}$ be a coherent sequence not equal to $\sequence {\alpha_j}$.
From Representatives of same P-adic Number iff Difference is Null Sequence, it needs only to be shown that $\sequence {\alpha_j - \alpha'_j}$ is not a null sequence.
Since $\sequence {\alpha'_j} \ne \sequence{\alpha_j}$ then:
:$\exists i_0 \in \N : \alpha'_{i_0} \ne \alpha_{i_0}$
By definition of coherent sequences:
:$0 \le \alpha_{i_0}, \alpha'_{i_0} < p^{i_0 + 1}$
From Reduced Residue System Modulo Prime:
:$\alpha_{i_0} \not \equiv \alpha'_{i_0} \pmod {p^{i_0 + 1} }$
By definition of a coherent sequence, for all $i > i_0$:
:$\alpha_i \equiv \alpha_{i_0} \pmod {p^{i_0 + 1} }$
:$\alpha'_i \equiv \alpha'_{i_0} \pmod {p^{i_0 + 1} }$
Then:
:$\forall i > i_0: \alpha_i \equiv \alpha_{i_0} \not \equiv \alpha'_{i_0} \equiv \alpha'_i \pmod {p^{i_0 + 1} }$
That is:
:$\forall i > i_0: p^{i_0 + 1} \nmid \alpha_i - \alpha'_i$
By definition of the $p$-adic norm on integers:
:$\forall i > i_0: \norm {\alpha_i - \alpha'_i} > \dfrac 1 {p^{i_0 + 1} }$
By definition of convergence, $\sequence {\alpha_j - \alpha'_j}$ is not a null sequence.
The result follows.
Category:P-adic Integer has Unique Coherent Sequence Representative
P-adic Integer has Unique Coherent Sequence Representative/P-adic Expansion

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $a$ be a $p$-adic number, that is left coset, in $\Q_p$ such that $\norm a _p \le 1$.
Then $a$ has exactly one representative that is a $p$-adic expansion of the form:
:$\ds \sum_{n \mathop = 0}^\infty d_n p^n$
== Proof ==
From P-adic Integer has Unique Coherent Sequence Representative, $a$ has exactly one representative coherent sequence.
Let $\sequence{\alpha_n}$ be the coherent sequence such that:
:$\sequence{\alpha_n}$ is a representative of $a$.
From Coherent Sequence is Partial Sum of P-adic Expansion, there exists a $p$-adic expansion of the form:
:$\ds \sum_{n \mathop = 0}^\infty d_n p^n$
such that:
:$\forall n \in \N: \alpha_n = \ds \sum_{i \mathop = 0}^n d_i p^i$
Then $\ds \sum_{n \mathop = 0}^\infty d_n p^n$ is a representative of $a$.
Let $\ds \sum_{n \mathop = 0}^\infty d'_n p^n$ be a $p$-adic expansion that is a representative of $a$.
From Partial Sums of P-adic Expansion forms Coherent Sequence, the partial sums:
:$\forall n \in \N: \alpha'_n = \ds \sum_{i \mathop = 0}^n d'_i p^i$
are a coherent sequence that is a representative of $a$.
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 $\ds \sum_{n \mathop = 0}^\infty d_n p^n$ is the only $p$-adic expansion that is a representative of $a$.
== Also see ==
* P-adic Integer is Limit of Unique P-adic Expansion
* P-adic Integer has Unique Coherent Sequence Representative
* P-adic Number has Unique P-adic Expansion Representative
== Sources ==
* : $\S 1.4$ The field of $p$-adic numbers $\Q_p$
Category:P-adic Integer has Unique Coherent Sequence Representative
P-adic Integer is Limit of Unique Coherent Sequence of Integers

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $\Z_p$ be the $p$-adic integers.
Let $x \in \Z_p$.
Then there exists a unique coherent sequence $\sequence {\alpha_n}$:
:$\ds \lim_{n \mathop \to \infty} \alpha_n = x$
== Proof ==
By the definition of a coherent sequence it needs to be proved that there exists a unique integer sequence $\sequence {\alpha_n}$:
:$(1): \quad \forall n \in \N: \alpha_n \in \Z$ and $0 \le \alpha_n \le p^{n + 1} - 1$
:$(2): \quad \forall n \in \N: \alpha_{n + 1} \equiv \alpha_n \pmod {p^{n + 1}}$
:$(3): \quad \ds \lim_{n \mathop \to \infty} \alpha_n = x$
By Integers are Arbitrarily Close to P-adic Integers then for all $n \in \N$ there exists $\alpha_n \in \Z$:
::$\text {(a)}: \quad 0 \le \alpha_n \le p^{n + 1} - 1$
::$\text {(b)}: \quad \norm {x - \alpha_n}_p \le p^{-\paren{n + 1} }$
Hence the sequence $\sequence {\alpha_n}$ satisfies $(1)$ above.
=== Lemma 1 ===
Hence the sequence $\sequence {\alpha_n}$ satisfies $(2)$ above.
=== Lemma 2 ===
Hence the sequence $\sequence {\alpha_n}$ satisfies $(3)$ above.
=== Lemma 3 ===
== Sources ==
* : $\S 3.3$ Exploring $\Q_p$: Lemma $3.3.4 \ \text{(iii)}$
Category:P-adic Number Theory
Category:P-adic Integer is Limit of Unique Coherent Sequence of Integers
P-adic Integer is Limit of Unique Coherent Sequence of Integers/Lemma 1

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $\Z_p$ be the $p$-adic integers.
Let $x \in \Z_p$.
Let $\sequence{\alpha_n}$ be an integer sequence:
:$(1): \quad 0 \le \alpha_n \le p^{n + 1} - 1$
:$(2): \quad \norm {x -\alpha_n}_p \le p^{-\paren{n + 1}}$
Then:
:$\forall n \in \N: \alpha_{n + 1} \equiv \alpha_n \pmod {p^{n + 1}}$
== Proof ==
For any $n \in \N$ then:
Hence:
:$p^{n + 1} \divides \paren {\alpha_{n + 1} - \alpha_n} $
or equivalently:
:$\alpha_{n + 1} \equiv \alpha_n \pmod {p^{n + 1} }$
The result follows.
== Sources ==
* : $\S 3.3$ Exploring $\Q_p$, Lemma $3.3.4 \ \text {(ii)}$
Category:P-adic Integer is Limit of Unique Coherent Sequence of Integers
P-adic Integer is Limit of Unique Coherent Sequence of Integers/Lemma 2

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $\Z_p$ be the $p$-adic integers.
Let $x \in \Z_p$.
Let $\sequence {\alpha_n}$ be a sequence of integers:
:$\norm {x -\alpha_n}_p \le p^{-\paren {n + 1} }$
Then:
:$\ds \lim_{n \mathop \to \infty} \alpha_n = x$
== Proof ==
From Sequence of Powers of Number less than One:
:$\ds \lim_{n \mathop \to \infty} p^{-n} = 0$
From Multiple Rule for Real Sequences:
:$\ds \lim_{n \mathop \to \infty} p^{-\paren {n + 1} } = 0$
By the Squeeze Theorem for Real Sequences :
:$\ds \lim_{n \mathop \to \infty} \norm {x - \alpha_n}_p = 0$.
Hence the limit of $\sequence {\alpha_n}$ is by definition:
:$\ds \lim_{n \mathop \to \infty} \alpha_n = x$
== Sources ==
* : $\S 3.3$ Exploring $\Q_p$, Lemma $3.3.4 \ \text {(ii)}$
Category:P-adic Integer is Limit of Unique Coherent Sequence of Integers
P-adic Integer is Limit of Unique Coherent Sequence of Integers/Lemma 3

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $\Z_p$ be the $p$-adic integers.
Let $x \in \Z_p$.
Let $\sequence {\alpha_n}$ be a sequence such that:
:$(1): \quad \forall n \in \N: \alpha_n \in \Z$ and $0 \le \alpha_n \le p^{n + 1} - 1$
:$(2): \quad \forall n \in \N: \alpha_{n + 1} \equiv \alpha_n \pmod {p^{n + 1} }$
:$(3): \quad \ds \lim_{n \mathop \to \infty} \alpha_n = x$
$\sequence {\alpha_n}$ is a unique sequence satisfying properties $(1)$, $(2)$ and $(3)$ above.
== Proof ==
Suppose that there exists a sequence $\sequence {\alpha'_n}$ with:
:$(1'): \quad \forall n \in \N: \alpha'_n \in \Z$ and $0 \le \alpha'_n \le p^{n + 1} - 1$
:$(2'): \quad \forall n \in \N: \alpha'_{n + 1} \equiv \alpha'_n \pmod {p^{n + 1} }$
:$(3'): \quad \ds \lim_{n \mathop \to \infty} \alpha'_n = x$
:
:$\alpha'_N \ne \alpha_N$ for some $N \in \N$
By Initial Segment of Natural Numbers forms Complete Residue System:
:$\alpha'_N \not \equiv \alpha_N \pmod {p^{N + 1} }$
Then for all $n > N$:
That is, for all $n > N$:
:$\alpha'_n \not \equiv \alpha_n \pmod {p^{N + 1} }$
Hence for all $n > N$:
:$\norm {\alpha'_n - \alpha_n}_p > p^{-\paren {N + 1} }$
By $(3)$ the limit of $\sequence {\alpha_n}$ is $x$:
:$\exists N_1 \in \N: \forall n \ge N_1: \norm {x - \alpha_n}_p \le p^{-\paren {N + 1} }$
Similarly for $\sequence{\alpha'_n}$:
:$\exists N_2 \in \N: \forall n \ge N_2: \norm {x - \alpha'_n}_p \le p^{-\paren {N + 1} }$
Let $M = \max \set {N + 1, N_1, N_2}$.
Then:
:$\norm {\alpha'_M - \alpha_M}_p > p^{-\paren {N + 1} }$
:$\norm {x - \alpha_M} _p\le p^{-\paren {N + 1} }$
:$\norm {x - \alpha'_M}_p \le p^{-\paren {N + 1} }$
But:
This contradicts the previous assertion that:
:$\norm {\alpha'_M - \alpha_M}_p > p^{-\paren {N + 1} }$
Hence:
:$\sequence {\alpha'_n} = \sequence {\alpha_n}$
The result follows.
== Sources ==
* : $\S 3.3$ Exploring $\Q_p$, Lemma $3.3.4 \, \text {(ii)}$
Category:P-adic Integer is Limit of Unique Coherent Sequence of Integers
P-adic Integer is Limit of Unique P-adic Expansion

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $\Z_p$ be the $p$-adic integers.
Let $x \in \Z_p$.
Then $x$ is the limit of a unique $p$-adic expansion of the form:
:$\ds \sum_{n \mathop = 0}^\infty d_n p^n$
== Proof ==
From P-adic Integer is Limit of Unique Coherent Sequence of Integers, there exists a unique coherent sequence $\sequence{\alpha_n}$ such that:
:$\ds \lim_{n \mathop \to \infty} \alpha_n = x$
From Coherent Sequence is Partial Sum of P-adic Expansion, there exists a unique $p$-adic expansion of the form:
:$\ds \sum_{n \mathop = 0}^\infty d_n p^n$
such that:
:$\forall n \in \N: \alpha_n = \ds \sum_{i \mathop = 0}^n d_i p^i$
That is:
:$\ds \lim_{n \mathop \to \infty} \sum_{i \mathop = 0}^n d_i p^i = x$
== Sources ==
* : $\S 3.3$ Exploring $\Q_p$: Corollary $3.3.11$
Category:P-adic Number Theory
P-adic Integers Form Integral Domain

== Theorem ==
Let $\Q_p$ be the $p$-adic numbers for some prime $p$.
Let $\Z_p$ be the $p$-adic integers.
Then:
:$\Z_p$ is an integral domain
== Proof ==
From Field is Integral Domain:
:$\Q_p$ is an integral domain
From P-adic Integers is Valuation Ring Induced by P-adic Norm/Corollary:
:$\Z_p$ is a local ring
By the definition of local ring:
:$\Z_p$ is a ring with unity
From Subdomain Test:
:$\Z_p$ is an integral domain
== Sources ==
* : $\S 1.8$ Algebraic properties of $p$-adic integers: Proposition $1.44$
Category:P-adic Integers