nyuuzyou/wikis huggingface dataset:
P-adic Integers is Metric Completion of Integers

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $d$ be the subspace metric of the $p$-adic metric on the $p$-adic integers $\Z_p$.
Then $\struct {\Z_p, d}$ is the metric completion of the integers $\Z$.
== Proof ==
The integers $\Z$ are a subring of the $p$-adic integers $Z_p$ by Integers form Subring of P-adic Integers.
Hence $\Z \subseteq \Z_p$.
The set of $p$-adic integers $\Z_p$ is closed in the $p$-adic metric by Set of P-adic Integers is Clopen in P-adic Numbers.
By Closure of Subset of Closed Set of Metric Space is Subset then the closure of $\Z$ is contained in $\Z_p$:
:$\map \cl \Z \subseteq \Z_p$
By P-adic Integer is Limit of Unique Coherent Sequence of Integers then:
:for all $x \in \Z_p$ there exists a sequence $\sequence {x_n}$ in $\Z$ such that $\ds \lim_{n \mathop \to \infty} x_n = x$
By Closure of Subset of Metric Space by Convergent Sequence then:
:$\Z_p \subseteq \map \cl \Z$ 
Hence $\map \cl \Z = \Z_p$.
By Metric Subspace Induces Subspace Topology then the topology induced by $d$ on $\Z_p$ is the subspace topology.
By Closure of Subset in Subspace then $\map \cl \Z = \Z_p$ in the subspace metric $d$ on $\Z_p$.
It follows that $\struct {\Z_p, d}$ is the metric completion of $\Z$.
== Sources ==
* : $\S 3.3$ Exploring $\Q_p$: Proposition $3.3.4$
Category:P-adic Number Theory
P-adic Integers is Valuation Ring Induced by P-adic Norm

== Theorem ==
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers for some prime $p$.
Then:
:the $p$-adic integers, $\Z_p$, is the valuation ring induced by the non-Archimedean norm $\norm {\,\cdot\,}_p$
=== Corollary ===
== Proof ==
By the definition of the $p$-adic integers:
:$\Z_p = \set {x \in \Q_p : \norm x_p \le 1}$
From P-adic Numbers form Non-Archimedean Valued Field:
:$\struct {\Q_p, \norm {\,\cdot\,}_p}$ is a non-Archimedean valued field.
By definition of the valuation ring induced by a non-Archimedean norm:
:$\Z_p$ is the valuation ring induced by $\norm {\,\cdot\,}_p$
== Sources ==
* : $\S 3.3$ Exploring $\Q_p$
Category:P-adic Integers
Category:P-adic Integers is Valuation Ring Induced by P-adic Norm
P-adic Integers is Valuation Ring Induced by P-adic Norm/Corollary

== Theorem ==
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers for some prime $p$.
Then:
:$(a): \quad$the $p$-adic integers, $\Z_p$, is a local ring
:$(b): \quad$the principal ideal $p\Z_p$ is the unique maximal ideal of $\Z_p$
== Proof ==
From P-adic Integers is Valuation Ring Induced by P-adic Norm:
:$\Z_p$ is the valuation ring induced by the non-Archimedean norm $\norm {\,\cdot\,}_p$
From Valuation Ideal of P-adic Numbers:
:$p\Z_p$ is the valuation ideal induced by the non-Archimedean norm $\norm {\,\cdot\,}_p$
From Corollary to Valuation Ideal is Maximal Ideal of Induced Valuation Ring:
:$(a):\quad \Z_p$ is a local ring
:$(b):\quad p\Z_p$ is the unique maximal ideal of $\Z_p$
== Sources ==
* : $\S 3.3$ Exploring $\Q_p$
* : $\S 1.8$ Algebraic properties of $p$-adic integers: Corollary $1.45$
Category:P-adic Integers is Valuation Ring Induced by P-adic Norm
P-adic Metric is Metric

== Theorem ==
Let $p \in \N$ be a prime.
Let $\norm {\,\cdot\,}_p: \Q \to \R_{\ge 0}$ be the $p$-adic norm on $\Q$.
Let $d_p$ be the $p$-adic metric on $\Q$:
:$\forall x, y \in \Q: \map {d_p} {x, y} = \norm{x - y}_p$
Then $d_p$ is a metric.
== Proof ==
The $p$-adic metric on $\Q$ is defined as the metric induced by the $p$-adic norm on $\Q$.
It follows from Metric Induced by Norm is Metric that $d_p$ is a metric.
Category:P-adic Metrics
P-adic Metric on P-adic Numbers is Non-Archimedean Metric

== Theorem ==
Let $p \in \N$ be a prime.
Let $\norm{\,\cdot\,}_p: \Q_p \to \R_{\ge 0}$ be the $p$-adic norm on the $p$-adic numbers $\Q_p$.
Let $d_p$ be the $p$-adic metric on $\Q_p$:
:$\forall x, y \in \Q_p: \map {d_p} {x, y} = \norm{x - y}_p$
Then $d_p$ is a non-Archimedean metric that extends the $p$-adic metric on the rationals $\Q$ to $\Q_p$.
=== Corollary ===
== Proof ==
The $p$-adic metric on $\Q_p$ is defined as the metric induced by the $p$-adic norm on $\Q_p$.
It follows from Metric Induced by Norm is Metric that $d_p$ is a metric.
By definition of the $p$-adic norm on $\Q_p$, $\norm{\,\cdot\,}_p$ is a non-Archimedean norm.
From Non-Archimedean Norm iff Non-Archimedean Metric, then $d_p$ is a non-Archimedean metric.
From Rational Numbers are Dense Subfield of P-adic Numbers:
:the $p$-adic norm $\norm{\,\cdot\,}_p$ on the $p$-adic numbers $\Q_p$ is an extension of the $p$-adic norm $\norm{\,\cdot\,}_p$ on $\Q$.
By the definition of the metric induced by the $p$-adic norm, it follows that $d_p$ is an extension of the $p$-adic metric on the rationals $\Q$.
Category:P-adic Metrics
P-adic Metric on P-adic Numbers is Non-Archimedean Metric/Corollary 1

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Then:
:$\forall x, y, z \in R: \norm {x - y}_p \le \max \set {\norm {x - z}_p, \norm {y - z}_p}$
== Proof ==
Let $d_p$ be the $p$-adic metric on $\Q_p$:
:$\forall x, y \in \Q_p: \map {d_p} {x, y} = \norm {x - y}_p$
From P-adic Metric on P-adic Numbers is Non-Archimedean Metric, $d_p$ is a non-Archimedean norm.
By definition of a non-Archimedean norm:
:$\forall x, y, z \in R: \norm {x - y}_p \le \max \set {\norm {x - z}_p, \norm {y - z}_p}$
Category:Normed Division Rings
Category:Non-Archimedean Norms
P-adic Norm Characterisation of Divisibility by Power of p

== Theorem ==
Let $p \in \N$ be a prime.
Let $\Q$ denote the rational numbers.
Let $\norm{\,\cdot\,}$ denote the $p$-adic norm on $\Q$.
Then:
:$\forall a, b \in \Z: a \equiv b \pmod {p^n} \iff \norm {a - b}_p \le p^{-n}$
== Proof ==
Let $a, b \in \Z$.
We have:
P-adic Norm and Absolute Value are Not Equivalent

== Theorem ==
Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rationals $\Q$ for some prime number $p$.
Let $\size{\,\cdot\,}$ be the absolute value on the rationals $\Q$.
Then $\norm {\,\cdot\,}_p$ and $\size{\,\cdot\,}$ are not equivalent norms.
That is, the topology induced by $\norm {\,\cdot\,}_p$ does not equal the topology induced by $\size {\,\cdot\,}$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 3.1.2$ Absolute Values on $\Q$, Problem $68$
Category:Norm Theory
Category:P-adic Number Theory
Category:P-adic Norm and Absolute Value are Not Equivalent
P-adic Norm and Absolute Value are Not Equivalent/Proof 1

== Theorem ==
Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rationals $\Q$ for some prime number $p$.
Let $\size {\,\cdot\,}$ be the absolute value on the rationals $\Q$.
Then $\norm {\,\cdot\,}_p$ and $\size {\,\cdot\,}$ are not equivalent norms.
That is, the topology induced by $\norm {\,\cdot\,}_p$ does not equal the topology induced by $\size {\,\cdot\,}$.
== Proof ==
By definition of the $p$-adic norm:
:$\norm p_p = \frac 1 p < 1$
By definition of the absolute value:
:$\size p = p > 1$
By definition of open unit ball equivalence, $\norm {\,\cdot\,}_p$ and $\size {\,\cdot\,}$ are not equivalent norms.
By Equivalence of Definitions of Equivalent Division Ring Norms and the definition of topologically equivalent norms then the topology induced by $\norm {\,\cdot\,}_p$ does not equal the topology induced by $\size{\,\cdot\,}$.
== Sources ==
* : $\S 3.1.2$ Absolute Values on $\Q$, Problem 68
Category:P-adic Norm and Absolute Value are Not Equivalent
P-adic Norm and Absolute Value are Not Equivalent/Proof 2

== Theorem ==
Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rationals $\Q$ for some prime number $p$.
Let $\size {\,\cdot\,}$ be the absolute value on the rationals $\Q$.
Then $\norm {\,\cdot\,}_p$ and $\size {\,\cdot\,}$ are not equivalent norms.
That is, the topology induced by $\norm {\,\cdot\,}_p$ does not equal the topology induced by $\size {\,\cdot\,}$.
== Proof ==
It is noted that:
:$\sup \set {\size n: n \in \Z} = +\infty$
By , $\size {\,\cdot\,}$ is Archimedean.
From P-adic Norm on Rational Numbers is Non-Archimedean Norm, $\norm {\,\cdot\,}_p$ is non-Archimedean.
By Equivalent Norms are both Non-Archimedean or both Archimedean, $\norm {\,\cdot\,}_p$ and $\size {\,\cdot\,}$ are not equivalent norms.
== Sources ==
* : $\S 3.1.2$ Absolute Values on $\Q$, Problem 68
Category:P-adic Norm and Absolute Value are Not Equivalent
P-adic Norm forms Non-Archimedean Valued Field/P-adic Numbers

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Then:
:$\struct{\Q_p, \norm {\,\cdot\,}_p}$ is a valued field
:$\norm {\,\cdot\,}_p$ is a non-Archimedean norm
That is, the $p$-adic numbers $\struct {\Q_p, \norm {\,\cdot\,}_p}$ form a valued field with a non-Archimedean norm.
== Proof ==
Let $\norm {\,\cdot\,}^\Q_p$ be the p-adic norm on the rationals $\Q$.
From P-adic Norm on Rational Numbers is Non-Archimedean Norm:
:$\struct{Q, \norm {\,\cdot\,}^\Q_p}$ is a valued field with non-Archimedean norm $\norm {\,\cdot\,}_p$ 
By definition of the $p$-adic numbers:
:$\Q_p$ is the quotient ring $\CC \, \big / \NN$
where:
:$\CC$ is the commutative ring of Cauchy sequences over $\struct {\Q, \norm {\,\cdot\,}^\Q_p}$.
and
:$\NN$ is the set of null sequences in $\struct {\Q, \norm {\,\cdot\,}^\Q_p}$.
From Corollary to Quotient Ring of Cauchy Sequences is Normed Division Ring:
:$\struct {\Q_p, \norm {\, \cdot \,}_p}$ is a valued field.
From Completion of Normed Division Ring:
:$\struct {\Q_p, \norm {\, \cdot \,}_p}$ is a normed division ring completion of $\struct {\Q, \norm {\,\cdot\,}^\Q_p}$
From Non-Archimedean Division Ring iff Non-Archimedean Completion:
:$\norm {\, \cdot \,}_p$ on $\Q_p$ is a non-Archimedean norm.
Category:P-adic Norm forms Non-Archimedean Valued Field
P-adic Norm forms Non-Archimedean Valued Field/Rational Numbers

== Theorem ==
The $p$-adic norm $\norm{\,\cdot\,}_p$ forms a non-Archimedean norm on the rational numbers $\Q$.
The rational numbers $\struct{\Q, \norm{\,\cdot\,}_p}$ with the $p$-adic norm is a valued field with a non-Archimedean norm.
== Proof ==
First we note that the $p$-adic norm is a norm.
Let $\nu_p$ denote the $p$-adic valuation on the rational numbers.
Recall the definition of the $p$-adic norm:
:$\forall q \in \Q: \norm q_p := \begin{cases}
 0 & : q = 0 \\
 p^{-\map {\nu_p} q} & : q \ne 0
\end{cases}$
We must show the following holds for all $x, y \in \Q$:
:$\norm {x + y}_p \le \max \set {\norm x_p, \norm y_p}$
If $x = 0$ or $y = 0$, or $x + y = 0$, the result is trivial, as follows:
Let $x = 0$.
Then:
and so $\norm {x + y}_p \le \max \left( \norm x_p, \norm y_p \right)$
The same argument holds for $y = 0$.
Let $x + y = 0$.
Let $x, y, x + y \in \Q_{\ne 0}$.
From $p$-adic Valuation is Valuation:
:$\map {\nu_p} {x + y} \ge \min \set {\map {\nu_p} x, \map {\nu_p} y}$
Then:
== Sources ==
* : $\S 2.1$: Absolute Values on a Field: Proposition $2.1.5$
* : $\S 1.4$ The field of $p$-adic numbers $\Q_p$: Proposition $1.26$
Category:P-adic Norm forms Non-Archimedean Valued Field
P-adic Norm is Norm/Proof 1

== Theorem ==
== Proof ==
Let $v_p$ be the $p$-adic valuation on the rational numbers.
Recall that the $p$-adic norm is defined as:
:$\forall q \in \Q: \norm q_p := \begin {cases} 0 & : q = 0 \\ p^{- \map {\nu_p} q} & : q \ne 0 \end {cases}$
We must show the following hold for all $x$, $y \in \Q$:
=== ===
By Power of Positive Real Number is Positive:
:$\ds \forall s \in \R: \frac 1 {p^s} > 0$
By definition of the $p$-adic norm it follows that:
:$\forall x \in \Q: \norm x_p = 0 \iff x = 0$
Thus the $p$-adic norm fulfils .
=== ===
Let $x = 0$ or $y = 0$.
Then $\norm x_p = 0$ or $\norm y_p = 0$ from , and:
Let $x, y\in \Q_{\ne 0}$.
Then:
Thus the $p$-adic norm fulfils .
=== ===
Let $x = 0$ or $y = 0$, or $x + y = 0$, the result is trivial.
Let $x = 0$.
Then:
and so $\norm {x + y}_p \le \norm x_p + \norm y_p$
The same argument holds for $y = 0$.
Let $x + y = 0$.
Let $x, y, x + y \in \Q_{\ne 0}$.
From $p$-adic Valuation is Valuation:
:$\map {\nu_p} {x + y} \ge \min \set {\map {\nu_p} x, \map {\nu_p} y}$
Then:
Thus the $p$-adic norm fulfils .
All norm axioms are seen to be satisfied.
Hence the result.
== Sources ==
* : $\S 2.1$: Absolute Values on a Field: Proposition $2.1.5$
* : $\S 1.4$ The field of $p$-adic numbers $\Q_p$: Proposition $1.26$
Category:P-adic Norm is Norm
P-adic Norm is Norm/Proof 2

== Theorem ==
== Proof ==
Recall that the $p$-adic norm is defined as:
:$\forall q \in \Q: \norm r_p := \begin {cases}
 0 & : r = 0 \\
 p^{- k} & : r \ne 0
\end {cases}$
where:
:$r = p^k \dfrac m n$
and:
:$k, n \in \Z, m \in \Z_{\ne 0} : p \nmid m, n$
where $\nmid$ stands for "does not divide".
We must show that the norm axioms for all $r_1$, $r_2 \in \Q$:
=== ===
Let $r \in \Q : r \ne 0$.
Let $k, m\in \Z, n \in \Z_{\ne 0} : p \nmid m, n$.
Suppose $r = 0$.
By definition:
:$\norm {r}_p = 0$
Suppose $r = p^k \dfrac m n \ne 0$
By definition:
:$\norm r_p = \dfrac 1 {p^k} > 0$
Suppose $\norm r_p = 0$.
By definition:
:$r = 0$
=== ===
Suppose $r_1 = 0$ or $r_2 = 0$.
From , $\norm {r_1}_p = 0$ or $\norm {r_2}_p = 0$.
Suppose $r_1 \ne 0 \ne r_2$.
Let $k_1, k_2, m_1, m_2 \in \Z, n_1, n_2 \in \Z_{\ne 0} : p \nmid n_1, n_2, m_1, m_2$
Let $r_1 = p^{k_1} \dfrac {m_1} {n_1}, r_2 = p^{k_2} \frac {m_2} {n_2}$
Then:
:$r_1 r_2 = p^{k_1 + k_2} \dfrac {m_1 m_2}{n_1 n_2}$
We have that $p \nmid m_1$, $p \nmid m_2$.
Since $p$ is prime:
:$p \nmid m_1 m_2$.
Similarly:
:$p \nmid n_1 n_2$.
Therefore:
=== ===
Suppose one of the following is true:
:$r_1 = 0$
:$r_2 = 0$
:$r_1 + r_2 = 0$
Then the result is straightforward.
Suppose $r_1 \ne 0$, $r_2 \ne 0$, $r_1 + r_2 \ne 0$.
Let $r_1 = p^{k_1} \dfrac {m_1} {n_1}, r_2 = p^{k_2} \dfrac {m_2} {n_2}$ where:
:$k_1, k_2, m_1, m_2 \in \Z, n_1, n_2 \in \Z_{\ne 0} : p \nmid m_1, m_2, n_1, n_2$
Then:
By Fundamental Theorem of Arithmetic:
:$\exists ! \tilde k \in \Z_{\ge 0} : \exists m \in \Z : p \nmid m : \tilde m = p^{\tilde k} m$
Obviously, $p \nmid n_1 n_2$
Hence:
All norm axioms are seen to be satisfied.
Hence the result.
== Sources ==
* : $\S 1.2$: Normed and Banach spaces. Normed spaces
Category:P-adic Norm is Norm
P-adic Norm is Well Defined

== Theorem ==
P-adic norm $\norm {\, \cdot \,}_p$ is well defined.
== Proof ==
 $\norm {\, \cdot \,}_p$ is not well defined.
Then, given $r \in \Q$, for two equivalent representations of $r$, $\norm r_p$ will yield two different results. 
Let $k_1, k_2, m_1, m_2 \in \Z, n_1, n_2 \in \Z_{\ne 0} : p \nmid m_1, m_2, n_1, n_2$.
Let $\ds r = p^{k_1} \frac {m_1} {n_1} = p^{k_2} \frac {m_2} {n_2}$, with $k_1 \ne k_2$.
Suppose $k_2 < k_1$.
Then:
:$p^{k_1 - k_2} m_1 n_2 = m_2 n_1$
Therefore:
:$p \divides m_2 n_1$
Since $p$ is prime, it cannot be expressed as a product of selected divisors of both $m_2$ and $n_1$.
Hence $p \divides m_2$ or $p \divides n_1$.
This is a contradiction.
Thus, $k_1 \le k_2$.
Similarly, assuming $k_1 < k_2$ leads to a contradiction.
Hence, $k_1 \ge k_2$.
Since $k_1 \ne k_2$, both $k_1$ and $k_2$ have to be such that:
:$k_1 < k_2$
:$k_1 > k_2$
are satisfied.
No integers satisfy this.
Hence, we reached a contradiction.
== Also see ==
* P-adic Valuation of Rational Number is Well Defined
== Sources ==
* : $\S 1.2$: Normed and Banach spaces. Normed spaces
Category:P-adic Number Theory
P-adic Norm not Complete on Rational Numbers

== Theorem ==
Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rationals $\Q$ for some prime $p$.
Then:
:the valued field $\struct {\Q, \norm {\,\cdot\,}_p}$ is not complete.
That is, there exists a Cauchy sequence in $\struct {\Q, \norm{\,\cdot\,}_p}$ which does not converge to a limit in $\Q$.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
Category:Metric Spaces
Category:Normed Spaces
Category:P-adic Number Theory
Category:P-adic Norm not Complete on Rational Numbers
P-adic Norm not Complete on Rational Numbers/Proof 1/Case 1

== Theorem ==
Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rationals $\Q$ for some prime $p > 3$.
 
Then:
:$\struct {\Q, \norm {\,\cdot\,}_p}$ is not a complete normed division ring.
That is, there exists a Cauchy sequence in $\struct {\Q, \norm{\,\cdot\,}_p}$ which does not converge to a limit in $\Q$.
== Proof ==
Let $p > 3$.
Then there exists $a \in \Z: 1 < a < p-1$.
Consider the sequence $\sequence {x_n} \subseteq \Q$ where $x_n = a^{p^n}$ for some $a \in \Z: 1 < a < p-1$.
Let $n \in \N$.
Then:
:$\norm {a^{p^{n + 1} } - a^{p^n} }_p = \norm {a^{p^n} (a^{p^n \paren {p - 1} } - 1) }_p$
From :
:$a^{p^n \paren {p - 1} } - 1 \equiv 0 \pmod {p^n}$
so:
:$\norm {a^{p^n} \paren {a^{p^n \paren {p - 1} } - 1} }_p \le p^{-n} \xrightarrow {n \to \infty} 0$
That is:
:$\ds \lim_{n \mathop \to \infty} \norm {x_{n + 1} - x_n}_p = 0$
By Characterisation of Cauchy Sequence in Non-Archimedean Norm
:$\sequence {x_n }$ is a cauchy sequence in $\struct {\Q, \norm {\,\cdot\,}_p }$.
 $\sequence {x_n}$ converges to some $x \in \Q$.
That is:
:$x = \ds \lim_{n \mathop \to \infty} x_n$
By Modulus of Limit on a Normed Division Ring:
:$\ds \lim_{n \mathop \to \infty} \norm {x_n }_p = \norm {x }_p$
Since $\forall n, p \nmid a^{p^n} = x_n$, then:
:$ \norm {x_n }_p = 1$
So:
:$\norm x_p = \ds \lim_{n \mathop \to \infty} \norm {x_n}_p = 1$ 
By Axiom (N1) of a norm on a division ring then:
:$x \ne 0$.
Since:
and $x \ne 0$ then:
:$x^{p-1} = 1$
So:
:$x = 1$ or $x = -1$
and so $a-x$ is an integer:
:$0 < a - x < p$
It follows that:
:$p \nmid \paren {a - x}$
and so:
:$\norm {x - a}_p = 1$
Since $x_n \to x$ as $n \to \infty$ then:
:$\exists N: \forall n > N: \norm {x_n - x}_p < \norm {x - a}_p$
That is:
:$\exists N: \forall n > N: \norm {a^{p^n} - x}_p < \norm {x - a}_p$
Let $n > N$:
As $\norm {x - a^{p^n} }_p < \norm {x - a}_p$:
This contradicts the earlier assertion that $\norm {x - a}_p = 1$.
In conclusion:
:$\sequence {x_n}$ is a Cauchy sequence that does not converge in $\struct {\Q, \norm {\,\cdot\,}_p }$.
Category:P-adic Norm not Complete on Rational Numbers
P-adic Norm not Complete on Rational Numbers/Proof 1/Case 2

== Theorem ==
Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rationals $\Q$ for $p = 2$ or $3$.
 
Then:
:$\struct {\Q, \norm {\,\cdot\,}_p}$ is not a complete normed division ring.
That is, there exists a Cauchy sequence in $\struct {\Q, \norm{\,\cdot\,}_p}$ which does not converge to a limit in $\Q$.
== Proof ==
Category:P-adic Norm not Complete on Rational Numbers
P-adic Norm not Complete on Rational Numbers/Proof 2

== Theorem ==
== Proof ==
Hensel's Lemma is used to prove the existence of a Cauchy sequence that does not converge.
==== Lemma 1 ====
Let $x_1 \in \Z_{>0}: p \nmid x_1, x_1 \ge \dfrac {p + 1} 2$
Let $k$ be a positive integer such that $k \ge 2, p \nmid k$.
Let $a = x_1^k + p$.
==== Lemma 2 ====
Let $\map f X \in \Z \sqbrk X$ be the polynomial:
:$X^k - a$
==== Lemma 3 ====
Let $\map {f'} X \in \Z \sqbrk X$ be the formal derivative of $\map f X$.
==== Lemma 4 ====
From Hensel's Lemma there exists a sequence of integers $\sequence {x_n}$ such that:
:$(1) \quad \forall n: \map f {x_n} \equiv 0 \pmod {p^n}$
:$(2) \quad \forall n: x_{n + 1} \equiv x_n \pmod {p^n}$
==== Lemma 5 ====
From Sequence of Consecutive Integers Modulo Power of p is Cauchy in P-adic Norm then:
:$\sequence {x_n}$ is a Cauchy sequence in $\struct {\Q, \norm {\,\cdot\,}_p}$
 $\sequence {x_n}$ is a sequence such that for some $c \in \Q$:
:$\ds \lim_{n \mathop \to \infty} x_n = c$
in $\struct {\Q, \norm {\,\cdot\,}_p}$
From Product Rule for Sequences in Normed Division Ring then:
:$\ds \lim_{n \mathop \to \infty} x_n^k = c^k$
Hence:
:$c^k = a \in \Z$.
From Nth Root of Integer is Integer or Irrational then:
:$c \in \Z$
This contradicts Lemma 2.
So the sequence $\sequence {x_n}$ does not converge in $\struct {\Q, \norm{\,\cdot\,}_p}$.
The result follows.
== Sources ==
* : $\S 3.2$ Completions: Lemma $3.2.3$
Category:P-adic Norm not Complete on Rational Numbers
P-adic Norm not Complete on Rational Numbers/Proof 2/Lemma 1

== Theorem ==
Let $p$ be a prime number.
Then:
:$\exists x \in \Z_{>0}: p \nmid x, x \ge \dfrac {p + 1} 2$
== Proof ==
Let $x = p + 1$.
Then $p \nmid x$ and:
:$x = p + 1 > p > 0$
Category:P-adic Norm not Complete on Rational Numbers
P-adic Norm not Complete on Rational Numbers/Proof 2/Lemma 2

== Theorem ==
Let $p$ be a prime number.
Let $x \in \Z_{> 0}: x \ge \dfrac {p + 1} 2$
Let $k \in \Z_{> 0}: k \ge 2$
Let $a = x^k + p$
 
Then:
:$a \in \Z_{> 0}: \nexists \,c \in \Z : c^k = a$
== Proof ==
Since $x, p > 0$ then $a > 0$.
 for some $c \in \Z:c^k = a$.
Since $c^k \in \Z$, by Nth Root of Integer is Integer or Irrational then:
:$c \in \Z$
Suppose $k$ is odd.
Since $a > 0$, by Odd Power Function is Strictly Increasing then $c > 0$
Hence $a = \size c^k$
On the other hand, suppose $k$ is even, that is $k = 2l$ for some $l \in Z_{> 0}$.
Then:
In either case $\size c \in \Z_{> 0}$ and $\size c^k = a$
Let $d = \size c$
 
By the definition of $a$ it follows that $d^k = x^k + p$
Hence:
Let $y = d^{k - 1} + d^{k - 2} x + d^{k - 3} x^2 + \dotsb + d x^{k - 2} + x^{k - 1}$
Since $d, x \in \Z_{> 0}$ then $d - x \in \Z$ and $y \in \Z$
So $d - x$ and $y$ are factors of $p$
The factors of $p$ by definition are:
:$\pm 1$ and $\pm p$
Since $d, x \in \Z_{> 0}$ then:
Hence $y = p$
Then:
It also follows that $d - x = 1$, that is, $d = x + 1$
Then
This contradicts the previous conclusion that $p \ge d + x$
So:
:$\nexists \,c \in \Z : c^k = a$
Category:P-adic Norm not Complete on Rational Numbers
P-adic Norm not Complete on Rational Numbers/Proof 2/Lemma 3

== Theorem ==
Let $x_1, p, k \in Z_{\gt 0}$ be any positive integers.
Let $a = x_1^k + p$
Let $\map f X \in \Z \sqbrk X$ be the polynomial:
:$X^k - a$
 
Then:
:$\map f {x_1} \equiv 0 \pmod p$
== Proof ==
Category:P-adic Norm not Complete on Rational Numbers
P-adic Norm not Complete on Rational Numbers/Proof 2/Lemma 4

== Theorem ==
Let $p, x_1, k \in \Z_{\gt 0}: p \nmid x_1, p \nmid k$
Let $a \in \Z$ be any integer.
Let $\map f X \in \Z \sqbrk X$ be the polynomial $X^k - a$
Let $\map {f'} X \in \Z \sqbrk X$ be the formal derivative of $\map f X$.
 
Then:
:$\map {f'} {x_1} \not \equiv 0 \pmod p$
== Proof ==
By Euclid's Lemma for Prime Divisors:
:$p \nmid k x_1^{k - 1}$
Hence:
:$k x_1^{k - 1} \not \equiv 0 \mod p$
The formal derivative $\map {f'} X \in \Z \sqbrk X$ of $\map f X$ is by definition:
:$k X^{k - 1}$
Then:
:$\map {f'} {x_1} = k x_1^{k - 1} \not \equiv 0 \pmod p$
Category:P-adic Norm not Complete on Rational Numbers
P-adic Norm not Complete on Rational Numbers/Proof 2/Lemma 5

== Theorem ==
Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rationals $\Q$ for some prime $p$.
Let $k$ be any positive integer.
Let $\map f X \in \Z \sqbrk X$ be the polynomial:
:$X^k - a$
for some $a \in \Z$.
Let $\sequence {x_n}$ be an integer sequence such that:
:$\forall n: \map f {x_n} \equiv 0 \pmod {p^n}$
Then:
:$\ds \lim_{n \mathop \to \infty} {x_n}^k = a$ in $\struct {\Q, \norm {\,\cdot\,}_p}$
== Proof ==
By assumption:
:$\forall n \in \N: p^n \divides \paren { {x_n}^k - a}$
By the definition of the $p$-adic norm:
:$\forall n \in \N: \norm { {x_n}^k - a}_p \le \dfrac 1 {p^n}$
By Sequence of Powers of Number less than One:
:$\ds \lim_{n \mathop \to \infty} \dfrac 1 {p^n} = 0$
By Squeeze Theorem for Real Sequences:
:$\ds \lim_{n \mathop \to \infty} \norm { {x_n}^k - a}_p = 0$
By the definition of convergence in $\struct {\Q, \norm {\,\cdot\,}_p}$:
:$\ds \lim_{n \mathop \to \infty} {x_n}^k = a$
Category:P-adic Norm not Complete on Rational Numbers