Integers are Arbitrarily Close to P-adic Integers

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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 for $n \in \N$ there exists unique $\alpha \in \Z$:

$(1): \quad 0 \le \alpha \le p^n - 1$
$(2): \quad \norm { x -\alpha}_p \le p^{-n}$


Proof

Let $n \in \N$.

From Rational Numbers are Dense Subfield of P-adic Numbers:

the rational numbers are dense in $\Q_p$.

So there exists:

$\dfrac a b \in \Q: \norm {x - \dfrac a b}_p \le p^{-n}$


From Unique Integer Close to Rational in Valuation Ring of P-adic Norm, there exists unique $\alpha \in \Z$ such that:

$\norm {\dfrac a b - \alpha}_p \le p^{-n}$
$0 \le \alpha \le p^n - 1$


Then:

\(\ds \norm {x - \alpha}_p\) \(=\) \(\ds \norm {\paren {x - \dfrac a b} + \paren {\dfrac a b - \alpha} }_p\)
\(\ds \) \(\le\) \(\ds \max \set {\norm {x - \dfrac a b}_p, \: \norm {\dfrac a b - \alpha}_p }\) Norm axiom (N4) (Ultrametric Inequality)
\(\ds \) \(\le\) \(\ds p^{-n}\)


Now suppose $\beta \in \Z$ also satisfies conditions $(1)$ and $(2)$, that is:

$0 \le \beta \le p^n - 1$
$\norm { x - \beta}_p \le p^{-n}$

Then:

\(\ds \norm{\dfrac a b - \beta}_p\) \(=\) \(\ds \norm{\paren{\dfrac a b - x} + \paren{x - \beta} }_p\)
\(\ds \) \(\le\) \(\ds \max \set{\norm{\dfrac a b - x}_p,\:\norm{x - \beta}_p}\) Norm axiom (N4) (Ultrametric Inequality)
\(\ds \) \(\le\) \(\ds \max \set{\norm{x - \dfrac a b}_p,\:\norm{x - \beta}_p}\) Norm of Negative
\(\ds \) \(\le\) \(\ds p^{-n}\)

From Unique Integer Close to Rational in Valuation Ring of P-adic Norm, $\alpha \in \Z$ was unique, so:

$\beta = \alpha$

The result follows.

$\blacksquare$


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