Integers are Arbitrarily Close to P-adic 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 for $n \in \N$ there exists unique $\alpha \in \Z$:
 * $0 \le \alpha \le p^n - 1$
 * $\norm { x -\alpha}_p \le p^{-n}$

Proof
Let $n \in \N$.

By definition of the $p$-adic numbers, the rational numbers are dense in $\Q_p$.

So there exists $\dfrac a b \in \Q: \norm {x - \dfrac a b}_p \le p^{-n} \lt 1$

we can assume that $\dfrac a b$ is in Canonical Form.

Consider:

Hence $\dfrac a b$ is an element of the valuation ring of $\struct {\Q, \norm{\,\cdot\,}_p}$.

By Valuation Ring of P-adic Norm on Rationals then:
 * $\dfrac a b \in \Z_{(p)} = \set{ \dfrac c d \in \Q : p \nmid d }$

Hence $p \nmid b$, or equivalently, $\norm {b}_p = 1$.

By Norm of Inverse then $\norm {\dfrac 1 b}_p = 1$.

Since $p \nmid b$, by Prime not Divisor implies Coprime then $p^n \perp b$.

By Integer Combination of Coprime Integers, there exists $m, l \in \Z: mb + lp^n = 1$.

Hence:

And:

By Integer is Congruent to Integer less than Modulus, then there exists $\alpha \in \Z$:
 * $\alpha \equiv am \pmod {p^n}$.
 * $0 \le \alpha \le p^n - 1$

Then $\norm {am - \alpha}_p \le p^{-n}$

Hence:

The result follows.