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

Proof
Let $n \in \N$.

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

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

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:

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:

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

The result follows.