Prove that $a \equiv b \mod{p^j} \iff |a-b|_{p}\leq p^{-j}$ Prove that $a \equiv b \mod{p^j} \iff |a-b|_{p}\leq p^{-j}$
Typically, when dealing with a congruence I go to the division statement. i.e 
$$a\equiv b\mod{p^j}\Rightarrow p^j|a-b \;\;\;(\star)$$
Moreover, I know that 
\begin{align}
|a-b|_{p}&=p^{-vp(a-b)}\\
&\leq\max{(p^{-vp(a)},p^{-vp(a)})}\;\;\; (\star\star)
\end{align}
However, I am having trouble with the key part that will connect $(\star)$ and $(\star\star)$.
I am hoping someone can help me fill in the missing parts. Thank you.

Instead of thinking of $a-b$ as a difference of two numbers, think of it as a single $p$-adic number. Then $a-b \equiv 0 \pmod {p^j}$ is exactly the statement that $p^j \mid (a-b)$, which is exactly the statement that the $p$-adic norm of $a-b$ is at least $j$ (and is more than $j$ if and only if additional factors of $p$ divide $a-b$, i.e. $a -b \equiv 0 \pmod{p^{j + \ell}}$).
