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.

it is like definition.
$p^j \mid a-b$ implies $a-b=p^tm$ where $t \geq j$ and $(m,p)=1$. Now $|a-b|_p=p^{-t}\leq p^{-j}$.
