Proof of $|a-b| < b \implies a > 0$

Question

$|a-b|< b \implies a > 0 $

This seems obvious and like common sense, but how can I prove this in a formally correct way?

Answer 1

$$|a-b|<b \iff -b<a-b<b \iff 0<a<2b.$$

Answer 2

Prove by contradiction. The first thing to notice is that $b\geq0$, because absolute values never go below zero. Assume, therefore, that $a<0$, then $a-b$ is going to be negative, so the value of the absolute value is $b-a$. Since $a<0$ in our assumption, then $b-a-b=-a>0$. This leads to a contradiction. Therefore, $a\geq0$. Finally, prove $a\neq0$.

Answer 3

$$|a-b|<b\implies a-b<b \text{ and } -(a-b)<b \text{ and } b>0$$ $$a-b<b \text{ and } b>0 \implies 0<a<2b$$ $$-(a-b)<b \implies b-a<b \implies -a<0 \implies 0<a$$

So by either case, $0<a$.