Absolute value of real numbers and inequality

Question

Let $a$ and $b$ be real numbers. Show that

$\vert a-b\vert < \epsilon \Rightarrow \vert a\vert<\vert b\vert+\epsilon$

for $\epsilon>0$.

Looks quite easy but I'm not getting it. I tried to use triangular inequality in many forms but it doensn't come.

Answer 1

You had the idea -- this is indeed the Triangle Inequality: $$|a| = |(a-b) + b| \leq |a-b| + |b| < \epsilon + |b|.$$

Answer 2

$$|a - b| < \epsilon $$ $$|a| = |(a-b) + b| < |a-b| + |b| $$ Therefore by using (1) in (2) $$|a| < |b| + \epsilon$$