Show that $|a-b|<\epsilon \Rightarrow |a|<|b|+\epsilon$

Question

For any $\epsilon>0$ and $a,b\in\mathbb{R}$ show that $$|a-b|<\epsilon\Rightarrow |a|<|b|+\epsilon$$

In some notes I found the following property $$|a|-|b|\leq ||a|-|b||\leq|a-b|$$ So $$|a-b|<\epsilon\Rightarrow |a|-|b|<\epsilon\Rightarrow |a|<|b|+\epsilon$$

I have two doubts

1) Is it right?

2)In triangular inequality I have that $$|a+b|\leq |a|+|b|$$ $$|a+(-b)|\leq |a|+|-b|\leq|a|+|b|$$

Is wrong do that?

Answer 1

Well, we can do it directly as follows: $$\begin{align} |a|&=|(a-b)+b|\quad\text{then use Triangle Inequality applied to $(a-b)$ and $b$ to get}\\ &\leq |a-b|+|b|\quad\text{use the hypothesis that $|a-b|<\epsilon$}\\ &<\epsilon+|b|. \end{align}$$

Answer 2

Yes, both of those are correct manipulations. The first proves the inequality, but you'll have a hard time getting from the second to a proof. The inequalities don't line up to combine the way one would like.