# Proofs Involving the Triangle Inequality

I have solved the Triangle Inequality and the Reverse Triangle Inequality (in the real numbers), but for some reason, these two parts are giving me trouble:

(a) If |x-y|< c, then |x|<|y|+c.

(b) If |x-y|< a for all a>0, then x=y.

For a) it suffices to show that $|x| - |y| \leq |x-y|$. By the triangle inequality, $|x| = |x-y+y| \leq |x-y| + |y|$. Hence $|x|- |y| \leq |x-y| + |y| - |y| = |x-y|$.

• How does that prove (a)? Feb 7 '18 at 1:41
• Because $c > |x-y| \geq |x|-|y|$ implies $|x| < |y| +c$. Feb 7 '18 at 1:43

For $b)$, using the squeeze theorem: $$|x-y|<a \Rightarrow -a<x-y<a$$ $$\lim_\limits{a\to 0} (x-y)=0 \Rightarrow x=y.$$

For b.) Without using the squeeze theorem, it suffices to show that

$$|x-y| < a \quad \forall a > 0 \implies |x-y| = 0$$

By Contradiction, suppose $|x-y| < a \$ for all $\ a > 0$ but $|x-y| \neq 0$,

then we must have $|x-y| > 0$, set $a = |x-y|$, then $$|x-y| < a = |x-y|$$

In particular, $$|x-y| < |x-y|$$

This violates trichotomy $($since also $|x-y| = |x-y|\ )$

and hence we have a contradiction.

We therefore conclude $|x-y| = 0$ and thus $x = y$