Prove that if $|x-y| < c$ then $|x| < |y| + c$

Workings:

I have to ideas of a proof

Proof 1:

$$|x| =|x-y+y|leq |x-y|+|y|leq c+|y|$$

Thus $|x| < |y| + c$

Proof 2:

From the triangle inequality:

$$-|x-y| leq |x| – |y| leq |x-y| < c$$

So $$|x| – |y| < c$$

Split this into two cases:

Case 1: When $|x| > |y|$

Case 2: When $|y| > |x|$

When $$|x| > |y|, |x-y| > 0$$ Meaning $c > 0$

So by (< preserved by addition) $|x|$ < $|y| + c$

When $|y|$ > $x$, $x-y < 0$, though $|x-y| > 0$. So $x-y < 0 < |x-y|$

And $c > |x-y|$

So $c > 0$

So $|x| < |y| + c$

I’m not sure if what one, both or neither of my proofs are correct. So any help will be appreciated.

## Answer

Both of your proofs look perfectly fine for me. Except for where you added cases. You’ll notice you get exactly the result you were trying to prove *before* you had to consider cases. Hence, there is no need for it. You also don’t need to worry about whether or not $c>0$. (Although by your assumption that $|x-y|<c$ you were already implicitly given that $c>0$.) To be slightly more thorough, you might want to add a line to the beginning of your second proof where you begin with $||x|-|y||leq |x-y|$ before moving to $-|x-y|leq |x|-|y|leq |x-y|$. But that depends on whoever will be reading your proof, and may be clear without the extra line.