Absolute Value Brackets Proof Code Answer

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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.

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