Prove: If $|x+y|<|x|+|y|$, then $x<0$ or $y<0$
This looks as though it's true from the start. Take $x=-4, y=4$.
$0<8$ is true.
The question is asking for a proof by contradiction or contrapositive. Which means I am going to negate some part of the ending in order to find a contradiction in the hypothesis.
This is of form: If $ P\implies Q$
So for a proof by contradiction I need:
$P \implies \lnot Q $
If pr the contrapositive: $ \lnot Q \implies \lnot P$
Will the following proof work? Also, is my proof formal enough? What can be done to improve it's form?
PF. (by contradiction)
If $|x+y|<|x|+|y|, \implies x \geq 0 \lor y \geq 0$
$x \geq 0, y \geq0$
since $x \geq 0$
$|x+y|<|x|+|y|$ is false proof by contradiction