Proof by counter-example

Question

As is probably obvious by the title I cannot work out any set of numbers which disporoves the following conjecture

$$ a^2 > b^2 => a > b $$

where A and B are real numbers.

Anyone to give a push in the right direction as I'm reaching the point of insanity with this probably simple question.

then you would get an imaginary number when you square root it — Abby, Jan 15 '15 at 09:16
So you can go from left to right even with the implication sign as it is — Abby, Jan 15 '15 at 09:19
You should go from left to right because of the implication,if you had $\iff$ you could've started from both sides. — kingW3, Jan 15 '15 at 09:22
Choose any $a,b$ such that $a<b<0$. For example, if $a=-2$ and $b=-1$ then $a^2=4>1=b^2$ but $a\not>b$. — barak manos, Jan 15 '15 at 09:34

score 1 · Answer 1 · answered Jan 15 '15 at 09:21

1

$a^2 > b^2 \implies a^2-b^2 =(a-b)(a+b) >0 $ and this last is true when either .....

answered Jan 15 '15 at 09:21

RikOsuave

1 Answers1