Proof by counter-example

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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.

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