If $a > b$, is $a^2 > b^2$? Given $a &gt; b$, where $a,b ∈ ℝ$, is it always true that $a^2 &gt; b^2$?
If $\: \color{#c00}{a &gt; b}\: $ then $\: a^2\! -\! b^2 = (\color{#c00}{a\!-\!b})(a\!+\!b) &gt; 0 \iff a\!+\!b &gt;0 $
Given that $ a &gt; b $, it is not always true that $ a^{2} &gt; b^{2} $. One counterexample would be that one of $ a $ or $ b $ is negative, say $ a = 1 $, and $ b = -1 $. Then $$ a^2 = 1^{2} = 1 $$ and $$ b^{2} = (-1)^{2} = 1 $$ making $ a^2 = b^2 $, a contradiction of our assumption that $ a^{2} &gt; b^{2} $.