If $a > b$, is $a^2 > b^2$?

Given $a > b$, where $a,b ∈ ℝ$, is it always true that $a^2 > b^2$?

If $\: \color{#c00}{a > b}\: $ then $\: a^2\! -\! b^2 = (\color{#c00}{a\!-\!b})(a\!+\!b) > 0 \iff a\!+\!b >0 $

no its not. When $a,b$ are positive it happens. Consider $a=-2$ and $b =-3$. notice that inequality reverses.

The correct statement is,$$|a|>|b|\iff a^2 > b^2 $$A counterexample of your hypothesis is $a = 7, b = -8.$

Yes, $a >b $, but $b^2 > a^2$, i.e.:$$ (-8)^2 > 7^2\\64 > 49$$

If $a > b > 0$ then $a^2 > b^2$.

$a > b$ means there is a positive $k$ such that $a = b + k$. Squaring this equation we have $a^2 = b^2 + (2bk + k^2)$ but $2bk + k^2$ is just another positive so $a^2 > b^2$.

The reason we know $2bk + k^2$ is positive is because of the ordered field axioms, one says if $x$ and $y$ are positive so is $xy$ and another says that $x+y$ is positive. That is why we need $b$ to be positive.