If $a > b$, is $a^2 > b^2$? Given $a > b$, where $a,b ∈ ℝ$, is it always true that $a^2 > b^2$?
 A: The conclusion is correct on $[0, +\infty)$ because that is precisely the interval over which the function $f(x) = x^2$ is an increasing function. No algebra is necessary. Draw the parabola and LOOK!
A: no its not. When $a,b$ are positive it happens. Consider $a=-2$ and $b =-3$. notice that inequality reverses.
A: Yes, when $a$ and $b$ are positive real numbers. In this case, we can write:
$a>b \implies a-b>0 \implies (a+b)(a-b)>0 \implies (a^2)-(b^2)>0$
A: 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.:$$ (-8)^2 > 7^2\\64 > 49$$
A: If $\: \color{#c00}{a > b}\: $ then $\: a^2\! -\! b^2 = (\color{#c00}{a\!-\!b})(a\!+\!b) > 0 \iff a\!+\!b >0 $
A: 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.
A: Given that $ a > b $, it is not always true that $ a^{2} > 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} > b^{2} $.
A: This answer is the same as Bill Dubuque's, but goes into a bit more detail. The question

Does $a>b$ imply that $a^2>b^2$?

is equivalent to

Does $a-b>0$ imply that $a^2-b^2>0$?

In other words, the answer to the bottom question is the same as the answer to the top question.
By factorising $a^2-b^2$, we see that no, $a-b>0$ does not imply that $a^2-b^2>0$:
$$
a^2-b^2=(a+b)(a-b) > 0 \text{ if, and only if, $a+b>0$} \, .
$$
We know that $a-b>0$, but that is not enough. We also require that $a+b>0$ so that $(a+b)(a-b)>0$. If we instead ask

Does $a>b$ and $a>-b$ imply that $a^2>b^2?$

then the answer is in the affirmative.
