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

Question

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

Answer 1

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

Answer 2

If $\rm\ a > b\ $ then $\rm\ a^2 - b^2 = (a-b)\:(a+b) > 0\ \iff\ a+b >0 $

Answer 3

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.

Answer 4

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.:$$ (-8)^2 > 7^2\\64 > 49$$

Answer 5

Yes, when $a$ and $b$ are positive real numbers. In this case, we can write:

$a>b \implies a-b>0 \implies (a+b)(a-b)>0 \implies (a^2)-(b^2)>0$