Prove that If $a$ and $b$ are real numbers with $a < b < 0$ then $a^2 > b^2$

Question

Suppose $a$ and $b$ are real numbers. Prove that if $a < b < 0$ then $a^2 > b^2$.

My attempt:

1. We know that if $x > y$, then $-x < -y$.

2. We also know that if $x > y\ge 0$, then $x^2 > y^2$.

Now consider our example:

Given that $a < b < 0$

$$\tag1 a<b$$

$$\tag2 -a > -b$$

$$\tag3(-a)^2 > (-b)^2$$

$$\tag4 a^2 > b^2$$

Is it correct? Any suggestions for improvement would be welcome.

Answer 1

The idea is fine, but look again at what you wrote; it is just a sequence of assertions, with no links between them. I suggest that you type it as follows:\begin{align}a<b&\implies-a>-b\\&\implies(-a)^2>(-b)^2\text{ (since $-a>-b\geqslant0$)}\\&\iff a^2>b^2.\end{align} Another possibility consists in noting that$$a^2-b^2=\overbrace{(a-b)}^{\phantom{0}<0}\,\overbrace{(a+b)}^{\phantom{0}<0}>0.$$

Answer 2

Your solution is right!

$f(x)=x^2$ increases on $[0,+\infty).$

By your work $-a>-b>0.$

Thus, indeed $(-a)^2>(-b)^2$ and we are done!

I think, it's better to use that $f$ decreases on $(-\infty,0].$

From this we'll obtain the statement immediately.