Absolute value property proof How can I proof that $|a| < |b|$ iff   $a^2 < b^2$.  I tried by $b^2-a^2$ = a positive and used difference of squares formula $(b-a)(b+a)$ = positive number. I know it's a simple proof but I still got stuck.
 A: Hint: (Assuming we can use the fact that $x^2=|x|^2$ for all $x\in\mathbb{R}$) observe that, $$b^2-a^2=|b|^2-|a|^2=(|b|+|a|)(|b|-|a|)$$
A: Using your idea you can conclude $(b−a)(b+a)>0$ iff $b-a,b+a>0$ or $b-a,b+a<0$.
In the first case you have $a<b$ and $a>-b$ so $|a|<|b|$.
In the second you have $b<a$ and $b<-a$ so $|b|>|a|$.
From both you have the equivalence.
A: Since absolute values are non-negative, it's easy to show that 
$$
|a| < |b| \iff |a|^2 < |b|^2
$$
From there, it suffices to note that $|a|^2 = a^2$, which is to say that $a^2 = (-a)^2$.
A: We have
$$|a|=|-a|$$ and $$a^2=(-a)^2$$
thus WLOG we can suppose
that $a>0 $ and $b >0$.
$x\mapsto x^2$ and $x\mapsto \sqrt {x} $ are strictly increasing  at $(0,+\infty) $.
Done.
A: Note that
$$
|x|=\sqrt{x^2}\quad (x\in\mathbb{R}).
$$
Let $a,b\in \mathbb{R}$. Then
$$
\sqrt{a^2} < \sqrt{b^2}\iff a^2 < b^2
$$
since the square root function is monotonic (for the $\impliedby$ direction) and the squaring function is monontonic on $[0,\infty)$ (for the $\implies$ direction)
A: It's an axiom that if $x >0$ and $a < b$ then $ax < bx$.  So.... 
So $ |a| < |b|\implies |b| > 0 \implies |a|^2 = |a|*|a| < |a|*|b| < |b|*|b| = |b|^2$.
So that (almost) proves the $\Rightarrow$ direction.
If $ |a| \not < |b|$ then either $|a| = |b|$ and $|a|^2 = |b|^2$ and $|a|^2 \not < |b|^2$, or $|b| < |a|$ and by the same reason as above $|b|^2 < |a|^2$ so $|a|^2 \not < |b|^2$.
So  $|a|^2 < |b|^2 \implies $ it is not the case that $|a| \not < |b|$ so $\implies |a| < |b|$.
So that (almost) proves the $\Leftarrow$ direction.
It only remains to show that $|x|^2 = x^2$.
Can you prove that?
Hint:  If $x \ge 0$ then $|x| =x$.  If $x < 0$ then $|x| = (-x)$.  So this comes down to proving $(-x)^2 = x^2$.  Which in turn comes down to proving $x(-y) = -(xy) = (-x)y$
Further hint:  Prove $-(-x) = x$. 
