# Prove $|a|=\sqrt{a^2}$

I am having difficulty showing that the last case holds and I also just wanted to make sure I am proving this correctly.

Case (i): If $a=0$ then we have $|a|=0=\sqrt{0^2}$ so trivially this is true.

Case (ii): If $a>0$ then we have $|a|=a=\sqrt{a^2}$

Case (iii): This is where I get stuck since if $a<0$ then we have that $|a|=-a$, so if I start from $\sqrt{a^2}=\sqrt{(-a)^2}$ I'm not really sure what to do since I need this being equal to $|a|=-a$, I would like to say $\sqrt{(-a)^2}=-a$ but this seems to be a problem since it could also be written as $\sqrt{(-a)^2}=\sqrt{a^2}=a$

• In the very last part, you said it can also be written as $\sqrt{a^2}=a$. Huh? Where did you get this? You were on the right path before that, so I think you're overthinking it. – Corellian Nov 4 '15 at 5:21
• I was thinking since the negatives from $(-a)(-a)$ cancel and this honestly makes sense to me since $(-2)^2=2^2=4$ but it appears when doing these proofs we only consider the positive root. – Craig Nov 4 '15 at 5:23
• Typically, $\sqrt{u}$ denotes the principal square root of $u$, i.e. the positive square root only. As a result, we have $\sqrt{x^2}=|x|$ for all real $x$. The principal square root is used very frequently just about everywhere (not just in these proofs). – Corellian Nov 4 '15 at 5:36

Your first instinct is right, you can rigorously say $\sqrt{(-a)^2}=-a$ because $a<0$. That is, you can show $-a\geq 0$ and $(-a)^2=\sqrt{(-a)^2}^2$ hence $-a$ is the only solution to $\sqrt{(-a)^2}$.
On the other hand $\sqrt{(-a)^2}=\sqrt{a^2}=a$ is wrong because $a<0$ and a square root cannot be less than zero.
Suppose $a<0$. Then let $b=-a$ and $b>0$. Then by case ii, $|b|=\sqrt{b^2}$, so $|-a|=\sqrt{(-a)^2}$, or $|a|=\sqrt{a^2}$.