4
$\begingroup$

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 $\sqrt{(-a)^2}=\sqrt{a^2}=a$

$\endgroup$
  • 2
    $\begingroup$ 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. $\endgroup$ – Corellian Nov 4 '15 at 5:21
  • 1
    $\begingroup$ 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. $\endgroup$ – Craig Nov 4 '15 at 5:23
  • 1
    $\begingroup$ 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). $\endgroup$ – Corellian Nov 4 '15 at 5:36
2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

Suppose $a<0$. Then let $b=-a$ and $b>0$. Then by case ii, $|b|=\sqrt{b^2}$, so $|-a|=\sqrt{(-a)^2}$, or $|a|=\sqrt{a^2}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.