# Thread: Prove equality (real numbers)

1. ## Prove equality (real numbers)

a is a real number.

Demonstrate the following equality.

$\surd(a^2) = |a|$

I tried in this way but i think it is wrong.

$\surd(a^2) = a~~~(a\geq0)$

$a\geq0~~so~~a = |a|$

Is it correct?

2. ## Re: Prove equality (real numbers)

You considered only the case $a\ge 0$. Also, since the problem is so basic, the proof has to be really detailed, and I am not sure you sufficiently explained why $\sqrt{a^2}=a$ when $a\ge 0$. So, what happen when $a < 0$? Recall that by definition $\sqrt{x}$ is a nonnegative number $y$ such that $y^2=x$.

3. ## Re: Prove equality (real numbers)

Originally Posted by emakarov
You considered only the case $a\ge 0$. Also, since the problem is so basic, the proof has to be really detailed, and I am not sure you sufficiently explained why $\sqrt{a^2}=a$ when $a\ge 0$. So, what happen when $a < 0$? Recall that by definition $\sqrt{x}$ is a nonnegative number $y$ such that $y^2=x$.

Thats the point the exercise is so basic. But hard to do.

4. ## Re: Prove equality (real numbers)

Originally Posted by Fabio010
Thats the point the exercise is so basic. But hard to do.
Any proof of this depends on what definitions you have and what other properties of real numbers you know.
If could be this simple:
1) if $a\ge 0$ then $\sqrt{a^2}=a=|a|$
2) if $a< 0$ then $\sqrt{a^2}=-a=|a|$.

5. ## Re: Prove equality (real numbers)

Originally Posted by Plato
Any proof of this depends on what definitions you have and what other properties of real numbers you know.
If could be this simple:
1) if $a\ge 0$ then $\sqrt{a^2}=a=|a|$
2) if $a< 0$ then $\sqrt{a^2}=-a=|a|$.
I found some properties in internet , and i tried:

$\sqrt{a^2}=b~~b\geq0~~so~~b^2=a^2~so~(b-a)(b+a)=0$

$b = a,~in~this~case~a\geq0$
$b = -a, in~this~case~a\leq0$

so $\sqrt{a^2}=b=|a|$