a is a real number.

Demonstrate the following equality.

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

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

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

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

Is it correct? :)

Printable View

- Oct 17th 2011, 10:55 AMFabio010Prove equality (real numbers)
a is a real number.

Demonstrate the following equality.

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

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

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

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

Is it correct? :) - Oct 17th 2011, 01:45 PMemakarovRe: Prove equality (real numbers)
You considered only the case $\displaystyle a\ge 0$. Also, since the problem is so basic, the proof has to be really detailed, and I am not sure you sufficiently explained why $\displaystyle \sqrt{a^2}=a$ when $\displaystyle a\ge 0$. So, what happen when $\displaystyle a < 0$? Recall that by definition $\displaystyle \sqrt{x}$ is a nonnegative number $\displaystyle y$ such that $\displaystyle y^2=x$.

- Oct 18th 2011, 10:58 AMFabio010Re: Prove equality (real numbers)
- Oct 18th 2011, 11:07 AMPlatoRe: Prove equality (real numbers)
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 $\displaystyle a\ge 0$ then $\displaystyle \sqrt{a^2}=a=|a|$

2) if $\displaystyle a< 0$ then $\displaystyle \sqrt{a^2}=-a=|a|$. - Oct 18th 2011, 11:42 AMFabio010Re: Prove equality (real numbers)