Numerical Identities Please can someone explain if this identity is correct:
|a| = $\sqrt{a^2} \ $ 
I thought it should be: |a| = $(\sqrt{a})^2\ $
being that the former would produce an answer that is either positive or negative.
Thank you for your help.
PS: The full question was comparing say:
|2x|/|3y| with $(\sqrt{(-2x/3y)} )^2 \ $
 A: Since $\sqrt{x}$ is defined in $\mathbb{R}$ only for $x\ge 0$ and it's always positive:
the first is correct and the absolute value is necessary , e.g. $\sqrt{(-2)^2}=|-2|=2$
the second is redundant since the square root exists only if $a>0$
An answer to the PS. require a discussion of the sign of $x/y$. Can you do this?
A: You are confusing the standard meaning of the square root symbol with a correct understanding of squares and square roots. There are in fact two solutions to the equation $x^2 = 9$. namely $3$ and $-3$. But the accepted convention is that $\sqrt{9} = 3$, not $\pm 3$.
A: If we examine your suggestion, we would have:
$$|(-3)|=\left(\sqrt{-3}\right)^{2}=\left(\sqrt{3}i\right)^{2} = 3i^{2} = -3 \neq 3$$
Moreover, the $\sqrt{\cdot}$ operator is defined to be non-negative for all non-negative real arguments, so $\sqrt{x^{2}}$ does give us a well defined value $\forall x \in \mathbb{R}$.
In response to your comment that $x \geq 0$ is not stated, we must remember that $x^{2}:\mathbb{R}\to \mathbb{R}^{*}$, where $\mathbb{R}^{*}$ is the set of non-negative real numbers. And so the square-root operator in $\sqrt{x^{2}}$ only ever has a non-negative argument.
