Why ${(a^2)}^{\frac 12}=\sqrt {a^2}=|a| \neq a$?

Question

Let $a\in \mathbb R$. It should be true that $\sqrt {a^2}=|a|$, since $\sqrt {(-2)^2}=\sqrt{2^2}=2$ and so on. But, it is also true that ${(a^2)}^{\frac 12}=a$, and by definition, ${(a^2)}^{\frac 12}=\sqrt {a^2}=|a|$. Therefore, $a=|a|$, which is not true. So where is the problem?

Answer 1

Yes, the equality $$ \sqrt{a^2}=|a| $$ is correct, because, by definition, the square root of $x$ is the (unique) non negative real number $y$ such that $y^2=x$. Of course we need $x\ge0$ as well, because if $x<0$ there's no number $y$ such that $y^2=x$. Now, $$ |a|^2=a^2 $$ is clear: if $a\ge0$ we have $|a|^2=a^2$; if $a<0$, then $|a|^2=(-a)^2=a^2$.

You can't argue that $-2=2$ from this fact, because $\sqrt{4}=2$, not $-2$.

Why is this the definition? Because we want to be able to do algebraic manipulations with square roots, so we need a definite result.

Don't confuse it with the search for solutions of the equation $x^2=2$; here we want to find the numbers that, squared, are $2$. So the solutions of the equations are $x=\sqrt{2}$ and $x=-\sqrt{2}$.