What is the correct solution of $x^{2/2}$?

Question

Ok I get that $$\left(\sqrt x\right)^2 = x$$ and $$\sqrt{x ^2} = |x|.$$

I can explain this to myself when I insert $-1$ for $x$. With complex number $i$ I obtain those solutions. Fine.

Now when we have the power rules $$\left(b^m\right)^n = b^{m n}.$$

Now we can write $\sqrt x = x^{1/2}$. Thus, $\left(\sqrt x\right)^2 = x^{1/2 \ 2}$ but then $\sqrt{x ^2} = x^{2 \ 1/2}$.

So my question is, is $x^{2/2}$ equal $x$ or $|x|$ and why?

Answer 1

Let us start with the final question. The answer is $x^\frac22=x^1=x$.

However, you should not use the notation $\sqrt x$ when $x\in\mathbb{C}\setminus(0,+\infty)$, because then $x$ has two square roots and it is, in general, not clear which one is the one that you have in mind. On the other hand, if we accept that $\sqrt x$ is some square root of $x$, then it is trivial that $\sqrt x^2=x$.

But if $x\notin\mathbb R$ then no square root of $x^2$ is equal to $\lvert x\rvert$. So, I suggest that you don't write that $\sqrt{x^2}=\lvert x\rvert$, unless $x\in\mathbb R$.

Answer 2

If $x\in [0,\infty)$ then it is $$\left(\sqrt x\right)^2 = x$$ and $$\sqrt{x ^2} = |x|.$$

If $x\in (-\infty,0)$ then it is $$\sqrt{x ^2} = |x|$$ and the LHS of $$\left(\sqrt x\right)^2 = x$$ doesn't exist unless you consider complex numbers.

Answer 3

It is a common misconception that the power rule

$$(b^n)^m=b^{nm}$$

holds for any base and any exponent.

For example, $$((-1)^2)^{1/2}\ne(-1)^{2\cdot1/2}.$$

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There is no doubt that $$x^{2/2}=x^1=x.$$