Prove that $\sqrt[n]{x} \cdot \sqrt[n]{y} = \sqrt[n]{x \cdot y}$

Question

I am currently learning about properties of radicals in a high school math class, and I am curious as to why this property holds true?

$\sqrt[n]{x} \cdot \sqrt[n]{y} = \sqrt[n]{x \cdot y}$

Could anyone intuitively explain this to me?

Answer 1

If $\sqrt[n]{a}$ and $\sqrt[n]{b}$ are defined and real for real $a,b$, this is true.

Note that $(xy)^n = x^ny^n$. So if $x=\sqrt[n]{a}$ and $y=\sqrt[n]{b}$ then $$(xy)^n=\left(\sqrt[n]{a}\sqrt[n]{b}\right)^n = (\sqrt[n]{a})^n(\sqrt[n]{b})^n=ab$$

So $z=\sqrt[n]{a}\sqrt[n]{b}$ satisfies $z^n= ab$. So $z=\sqrt[n]{ab}$ by definition. (You'll need some additional arguments that $z$ is the right sign when $n$ is even, but that is easy.)