How to prove that square root of $x$ times $y$ is square root of $x$ times square root of $y$
Simple question - how to prove that:
$\sqrt {x\times y} = \sqrt x \times \sqrt y$ ?
If I use the exponentation the answer seems easy, because
$(x\times y)^n = x^n \times y^n$ because I get
$(x\times y) \times\cdots\times (x\times y)$ (where $x$ occurs $n$ times and $y$ occurs $n$ times) can be rewritten as: $x \times\cdots\times x \times y \times \cdots \times y$.
But in case of square roots that's not so obvious, because I can't rewrite the it the same way. I can of course reason that $\sqrt x$ is $x^{\frac12}$ and $\sqrt y$ is $y^{\frac12}$ and think using induction, but that seems to not satisfy me.
$\endgroup$ 02 Answers
$\begingroup$If $z:=\sqrt{x}\sqrt{y}\ge0$ then $z^2=\sqrt{x}\sqrt{y}\sqrt{x}\sqrt{y}=\sqrt{x}\sqrt{x}\sqrt{y}\sqrt{y}=xy$ so $z=\sqrt{xy}$.
$\endgroup$ 1 $\begingroup$Since $n$ isn't a natural number, we can approach the proof from a different angle.
Change the square root and rewrite it as a power of a $0.5 $, then use the power of a product rule to get:
$$\sqrt{xy} = (xy)^{0.5} = x^{0.5} \cdot y^{0.5} = \sqrt x \cdot \sqrt y$$
This is true, of course, as long as $x$ and $y$ aren't negative numbers.
$\endgroup$ 6
