Why $\sqrt {xy} = \sqrt x \sqrt y$ and same for division

Question

Why $\sqrt {xy} = \sqrt x \sqrt y$ and same for division? I found question like these on website but I don't anything about precalculus and not even whole algebra so I want to know the proof with basic concepts that can explain it.

Answer 1

$\sqrt x$ is defined to be the unique non-negative number such that $\sqrt x^2=x$. The same goes for $\sqrt y$ and $\sqrt{xy}$. They, too, are the unique non-negative numbers whose square is $y$ or $xy$, respectively. Now since $\sqrt x$ and $\sqrt y$ are non-negative, their product $\sqrt x\sqrt y$ is also non-negative. In addition, we have that $$\left(\sqrt x\sqrt y\right)^2=\sqrt x^2\sqrt y^2=xy.$$ So $\sqrt x\sqrt y$ is a non-negative number whose square is $xy$. So by definition, $\sqrt{xy}$ must be this number. Or more concisely: $$\sqrt x\sqrt y=\sqrt{xy}.$$ For division, we must first show that $\sqrt{\frac{1}{x}}=\frac{1}{\sqrt x}$. This works exactly the same: $\frac{1}{\sqrt x}$ is non-negative and its square is $\frac{1}{x}$, so by definition it is the square root of $\frac{1}{x}$. Now apply both rules to get $$\sqrt{\frac{x}{y}}=\sqrt{x\frac{1}{y}}=\sqrt x\sqrt{\frac1y}=\sqrt x\frac{1}{\sqrt y}=\frac{\sqrt x}{\sqrt y}.$$

Answer 2

It follows from laws of exponents.

$$(ab)^m=a^mb^m$$

Since $\sqrt x=x^{1/2}$, it follows that $$\sqrt{xy}=(xy)^{1/2}=x^{1/2}y^{1/2}=\sqrt x\sqrt y$$ for non-negative $x,y$.

As for division, simply by understanding that $$\frac{x}{y}=x\cdot\frac1y$$ and following the above reasoning should be sufficient in showing that the same applies.