Why does $\sqrt{ab} = \sqrt{a}\sqrt{b}$? ( Considering that they are both surds)

algebra-precalculus

Why does $\sqrt{ab} = \sqrt{a}\sqrt{b}$? ( Considering that they are both surds)

From the definition of a nth root that means "what number multiplied by itself n times will get you back to the original number under the root", only justifies for single nth roots. Eg. square root of $49$ is $7$; because $7$ times $7$ gets you $49$. However, this doesn't explain why you can combine two surds together by just multiplying the numbers beneath the roots. Is there a logical proof of this that can make me understand this better?

Best Answer

If the definition of $y = \sqrt x$ is "that number $y$ for which $y^2 = x$", and if you also agree that (positive) square roots are unique (there is only one; any two positive numbers $y_1$, $y_2$ with the property that $y_1^2 = x = y_2^2$ must in fact be the same, $y_1 = y_2$), then it is not hard to prove that $\sqrt a \sqrt b = \sqrt {ab}$.

Here is a proof: if you claim that $y_1 = \sqrt a \sqrt b$ and $y_2 = \sqrt{ab}$ are in fact the same, you can demonstrate it by showing that they satisfy the same property; namely, $y_1^2 = y_2^2 = ab$.

Indeed, $y_1^2 = (\sqrt a \sqrt b)^2 = \sqrt a^2 \sqrt b^2$ (by commutativity of multiplication), and $\sqrt a^2 \sqrt b^2 = ab$ by definition of square root.

Likewise, $y_2^2 = \sqrt{ab}^2 = ab$ by definition of square root.

Hence $y_1^2 = y_2^2$, ie each of $y_1$ and $y_2$ satisfy the property of "is a square root of $ab$"; since (positive) square roots are unique, then $y_1 = y_2$.

You can present an even shorter proof once you are savvy with definitions; observe $(\sqrt a \sqrt b)^2 = \sqrt a^2 \sqrt b^2 = ab$, then immediately by definition we have $\sqrt{ab} = \sqrt a \sqrt b$.