# Proof that the nth root of a times nth root of b is equal to nth root of ab

Consider the following multiplications.

$\sqrt{4} \times \sqrt{16} = 2 \times 4 = 8$
$\sqrt{4 \times 16} = \sqrt{64} = 8$.

Also

$\sqrt[3]{27} \times \sqrt[3]{64} = 3 \times 4 = 12$
$\sqrt[3]{27} \times \sqrt[3]{64} = \sqrt[3]{1728} = 12$.

Try also several calculations like these and observe what happens.

From these calculations, we can observe that

$\sqrt{ab} = \sqrt{a} \times \sqrt{b}$

and

$\sqrt[3]{ab} = \sqrt[3]{a} \times \sqrt[3]{b}$

or, in general

$\sqrt[n]{ab} = \sqrt[n]{a} \times \sqrt[n]{b}$.

We now, prove our conjecture.

Theorem: $\sqrt[n]{a} \sqrt[n]{b} = \sqrt[n]{ab}$

Proof

If we multiply the expression on the left hand side by itself n times, we have

$(\sqrt[n]{a} \sqrt[n]{b})^n = (\sqrt[n]{a})^n (\sqrt[n]{b})^n = ab$.

On the other hand, a positive number ab has only a single positive root, $\sqrt[n]{ab}$.

Therefore,

$\sqrt[n]{a} \sqrt[n]{b} = \sqrt[n]{ab}$.

This completes our proof.