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.

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