Root of Quotient equals Quotient of Roots

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Theorem

Let $a, b \in \R_{>0}$ be (strictly) positive real numbers.

Let $n \in \Z_{>0}$ be a strictly positive integer


Then:

$\sqrt [n] {\dfrac a b} = \dfrac {\sqrt [n] a} {\sqrt [n] b}$


where $\sqrt [n] {}$ denotes the $n$th root.


Proof

\(\ds \sqrt [n] {\dfrac a b}\) \(=\) \(\ds \paren {\dfrac a b}^{1 / n}\) Definition of $n$th Root
\(\ds \) \(=\) \(\ds \paren {a \times \dfrac 1 b}^{1 / n}\)
\(\ds \) \(=\) \(\ds a^{1 / n} \times \paren {\dfrac 1 b}^{1 / n}\) Power of Product
\(\ds \) \(=\) \(\ds a^{1 / n} \times b^{-1 / n}\) Exponent Combination Laws/Negative Power
\(\ds \) \(=\) \(\ds a^{1 / n} \times \dfrac 1 {b^{1 / n} }\) Exponent Combination Laws/Negative Power
\(\ds \) \(=\) \(\ds \sqrt [n] a \times \dfrac 1 {\sqrt [n] b}\) Definition of $n$th Root
\(\ds \) \(=\) \(\ds \dfrac {\sqrt [n] a} {\sqrt [n] b}\)

$\blacksquare$


Sources

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