Proofs of the radical properties

Rational exponents

$(m^{\frac {1}{n}})^{n}=m^{({\frac {1}{n}}\cdot n)}=m$

We know $({\sqrt[{n}]{m}})^{n}=m$ and $(m^{\frac {1}{n}})^{n}=m$, so we can establish a relationship between them.

$({\sqrt[{n}]{m}})^{n}=(m^{\frac {1}{n}})^{n}$ from the transitive property, raising both sides to the power of n keeps equality.

Taking the nth root of both sides yields ${\sqrt[{n}]{m}}=m^{\frac {1}{n}}$our final result.

${\sqrt[{x}]{m^{n}}}=(m^{n})^{\frac {1}{x}}=(m^{n})^{\frac {1}{x}}=m^{n\cdot {\frac {1}{x}}}=m^{\frac {n}{x}}$

Product and quotient properties

${\sqrt[{x}]{mn}}=(mn)^{\frac {1}{x}}=m^{\frac {1}{x}}\cdot n^{\frac {1}{x}}={\sqrt[{x}]{m}}\cdot {\sqrt[{x}]{n}}$ where n > 0 and m > 0

${\sqrt[{x}]{\frac {m}{n}}}={\Bigl (}{\frac {m}{n}}{\Bigr )}^{\frac {1}{x}}={\frac {m^{\frac {1}{x}}}{n^{\frac {1}{x}}}}={\frac {\sqrt[{x}]{m}}{\sqrt[{x}]{n}}}$where n > 0

Root of a root property

${\sqrt[{m}]{\sqrt[{n}]{x}}}={\sqrt[{m}]{x^{\frac {1}{n}}}}={\Bigl (}x^{\frac {1}{n}}{\Bigr )}^{\frac {1}{m}}=x^{{\frac {1}{n}}\cdot {\frac {1}{m}}}=x^{\frac {1}{mn}}={\sqrt[{mn}]{x}}$