Root of Reciprocal is Reciprocal of Root

Theorem

Let $x \in \R_{\ge 0}$.

Let $n \in \N$.

Let $\sqrt [n] x$ denote the $n$th root of $x$.

Then:

$\sqrt [n] {\dfrac 1 x} = \dfrac 1 {\sqrt [n] x}$

Proof

Let $y = \sqrt [n] {\dfrac 1 x}$.

Then:

\(\ds \sqrt [n] {\dfrac 1 x}\)

\(=\)

\(\ds y\)

\(\ds \leadstoandfrom \ \ \)

\(\ds \dfrac 1 x\)

\(=\)

\(\ds y^n\)

Definition of $n$th root

\(\ds \leadstoandfrom \ \ \)

\(\ds x\)

\(=\)

\(\ds \dfrac 1 {y^n}\)

Reciprocal of Real Number is Unique

\(\ds \leadstoandfrom \ \ \)

\(\ds x\)

\(=\)

\(\ds \paren {\dfrac 1 y}^n\)

Powers of Group Elements/Negative Index

\(\ds \leadstoandfrom \ \ \)

\(\ds \sqrt [n] x\)

\(=\)

\(\ds \dfrac 1 y\)

Definition of $n$th root

\(\ds \leadstoandfrom \ \ \)

\(\ds \dfrac 1 {\sqrt [n] x}\)

\(=\)

\(\ds y\)

Reciprocal of Real Number is Unique

\(\ds \leadstoandfrom \ \ \)

\(\ds \dfrac 1 {\sqrt [n] x}\)

\(=\)

\(\ds \sqrt [n] {\dfrac 1 x}\)

Definition of $y$

Hence the result.

$\blacksquare$