# Nth Root of Integer is Integer or Irrational

## Contents

## Theorem

Let $n$ be a natural number.

Let $x$ be an integer.

If the $n$th root of $x$ is not an integer, it must be irrational.

## Proof

Suppose that $x^{1/n}$ is not an integer.

Aiming for a contradiction, suppose the $n$th root of $x$ is rational.

Then by Existence of Canonical Form of Rational Number, there exist an integer $a$ and a natural number $b$ which are coprime such that:

\(\displaystyle x^{1/n}\) | \(=\) | \(\displaystyle \frac a b\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle x\) | \(=\) | \(\displaystyle \frac {a^n} {b^n}\) |

Since $x$ is an integer, $a^n$ and $b^n$ must share a common factor if $b \ne 1$.

Let $a^n$ and $b^n$ be coprime.

Let $b \ne 1$.

Then $\dfrac {a^n} {b^n}$ is by definition in canonical form.

As the denominator of $\dfrac {a^n} {b^n}$ is not $1$, $\dfrac {a^n} {b^n}$ is not an integer.

However, since $a$ and $b$ are coprime, $a^n$ and $b^n$ are coprime because Natural Number has Same Prime Factors as Integer Power.

Thus, $b$ must equal $1$.

\(\displaystyle x\) | \(=\) | \(\displaystyle a^n\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle x^{1/n}\) | \(=\) | \(\displaystyle a\) |

Thus, $x^{1/n}$ must be an integer, which is a contradiction.

Therefore the $n$th root of an integer must be irrational if it is not an integer.

$\blacksquare$

## Historical Note

The fact that the Square Root of 2 is Irrational was known to Pythagoras of Samos.

Theodorus of Cyrene proved that the square roots of the natural numbers from $3$ to $17$, except for $4$, $9$ and $16$, are irrational.

He clearly did not have a general proof of this phenomenon.

## Sources

- 1982: P.M. Cohn:
*Algebra Volume 1*(2nd ed.) ... (previous) ... (next): $\S 2.2$: Divisibility and factorization in $\mathbf Z$: Exercise $6$