Nth Root of Integer is Integer or Irrational

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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.


We prove the contrapositive: if the $n$th root of $x$ is rational, it must be an integer.

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

\(\ds x^{1/n}\) \(=\) \(\ds \frac a b\)
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds \frac {a^n} {b^n}\)

Since $a$ and $b$ are coprime, $a^n$ and $b^n$ are coprime by Powers of Coprime Numbers are Coprime.

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

Suppose $b \ne 1$.

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

This is a contradiction.

Thus $b = 1$, and thus:

$x^{1/n} = a$

which is an integer.


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.