Nth Root of Integer is Integer or Irrational

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

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.

$\blacksquare$

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 $6$