Nth Root of Integer is Integer or Irrational

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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.

Suppose for the sake of contradiction that 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 no new prime factors are introduced.

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