# Integer to Rational Power is Irrational iff not Integer or Reciprocal

## Theorem

Let $m$ be a rational number.

Let $n$ be a positive integer.

Then $n^m$ is an irrational number if and only if $n^{\left\lvert{m}\right\rvert}$ is not an integer.

## Proof

### Necessary Condition

Let $n^{\left\lvert{m}\right\rvert} \notin \Z$.

We have that:

- $n^m = n^{u/v}$ for some $u \in\Z$ and $v \in \Z_{\ne 0}$

Then it follows from the definition of a rational power and the existence of a real $v$th root of $n^u$ that:

- $n^m = n^{u/v} = \paren {n^u}^{1/v} \in \R$

Aiming for a contradiction, suppose $n^m \in \Q$.

We have that:

- $\left\lvert{m}\right\rvert = 0 \implies n^{\left\lvert{m}\right\rvert} \in \Z$

It follows that:

- $\left\lvert{m}\right\rvert > 0$

Then:

- $\left\lvert{m}\right\rvert>0$
- $m \in \Q$
- $n^{\left\lvert{m}\right\rvert} > 0$
- $n^m \in \Q$

imply that

- $n^{\left\lvert{m}\right\rvert} = n^{p/q} = \dfrac r s$

for some $\left({p, q, r, s}\right) \in \Z_{>0}^4$ where $r$ and $s$ have no common prime factors.

Raising both sides of $n^{p/q} = \dfrac r s$ to the power of $q$ yields:

- $n^p = \dfrac {r^q} {s^q}$

We have that:

- $\dfrac {r^q} {s^q} = n^p \in \Z$

Hence $r^q$ is divisible by $s^q$.

By the Fundamental Theorem of Arithmetic, $r^q$ and $s^q$ have the same prime factors as $r$ and $s$ respectively.

But we have that:

- $s^q \mathrel \backslash r^q$

and:

- $s \ne 1$

imply that $r$ and $s$ have a common prime factor.

But this would contradict the fact that $r$ and $s$ have no common prime factors.

Therefore $s = 1$.

Then:

- $n^{\left\lvert{m}\right\rvert} = \dfrac r s$

and:

- $s = 1$

imply that:

- $r = n^{\left\lvert{m}\right\rvert} \notin \Z$

From this contradiction it follows that:

- $n^m \notin \Q$

and since $n^m \in \R$:

- $n^m \in \R \setminus \Q$

That is, $n^m$ is an irrational number.

$\Box$

### Sufficient Condition

Now let $n^m$ be an irrational number.

That is:

- $n^m \in \R \setminus \Q$

Then because $n^{-m}$ is the reciprocal of $n^m$:

- $n^{-m} \in \R \setminus \Q$

So:

- $n^{\left\lvert{m}\right\rvert} \in \R \setminus \Q$

and so since $\Z \subseteq \Q$:

- $n^{\left\lvert{m}\right\rvert} \notin \Z$

$\blacksquare$