Integer to Rational Power is Irrational iff not Integer or Reciprocal

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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 $\Z \subseteq \Q$:

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

$\blacksquare$