For every positive integer $n$, $\log_7(n)$ is either integer or irrational.
I have seen different kinds of proof for this statement, this was my approach kindly verify if I am correct.
$\text{Proof - } \text{ This proof is by contradiction.}$
$\text{Let $log_7(n)$ be rational number which is not an integer.}$
Which is the same thing as saying,
$$log_7(n) = p/q$$
$\text{Where $p,q$ don't have common divisor and $q>1$}$
$\begin{align*} log_7(n) = p/q &\Rightarrow n = 7^{p/q} \\ &\Rightarrow n = 7^p \cdot 7^{1/q} \\ &\Rightarrow (7^p \in \mathbb{Z}) \wedge (7^{1/q} \text{ is irrational}) \end{align*}$
as $7^p$ is a product of integers it is an integer and $\forall(q>1) 7^{1/q}$ is irrational, I have proved it but did not include it to keep this question short.
Is irrational times rational always irrational?
Theorm. Any nonzero rational number times an irrational number is irrational.
Proof - Let $r$ be nonzero and rational and x irrational. If $rx=q$ and $q$ is rational, then $x=q/r$, which is rational. This is a contradiction.
It follows from this that $n$ is an irrational number which is a contradiction as we started with n as integer, hence $log_7(n)$ must be irrational or Integer.
Is this proof sound? I think so my conclusion is not sound enough to conclude that $log_7(n)$ is irrational or Integer. I think so I am not contradicting the assumption; which is what we must do in proof by contradiction?
