I have been trying to prove a certain claim and have hit a wall. Here is the claim…
Claim: If $n$ is a positive integer then
$\log_{7}n$ is an integer or it is irrational
Proof (so far): Let $y=\log_{7}n$. Note that to say
$n$ is a positive integer is equivalent
to saying that n is a non-zero natural
number. We will proceed by trying
to prove the contrapositive.
Claim (alternate version): If $y$ is
a rational number and is not an integer,
then either $n$ is zero or it is not a
natural number.
Given the above we can assume that there
exist integers $a$ and $b$ such that $y$ equals
the quotient of $a$ over $b$. We can also
assume from the premises that $b$ does not
equal one and that $a$ and $b$ are relatively
prime. Note thus that $n$ may be considered
equal to seven raised to the power of $a$
over $b$. Further note that because of this
$n$ cannot be zero or negative. To prove
the claim, one must prove that $n$ is not
a natural number.
Where I am stuck: How can I guarantee from here that $n$
is not a natural number? Is there any
way to concretely ensure that there are
no integers $a$ and $b$ such that the
fractional exponent above will never give
an integer when raising seven to its power?
I have been trying to play around with
a proof that there is no such thing
as a rational root of a prime number, but
that hasn't shook anything loose so far.
Best Answer
Suppose $\log_7 n =\frac{p}{q}$ is rational, then $7^{p/q}=n$, raising both sides to the $q^{\text{th}}$ power, we see that $7^p=n^q$. Now we have by unique prime factorization that $n=7^k$ for some integer $k$, since it divides $7^p$. But then $7^p=7^{kq}$, or $p=kq$, but then $\frac{p}{q}=k$ is an integer as desired.