Prove $\log_7 n$ is either an integer or irrational 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.
 A: Usually negative statements are proven by contradiction.
The definition of an irrational number is a number which is not rational. This suggest that you should try contradiction.
Proof start
Assume by contradiction that $\log_7 (n) =\frac{a}{b}$ with $a,b \in \mathbb Z$ and $b \nmid a$.
This implies that 
$$7^{\frac{a}{b}}=n$$
or 
$$n^b=7^a$$
What does this tell you about the prime factorisation of $n$? From here it should be easy.
A: 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.
A: The contrapositive is an awkward approach to take in this problem, because the alternatives are $\log_7 n$ is either an integer or irrational in the conclusion.
So the contrapositive would be to assume $\log_7 n$ is a rational number which is not an integer, and then try to prove $n$ is not a positive integer.
An approach using the Rational Root Theorem seems easier to explain.
Let's assume that $n$ is a positive integer and that $x = k/m$ is rational, i.e. the ratio of two coprime integers, say $m \gt 0$, and then show that in fact $x$ would have to be an integer.  Note that:
$$ x = \log_7 n \implies 7^x = 7^{k/m} = n \implies n^m - 7^k = 0 $$
Now $n$ satisfies a polynomial.  If $k \ge 0$, then this polynomial has integer coefficients, and since the leading coefficient is one, any rational root $n$ would have to be an integer of the form $n \mid 7^k$.  Since $n$ is known to be a positive integer, then $n = 7^d$ for some integer $d$ between $0$ and $k$, and thus $x = \log_7 n = d$ is an integer.
On the other hand, suppose $k \lt 0$.  Then we convert the above to an integer polynomial by multiplying by $7^{-k}$:
$$  7^{-k} n^m - 1 = 0 $$
Now the Rational Roots Theorem says $n$ must have the form $\frac{\pm 1}{7^d}$ where $d$ is an integer between $0$ and $-k$.  But we have assumed $n$ is a positive integer (in this proof by contradiction), so the sign of $n$ must be positive and $d$ must be zero.  That is, $n=1$ and $x = \log_7 n = 0$, which is an integer.
Therefore if $n$ is a positive integer, then $x = \log_7 n$ is either irrational or it is an integer.  QED
