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.

  • $\begingroup$ Is \log_7n meant to be $\log_7 n$? $\endgroup$ – 6005 Aug 11 '15 at 16:28
  • $\begingroup$ Yes. I am having a little trouble with the LaTeX. $\endgroup$ – Johnq Aug 11 '15 at 16:28
  • $\begingroup$ Johnq, I hope your question looks okay now. Sorry for all the editing. $\endgroup$ – 6005 Aug 11 '15 at 16:36

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.

  • $\begingroup$ Are you starting from my original claim? $\endgroup$ – Johnq Aug 11 '15 at 17:07
  • $\begingroup$ @Johnq yes, I'm showing that if $\log_7 n$ is rational, it must be an integer. $\endgroup$ – jgon Aug 12 '15 at 7:43

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.

  • $\begingroup$ So would you say that the contrapositive in this case was the wrong way to go? $\endgroup$ – Johnq Aug 11 '15 at 16:57
  • $\begingroup$ I am a bit stumped at what you mean by "what does this tell you about the prime factorization of n". I see that 7 divides $n^b$ ,but from there I don't see what that says about n, a, or b. $\endgroup$ – Johnq Aug 11 '15 at 17:20
  • $\begingroup$ Would it suggest that n can only be comprised of roots of 7? $\endgroup$ – Johnq Aug 11 '15 at 17:25
  • $\begingroup$ @Johnq Contrapositive probably also works. But most of the times if contrapositive works, so does contradiction. $\endgroup$ – N. S. Aug 11 '15 at 17:53
  • $\begingroup$ @Johnq If $p |n$ then $p|n^b=7^a$. This shows that the only prime dividing $n$ is $7$. $\endgroup$ – N. S. Aug 11 '15 at 17:54

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

  • $\begingroup$ What part of this serves as the proof that n could be irrational? $\endgroup$ – Johnq Aug 11 '15 at 18:28
  • $\begingroup$ As I explain in the opening paragraph, my preferred approach is not a strict contrapositive as you tried to pursue. I agree with what you state as the strict contrapositive, restating this in the second paragraph. In the fourth paragraph I give the logically equivalent statement, in which we will assume both $\log_7 n$ is rational and $n$ is a positive integer, and prove that then $\log_7 n$ is an integer. $\endgroup$ – hardmath Aug 11 '15 at 18:33

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