What is the simplest way to prove that the logarithm of any prime is irrational?

Question

What is the simplest way to prove that the logarithm of any prime is irrational?

I can get very close with a simple argument: if $p \ne q$ and $\frac{\log{p}}{\log{q}} = \frac{a}{b}$, then because $q^\frac{\log{p}}{\log{q}} = p$, $q^a = p^b$, but this is impossible by the fundamental theorem of arithmetic. So the ratio of the logarithms of any two primes is irrational. Now, if $\log{p}$ is rational, then since $\frac{\log{p}}{\log{q}}$ is irrational, $\log{q}$ is also irrational. So, I can conclude that at most one prime has a rational logarithm.

I realize that the rest follows from the transcendence of $e$, but that proof is relatively complex, and all that's left to show is that no integer power of $e$ is a prime power (because if $\log p$ is rational, then $e^a = p^b$ has a solution). It is easy to prove that $e$ is irrational ($e = \frac{a}{b!} = \sum{\frac{1}{n!}}$, multiply by $b!$ and separate the sum into integer and fractional parts) but I can't figure out how to generalize this simple proof to show that $e^x$ is irrational for all integer $x$; it introduces a $x^n$ term to the sum and the integer and fractional parts can no longer be separated. How to complete this argument, or what is a different elementary way to show that $\log{p}$ is always irrational?

Answer 1

A proof of the irrationality of rational powers of $e$ is given on page 8 of Keith Conrad's notes.

Answer 2

I figured out an elementary proof that $e^2$ is irrational, although it doesn't seem to generalize to other powers:

Suppose $e^2$ is rational. Then there exist $a, b, k, n$ such that $e^2 \cdot 2^k = \frac{a}{b}$, $b = \frac{(2^n)!}{2^{2^n-1}}$, and $b$ is an odd integer. By definition (Taylor series for $e^x$) we have $2^k \cdot e^2 = 2^k \cdot \sum_{j \ge 0}{\frac{2^j}{j!}} = H_n + T_n$, $H_n = 2^{k} \cdot \sum_{0 \le j \le 2^n}{\frac{2^{j}}{j!}}$, $T_n = 2^k \cdot \sum_{j \gt 2^n}{\frac{2^j}{j!}}$. Since every term of the sum in the definition of $H_n$ can be reduced to an odd denominator, and $b$ is an integer divisible by every such odd denominator, $b \cdot H_n$ is an integer. And $0 \lt b \cdot T_n \lt \frac{2^{k+1}}{2^n}$, so for sufficiently large $n$, $b \cdot T_n$ is not an integer. But $2^k \cdot b \cdot e^2 = b \cdot H_n + b \cdot T_n = a$ is an integer, contradicting the assumption that $e^2$ is rational.

Answer 3

I'm not sure if this qualifies as a very elementary proof (and I have not worked out all the details myself), but there appears to be quite a nice proof that any rational power of $e$ is irrational in Chrystal's Algebra, which might well be available online, as it is certainly old enough to be out of copyright. Interestingly, Chrystal refers to his "Algebra" as "An Elementary Text-Book".

In my copy (Vol II. published in 1889) the result appears as Corollary 3, on page 495, in chapter XXXIV on "General Continued Fractions". He first proves a result about general continued fractions, to the effect that:

If $a_2, a_3, \dots$ and $b_2, b_3, \dots$ are all positive integers, then the continued fraction $$\cfrac{b_2}{a_2 + \cfrac{b_3}{a_3 + \cfrac{b_4}{a_4 + \dots}}}$$

converges to an irrational limit provided that after some value of $n$ the condition $a_n\nless b_n$ be always satisfied.

He then deduces the above Corollary 3, from the expansion

$$\tanh x = \cfrac{x}{1 + \cfrac{x^2}{3 + \cfrac{x^2}{5 + \dots}}},$$

although he does not explain the deduction in details, but simply states that it can be deduced in a similar way to a previous result he gives concerning $\pi$ and $\pi^2$.