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?

  • $\begingroup$ You can prove that if $a$ is an integer, then $e^a$ is irrational using, e.g., Niven polynomials. Then if $\log p = \frac{a}{b}$, it follows that $e^a = p^b$, a contradiction. Don't know if there is an easier way of showing $e^a$ irrational when $a$ is an integer, though. $\endgroup$ Jul 4, 2012 at 7:08
  • $\begingroup$ More generally: The natural log of any integer ≥2 is irrational. $\endgroup$
    – PAD
    Jul 4, 2012 at 9:37
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    $\begingroup$ Keith Conrad's page is excellent so I can't add to that. To reach a contradiction to $e^2$ being rational, note that if $E$ denotes the sum of even elements of the power series for $e$ and $O$ denotes the odd, that statement would imply that there exists a rational number $r$ such that $E = rO$. $\endgroup$ Jul 4, 2012 at 23:31
  • $\begingroup$ @Eugene, that is interesting but I can't follow the argument, can you explain in more detail why $e^2$ is irrational? $\endgroup$ Jul 5, 2012 at 5:44
  • $\begingroup$ Oh, my apologies, I didn't walk it all the way through myself -- it was more of an attempt to stir more thought. Note that, using the aforementioned notation, $e = E + O$ and $1/e = E - O$, so that if $e^2 = a/b$, $E+O = a/b(E-O)$ and $E = rO$, with $r$ a rational I am too lazy to compute. Proceeding to a contradiction from here requires some further cute insight into the power series I don't immediately see. $\endgroup$ Jul 5, 2012 at 9:23

3 Answers 3


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

  • $\begingroup$ Thanks, this is helping me understand irrationality proofs much better. $\endgroup$ Jul 5, 2012 at 5:41
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    $\begingroup$ On page 3, why did he write that $$e = 1 + \frac{1}{2!}+\frac{1}{3!}+\cdots$$ Isn't that a series for $e-1$? $\endgroup$
    – Argon
    Sep 24, 2012 at 22:03
  • $\begingroup$ @Argon, you're absolutely right - good catch! $\endgroup$ Sep 24, 2012 at 23:25

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.

  • $\begingroup$ +1 nice. Does this also prove that $e^2$ is not a square root $\sqrt{a/b}$? Since otherwise $e^2e^2=\sqrt{a/b}\sqrt{a/b}=a/b$; and I expect $e^4$, to be irrational as well... $\endgroup$
    – draks ...
    Jul 10, 2012 at 22:38
  • $\begingroup$ It does not. Nor can I make it work for $e^3$. $\endgroup$ Aug 4, 2012 at 6:43

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 $\pi^2$.


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