7
$\begingroup$

It's famously unknown whether the natural log of 2 is rational or not.

How about the natural log of other numbers. Is it known/unknown whether these are rational?

Obviously ln(1) is 0, and ln(2^n) is n*ln(2) (and is thus rational iff ln(2) is rational), but how about other cases?

  • 0
    @Jonas Meyer: Damn, you're good. Yes, that's exactly what I was thinking of!2010-12-23

2 Answers 2

17

To shoot a bird with a cannon...

By the Lindemann-Weierstrass theorem, $e^a$ is transcendental for all $a$ algebraic and non-zero. In particular if $a$ is rational, $e^a$ cannot be rational. Hence $\ln(n)$ is always irrational.

  • 6
    Sometimes, it's fun to nuke mosquitoes... :D2010-12-23
8

We can also use a non-simple continued fraction expansion of $\displaystyle e^{2x/y}$ to prove the irrationality of $\displaystyle e^{2x/y}$ when $\displaystyle x,y$ are positive integers. Thus if $\displaystyle \log n = x/y$, then $\displaystyle e^{2x/y} = n^2 $ is rational, contradicting irrationality of $\displaystyle e^{2x/y}$.

Incidentally, the first proof of irrationality of $\pi$ by Lambert used a continued fraction expansion (of $\tan x$, I believe).

The expansion we use:

alt text

and the theorem we use to prove irrationality is quoted in the wiki page for Generalized Continued Fractions here: Conditions of Irrationality.

By this theorem, it is enough that for all sufficiently large positive integers $\displaystyle m$ we have that $\displaystyle (2m+1)y \gt x^2$, which is true for fixed $\displaystyle x,y$.