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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?

NebulousReveal
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  • Yes, Virginia, this is related. – J. M. ain't a mathematician Dec 23 '10 at 03:38
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    $\log(2)=\frac{p}{q}\Rightarrow e^p=2^q$. Because $e$ is not algebraic, $\log(2)$ is not rational. (Willie Wong's answer includes this and much more, but I thought a comment up here countering the first sentence might help.) – Jonas Meyer Dec 23 '10 at 03:40
  • Thanks Jonas, I was a little confused by that statement. – picakhu Dec 23 '10 at 03:43
  • For some reason, I thought no one knew whether the sum of the alternating harmonic sequence (ln(2)) was rational or not. Guess I was wrong! –  Dec 23 '10 at 04:39
  • This question begs me to ask if there is a trivial method to show that e is irrational. – picakhu Dec 23 '10 at 04:47
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    @barrycarter: Perhaps you were thinking of http://en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant. @picakhu: Proving that $e$ is irrational isn't too hard. Concerning the difficulty of the proof that it is transcendental, you may be interested in the following if you haven't seen it: http://math.stackexchange.com/questions/12872/how-hard-is-the-proof-of-pi-or-e-being-transcendental – Jonas Meyer Dec 23 '10 at 05:09
  • @picakhu This is an easy proof that e is irrational. I'm not sure it qualifies as trivial but is certainly not that involved. – Adrián Barquero Dec 23 '10 at 05:11
  • @Jonas Meyer: Damn, you're good. Yes, that's exactly what I was thinking of! –  Dec 23 '10 at 05:30

2 Answers2

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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.

J. M. ain't a mathematician
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Willie Wong
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    (Yes, that answer is given somewhat tongue in cheek. If there exists a $r\in\mathbb{Q}$ such that $e^r = n$, then raising to a higher power we have that there exists $p,q\in \mathbb{N}$ such that $e^p = q$. This will imply that $e$ is algebraic. But we know $e$ is transcendental.) – Willie Wong Dec 23 '10 at 03:32
  • why is that shooting a bird with a canon? Answering a whole class of problems in one go and not to bother to deal with each individual case should be applauded. So what if some people's elephant is reduced to a fly? – jimjim Dec 23 '10 at 03:33
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    @Arjang: see my comment above. It suffices in reality to know that $e$ is transcendental, a result which predates the Lindemann-Weierstrass theorem. I just like to mention the LW theorem when there is a chance (as if that is not common enough on math websites). – Willie Wong Dec 23 '10 at 03:37
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    Sometimes, it's fun to nuke mosquitoes... :D – J. M. ain't a mathematician Dec 23 '10 at 03:57
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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 $\displaystyle x,y$.

Aryabhata
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