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Are $\pi$ and $\ln(2)$ linearly independent over rational numbers? Are there any proofs either way, or partial results?

user26857
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Andrew Frock
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1 Answers1

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Yes, $\{\pi,\ln(2)\}$ is linearly independent over $\mathbb Q$, because if $r\pi = s\ln(2)$ were true for rational $r$ and $s$, then $e^\pi = 2^{s/r}$ would be algebraic, but it isn't.

Jonas Meyer
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    Doesn't "algebraic over the rationals" allow for expressions of the form $r\pi=s\mathrm{ln}(2)+t$ where $t$ is also rational? – Oscar Cunningham Apr 05 '17 at 15:01
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    @Oscar: You put in quotes "algebraic over the rationals" but I don't know what you're quoting. I interpreted the question as asking whether ${\pi,\ln(2)}$ is a linearly independent set in $\mathbb R$ as a $\mathbb Q$-vector space. You are saying you would interpret it as linear independence of ${\pi,\ln(2),1}$? On the other hand, if algebraic independence was desired then it's much more than you say. – Jonas Meyer Apr 05 '17 at 15:04
  • @OscarCunningham $\pi, \ln{2}$ are independent as vectors over $\mathbb{Q}$ iff $\frac{\ln{2}}{\pi}$ is rational, but as Jonas said, it cannot be. – Joshhh Apr 05 '17 at 15:11