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I’m stuck with the following problem:

Prove that $\log_{2} 3 \in \mathbb{R} - \mathbb{Q} $ .

Thanks in advance!

Constantin Asavinei-Apostol
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  • Is it really correct to denote $\mathbb R\setminus \mathbb Q$ as $\mathbb R- \mathbb Q$? – user26486 Oct 03 '14 at 20:20
  • Prove?? Heck, I've known they were irrational ever since Mr Jesse first talked about them! – Daniel R Hicks Oct 03 '14 at 21:02
  • @mathh Yes. Those are two different notations for the same thing. I prefer $\mathbb R-\mathbb Q$, because it is clearer to me, but other people prefer $\mathbb R\setminus\mathbb Q$. – Justin Oct 04 '14 at 03:56

2 Answers2

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Suppose that $\log_23$ is rational, i.e. $$\log_23=\frac{n}{m}$$ where $m,n$ are positive integers.

Then, we have $$\log_23=\log_22^{\frac{n}{m}}\Rightarrow 3=2^{\frac nm}\Rightarrow 3^m=2^n.$$ Here, the LHS is odd and the RHS is even, which is a contradiction.

mathlove
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Assuming that $$\log_2(3)=\frac{p}{q},$$ you get: $$ 3 = 2^{p/q}$$ or: $$ 3^q = 2^p, $$ contradiction.

Jack D'Aurizio
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