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I know that the natural log of any positive algebraic number is transcendental, as a consequence of the Lindemann-Weierstrass theorem, but what about the natural log of the square root of two (which is irrational).

Is this rational or irrational?

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    $\begingroup$ $\sqrt2$ is algebraic: it is the root of $x^2-2=0$ $\endgroup$ – robjohn Aug 25 '14 at 12:29
  • $\begingroup$ $\ln 1=0$.${}{}$ $\endgroup$ – Andrés E. Caicedo Aug 25 '14 at 13:17
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Not only is $\ln(\sqrt{2})$ irrational, but it's also transcendental!

Proof: $$\Large \ln(\sqrt{2})=\ln(2^{1/2})=\frac{1}{2} \underbrace{\ln(2)}_{\in \mathbb{T}}$$ which is transcendental. $\square$

To see why the product of a transcendental number and a non-zero algebraic number is transcendental, see this .


For reference, $\mathbb{T}$ is the set of transcendental numbers.

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    $\begingroup$ How do they know that $\log(2)$ is transcendental? Because $2$ is a positive algebraic number. However, $\sqrt2$ is also a positive algebraic number. $\endgroup$ – robjohn Aug 25 '14 at 12:41
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    $\begingroup$ @robjohn Ah, yes. I suppose I could have mentioned that $\sqrt{2} \in \mathbb{A},$ but it seems you've beaten me to it. $\endgroup$ – beep-boop Aug 25 '14 at 12:44
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If you already know that the log of a positive algebraic number is transcendental, then all you need to realize is that $\sqrt2$ is a positive algebraic number. $\sqrt2$ is a root of $x^2-2=0$.

Therefore, $\log(\sqrt2)$ is transcendental $\implies$ $\log(\sqrt2)$ is irrational.

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IRRATIONAL

because $\sqrt{2} = 2^{1/2}$ and hence $\ln(\sqrt{2} )= 1/2 \ln(2)=$ an irrational number

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Think that $\log\sqrt2=\frac12\log2$ !

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