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Is $\ln(n)$ transcendental for all $n \in \mathbb{N} \setminus \{0, 1\}$? Is the answer even known?

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A related question:… – Jonas Meyer Jun 20 '11 at 20:24
up vote 5 down vote accepted

From the list of known transcendental numbers in the relevant Wikipedia article:

$\ln(a)$ if $a$ is algebraic and not equal to $0$ or $1$

So, in particular, $\ln(1)=0$ is not transcendental (although the numbers $2\pi ik$ for $k\in\mathbb{Z}\setminus\{0\}$ are, and these numbers have just as much right to be called $\ln(1)$; this is what they are referring to when they talk about a "branch of the logarithm function"), but $\ln(n)$ is transcendental for all $n\in\mathbb{N}\setminus\{0,1\}$.

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Yes, $\ln(\alpha)$ is transcendental for any positive algebraic number $\alpha \ne 1$, as a special case of the Lindemann–Weierstrass theorem

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(except for $\alpha = 1$) – Robert Israel Jun 20 '11 at 16:26
@Robert: Good point -- I have fixed it. – Jim Belk Jun 20 '11 at 17:17

Note that $x = \log{n} \implies e^x = n$ and it is well known that $e^x$ is transcendental if $x$ is algebraic and nonzero, which would give a contradiction if $x$ were algebraic. So $\log{n}$ is transcendental for all $n \geq 2$.

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Lindemann proved $e^α$ is transcendental for every non-zero algebraic number $α$. Let $n$ be a natural number.

Assume that $\log(n)$ is algebraic. It is non-zero if $n>1$. Thus, for $n>1$, we conclude that $e^{\log n}=n$ is transcendental. Contradiction.

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