Is it correct: natural-logarithm maps algebraic numbers to transcendentals and vice-verse, over the domain it is defined?

Question

Is it correct that the natural logarithm function maps algebraic numbers to transcendental and transcendental numbers to algebraic, other than 1? Of course, over the domain natural log is defined i.e. $(0,\infty)$?

i.e.

$$\ln:A^+ \rightarrow T \hspace{5 mm};\hspace{5 mm} \ln:T^+ \rightarrow A,$$

and$$\ln:A^+ \nrightarrow A \hspace{5 mm};\hspace{5 mm} \ln:T^+ \nrightarrow T.$$

where;

$A^+$ is the set of positive algebraic numbers except $1$.

$T^+$ is the set of positive transcendental numbers.

$T$ is the set of transcendental numbers.

$A$ is the set of algebraic numbers.

Yes. And you are aware that $\ln$ is injective on $(0,\infty)$? — Jonas Meyer, Aug 18 '14 at 04:58
Then, it is possible that $ln(t_1)=t_2$; where $t_1,t_2 \in {Transcendental}$. And my assertion in question is incorrect. — kaka, Aug 18 '14 at 05:14

score 6 · Accepted Answer · answered Aug 18 '14 at 05:15

6

It is impossible for cardinality reasons alone, since $\ln$ is injective, so it cannot map an uncountable set into a countable one.

It is true that $\ln$ maps positive algebraic numbers (except 1) to transcendental numbers, by the Lindemann–Weierstrass theorem. Every algebraic real number except $0$ is the logarithm of a positive transcendental number, but so are most transcendental numbers. For a concrete example, $e^\pi$ is a transcendental number whose logarithm is transcendental.

answered Aug 18 '14 at 05:15

Jonas Meyer

53,602

Well, if I say instead: $Ln:A^+ \rightarrow T$ and $Ln:A^+ \nrightarrow T$, would always hold true? – kaka Aug 18 '14 at 05:45
2

If $A^+$ is the set of positive algebraic real numbers except $1$, then the logarithm restricted to $A^+$ maps into $T$: $\ln(A^+)\subseteq T$. – Jonas Meyer Aug 18 '14 at 05:47
thanks, for detailed clarification. – kaka Aug 18 '14 at 05:48
1

You're welcome. As a side note, some of your notation isn't standard. E.g. "$\in$" is used to show containment of an element in a set, not set equality. I edited to adjust the definitions at the end of your answer. – Jonas Meyer Aug 18 '14 at 05:53
I thought $e^\pi$ being transcendental was an open problem. When was that solved? – zibadawa timmy Aug 18 '14 at 06:18
1

@zibadawatimmy http://en.wikipedia.org/wiki/Gelfond%27s_constant – kaka Aug 18 '14 at 06:45
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@zibadawatimmy: $e^\pi$ is a hyperlink in my answer. (Same link as in kaka's comment). – Jonas Meyer Aug 18 '14 at 12:05
Ah, so it is. It does not render in a visible way in the SE android app. – zibadawa timmy Aug 18 '14 at 12:07

Is it correct: natural-logarithm maps algebraic numbers to transcendentals and vice-verse, over the domain it is defined?

1 Answers1