Number theory Diophantine approximation and transcendence Theorem

Lindemann–Weierstrass theorem

Builds on Algebraic number Transcendental number

Statement

Let $\alpha_1, \dots, \alpha_n$ be distinct algebraic numbers. Then $e^{\alpha_1}, \dots, e^{\alpha_n}$ are linearly independent over the field of algebraic numbers. That is, if $\beta_1, \dots, \beta_n$ are algebraic numbers with $\beta_1 e^{\alpha_1} + \dots + \beta_n e^{\alpha_n} = 0$ , then $\beta_1 = \dots = \beta_n = 0$ .

Remarks

A consequence: $e^{\alpha}$ is transcendental for each nonzero algebraic number $\alpha$ . In particular, $e$ is transcendental. Another consequence: $\pi$ is transcendental. Otherwise $i\pi$ would be algebraic, and $e^{i\pi} + 1 = 0$ would contradict the theorem. So the circle cannot be squared with ruler and compass.

History

Ferdinand von Lindemann proved in 1882 that $e^{\alpha}$ is transcendental for each nonzero algebraic $\alpha$ . Karl Weierstrass proved the general statement in 1885.

Reading path

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Where this leads

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Impossibility of the classical construction problems

Tag	`0JNCBFMPXR`
URI	`m2://0JNCBFMPXR`
Kind	statement, theorem
Writing	W1 stated
MSC	11J81
Status	This copy of the site has no build data. Run `m2 status 0JNCBFMPXR`.
Uses	Algebraic number, Transcendental number
Same as	Wikidata Q1572474 Wikipedia Lindemann–Weierstrass theorem

Lindemann–Weierstrass theorem

Statement

Remarks

History

Reading path

Where this leads

Further reading

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