[過去ﾛｸﾞ] ガロア第一論文と乗数イデアル他関連資料スレ13 (1002ﾚｽ)
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721(1): 現代数学の系譜 雑談 ◆yH25M02vWFhP 02/11(火)19:45 ID:zr+dFWV7(12/15) AAS
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外部ﾘﾝｸ:en.wikipedia.org
Pi
The number π (/paɪ/ ⓘ; spelled out as "pi") is a mathematical constant, approximately equal to 3.14159, that is the ratio of a circle's circumference to its diameter.
Irrationality and normality
π is an irrational number, meaning that it cannot be written as the ratio of two integers. Fractions such as
22/7 and 355/113
are commonly used to approximate π, but no common fraction (ratio of whole numbers) can be its exact value.[21] Because π is irrational, it has an infinite number of digits in its decimal representation, and does not settle into an infinitely repeating pattern of digits. There are several proofs that π is irrational; they generally require calculus and rely on the reductio ad absurdum technique.
（Proof that π is transcendental から下記へ）
外部ﾘﾝｸ:en.wikipedia.org
Lindemann–Weierstrass theorem — if α1, ..., αn are algebraic numbers that are linearly independent over the rational numbers
Q, then eα1, ..., eαn are algebraically independent over Q.
Transcendence of e and π
See also: e (mathematical constant) and Pi
The transcendence of e and π are direct corollaries of this theorem.
To prove that π is transcendental, we prove that it is not algebraic. If π were algebraic, πi would be algebraic as well, and then by the Lindemann–Weierstrass theorem eπi = −1 (see Euler's identity) would be transcendental, a contradiction. Therefore π is not algebraic, which means that it is transcendental.
A slight variant on the same proof will show that if α is a non-zero algebraic number then sin(α), cos(α), tan(α) and their hyperbolic counterparts are also transcendental.
Lindemann–Weierstrass theorem
Lindemann–Weierstrass Theorem (Baker's reformulation). — If a1, ..., an are algebraic numbers, and α1, ..., αn are distinct algebraic numbers, then[10]
a1e^α1+a2e^α2+・・・ +ane^αn =0
has only the trivial solution
ai=0 for all i=1,・・・ ,n.
Proof
略
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