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ガロア第一論文と乗数イデアル他関連資料スレ13
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>>721 > >>680 追加 > > https://en.wikipedia.org/wiki/Pi > 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) > https://en.wikipedia.org/wiki/Lindemann%E2%80%93Weierstrasstheorem > 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 > > つづく
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