파이 π는 초월수이다

증명의 개요

• 린데만-바이어슈트라스 정리 를 써서 증명 가능

 

 

증명

먼저 린데만-바이어슈트라스 정리 에 의해, \(e\)는 초월수이다. 더 일반적으로 대수적수 \(\alpha\) 에 대하여, \(e^{\alpha}\) 는 초월수이다.

we can argue that if α is a nonzero algebraic number, then {0, α} is a set of distinct algebraic numbers, and so the set {e⁰, e^α} = {1, e^α} is linearly independent over the algebraic numbers and in particular e^α cannot be algebraic and so it is transcendental.

 

Now, we prove that π is transcendental. If π were algebraic, 2πi would be algebraic too (since 2i is algebraic), and then by the Lindemann–Weierstrass theorem e^2πi = 1 (see Euler's formula) would be transcendental, which is absurd.

 

 

 

 

 

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참고할만한 자료

• http://ko.wikipedia.org/wiki/초월수
• http://en.wikipedia.org/wiki/Proof_that_π_is_irrational
• http://en.wikipedia.org/wiki/transcendental_number
• http://en.wikipedia.org/wiki/Lindemann–Weierstrass_theorem
• http://www.wolframalpha.com/input/?i=
• http://front.math.ucdavis.edu/search?a=&t=&c=&n=40&s=Listings&q=
• http://www.ams.org/mathscinet/search/publications.html?pg4=AUCN&s4=&co4=AND&pg5=TI&s5=&co5=AND&pg6=PC&s6=&co6=AND&pg7=ALLF&co7=AND&Submit=Search&dr=all&yrop=eq&arg3=&yearRangeFirst=&yearRangeSecond=&pg8=ET&s8=All&s7=
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