"무리수와 초월수"의 두 판 사이의 차이

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* In general, <math>\alpha^{\beta} = \exp\{\beta \log \alpha\}</math> is [http://en.wikipedia.org/wiki/Multivalued_function multivalued], where "log" stands for the [http://en.wikipedia.org/wiki/Complex_logarithm complex logarithm]. This accounts for the phrase "any value of" in the theorem's statement.
 
* In general, <math>\alpha^{\beta} = \exp\{\beta \log \alpha\}</math> is [http://en.wikipedia.org/wiki/Multivalued_function multivalued], where "log" stands for the [http://en.wikipedia.org/wiki/Complex_logarithm complex logarithm]. This accounts for the phrase "any value of" in the theorem's statement.
 
* An equivalent formulation of the theorem is the following: if<math>\alpha</math>and <math>\gamma</math> are nonzero algebraic numbers, and we take any non-zero logarithm of<math>\alpha</math>, then<math>(\log \gamma)/(\log \alpha)</math>is either rational or transcendental.
 
* An equivalent formulation of the theorem is the following: if<math>\alpha</math>and <math>\gamma</math> are nonzero algebraic numbers, and we take any non-zero logarithm of<math>\alpha</math>, then<math>(\log \gamma)/(\log \alpha)</math>is either rational or transcendental.
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* If the restriction that<math>\beta</math>be algebraic is removed, the statement does not remain true in general (choose <math>\alpha=3</math> and <math>\beta=\log 2/\log 3</math>, which is transcendental, then <math>\alpha^{\beta}=2</math> is algebraic). A characterization of the values for α and β which yield a transcendental α<sup>β</sup> is not known.
+
* If the restriction that<math>\beta</math>be algebraic is removed, the statement does not remain true in general (choose <math>\alpha=3</math> and <math>\beta=\log 2/\log 3</math>, which is transcendental, then <math>\alpha^{\beta}=2</math> is algebraic). A characterization of the values for<math>\alpha</math> and <math>\beta</math>which yield a transcendental <math>\alpha^{\beta}</math> is not known.
  
 
 
 
 

2009년 6월 25일 (목) 18:29 판

간단한 소개
  • 복소수 중에서 어떠한 유리수 계수방정식도 만족시킬 수 없는 수를 초월수라 함
    • 유리수 계수방정식은 적당한 정수를 곱하여 다음과 같은 형태의 정수계수방정식으로 표현할 수도 있음.
      \(a_n x^n + a_{n-1} x^{n-1} + a_{n-2} x^{n-2} + \cdots + a_1 x + a_0 = 0, a_i \in \mathbb{Z}\)
    • 복소수 중에서 어떠한 정수계수방정식도 만족시킬 수 없는 수를 초월수라 해도 무방
  • 대수적수론 에 비해 훨씬 어렵고, 체계적인 이론이 확립되어 있지 않음.

 

 

린데만-바이어슈트라스 정리

 

겔퐁드-슈나이더 정리

If \(\alpha\) and \(\beta\) are algebraic numbers (with \(\alpha \ne 0\) and \(\log \alpha\) any non-zero logarithm of \(\alpha\)), and if \(\beta\) is not a rational number, then any value of \(\alpha^{\beta} =\exp\{\beta \log \alpha\}\)is a transcendental number.

Comments

  • The values of\(\alpha\) and \(\beta\)are not restricted to real numbers; all complex numbers are allowed.
  • In general, \(\alpha^{\beta} = \exp\{\beta \log \alpha\}\) is multivalued, where "log" stands for the complex logarithm. This accounts for the phrase "any value of" in the theorem's statement.
  • An equivalent formulation of the theorem is the following: if\(\alpha\)and \(\gamma\) are nonzero algebraic numbers, and we take any non-zero logarithm of\(\alpha\), then\((\log \gamma)/(\log \alpha)\)is either rational or transcendental.
  • If the restriction that\(\beta\)be algebraic is removed, the statement does not remain true in general (choose \(\alpha=3\) and \(\beta=\log 2/\log 3\), which is transcendental, then \(\alpha^{\beta}=2\) is algebraic). A characterization of the values for\(\alpha\) and \(\beta\)which yield a transcendental \(\alpha^{\beta}\) is not known.

 

 

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