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Math Help - Powers of transcendental numbers

  1. #1
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    Powers of transcendental numbers

    Hi guys,

    I'm having trouble proving that if \pi is transcendent, then \pi^{2k} and \beta\pi (where \beta is some non-zero rational constant) are transcendental.

    I know that if \beta\pi is algebraic, then it is the root of some polynomial, X-\beta\pi \in \mathbb{Q}[X]. Then, if this is a linear polynomial in \mathbb{Q}[X], is it right to conclude that any roots will be in \mathbb{Q}? If that is the case, can we then set some other polynomial, \beta X - \beta\pi \in \mathbb{Q}[X] and conclude a contradiction?

    For the other one, if we have \pi^{2k} is algebraic, then it is the root of X-\pi^{2k} \in \mathbb{Q}[X]. I don't know where to proceed from here, really.

    Thank you in advance,

    HTale.
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  2. #2
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    Quote Originally Posted by HTale View Post
    Hi guys,

    I'm having trouble proving that if \pi is transcendent, then \pi^{2k} and \beta\pi (where \beta is some non-zero rational constant) are transcendental.

    I know that if \beta\pi is algebraic, then it is the root of some polynomial, X-\beta\pi \in \mathbb{Q}[X]. Then, if this is a linear polynomial in \mathbb{Q}[X], is it right to conclude that any roots will be in \mathbb{Q}? If that is the case, can we then set some other polynomial, \beta X - \beta\pi \in \mathbb{Q}[X] and conclude a contradiction?

    For the other one, if we have \pi^{2k} is algebraic, then it is the root of X-\pi^{2k} \in \mathbb{Q}[X]. I don't know where to proceed from here, really.

    Thank you in advance,

    HTale.
    The algebraic numbers \widehat{\mathbb{Q}} are an example of an algebraically closed field. If \beta \pi is algebraic then \tfrac{\beta \pi}{\beta} = \pi is algebraic, since \beta \in \widehat{\mathbb{Q}}. Also, algebraic numbers contain all n-th roots of an algebraic number. Therefore, if \pi^n was algebraic then \pi would be algebraic. Another way to see this is if \pi^n was algebraic then consider the polynomial X^n - \pi^n \in \widehat{\mathbb{Q}}[X]. This polynomial splits over the algebraic numbers and so \pi \zeta^j are contained in \widehat{\mathbb{Q}} where 0\leq j\leq n-1, \zeta = e^{2\pi i/n}, but this is a contradiction.
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  3. #3
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    Here is an elementary proof. You need just to know the definition of the algebraic element

    Let n be a natural number and t be a trancendent element. If t^n were an algebraic element;

    there exists a non zero polynomial, say f(x), over rational numbers field Q such that f(t^n)=0


    Say g(x):=f(x^n) (is a non zero polynomial over Q, because f(x) is non zero)

    g(t)=f(t^n)=0. But this is a contradiction ( t is trancendent). Then t^n is not algebraic.....
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