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Math Help - Rational number if n is even proof

  1. #1
    Member Jason Bourne's Avatar
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    Rational number if n is even proof

    Prove that the number

    ( \sqrt{5} + 1)^n + (\sqrt{5} - 1)^n , (n \in \mathbb{N})

    is rational if and only if n is even.

    (I think going one way if n is even then its clear that its rational, what about assuming that n is even then the number is rational? Is it inductive?)
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  2. #2
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    Hi

    You can use Newton's formula to develop (\sqrt{5} + 1)^n and (\sqrt{5} - 1)^n then study the 2 cases (n even and n odd).

    When n is even all the terms involving \sqrt{5} disappear.
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  3. #3
    Member Jason Bourne's Avatar
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    Newton's formula?
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  4. #4
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    Sorry I don't know how it is called in English
    I am talking about this formula
    (a + b)^n = \sum_{k=0}^{n} \binom{n}{k}\: a^k \:b^{n-k}
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  5. #5
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    Quote Originally Posted by Jason Bourne View Post
    Prove that the number

    ( \sqrt{5} + 1)^n + (\sqrt{5} - 1)^n , (n \in \mathbb{N})
    Consider the sequence a_0,a_1,a_2,a_3,... defined as: \left\{ \begin{array}{c} a_0 = a_1 = 2 \\ a_{n+2} = 2a_{n+1} + 24a_n \text{ for }n\geq 0 \end{array} \right.

    Certaintly, the terms of the sequence \{ a_n \} are integers.
    Furthermore, the solution to this recurrence relation is given by: a_n = \left( 1 + \sqrt{5} \right)^n  + \left( 1 - \sqrt{5} \right)^n.
    If n is even then a_n = \left( 1 + \sqrt{5} \right)^n + \left( \sqrt{5} - 1 \right)^n \in \mathbb{N}

    Consider the Fibonacci sequence f_0,f_1,f_2,... it satifies f_n = \frac{1}{\sqrt{5}} \left[ \left( \frac{1+\sqrt{5}}{2} \right)^n - \left( \frac{1-\sqrt{5}}{2} \right)^n \right]

    Define b_n = \left( 1+\sqrt{5}\right)^n + \left( \sqrt{5}-1\right)^n.

    If n is odd then \frac{b_n}{2^n\sqrt{5}} = \frac{1}{\sqrt{5}} \left[ \left( \frac{1+\sqrt{5}}{2} \right)^n - \left( \frac{1-\sqrt{5}}{2} \right)^n \right] = f_n.

    Therefore for odd n we have b_n = \sqrt{5}\cdot 2^n \cdot f_n\not \in \mathbb{Q}.
    Last edited by ThePerfectHacker; January 16th 2009 at 10:00 AM.
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