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Thread: Induction

  1. #1
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    Induction

    Assume that x+(1/x) is an integer, how do I, by using induction, show that x^2 + (1/x^2) , x^3 + (1/x^3), .... , x^n + (1/x^n) are also integers?
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hello

    First, show that $\displaystyle x^2+\frac{1}{x^2}$ lies in $\displaystyle \mathbb{N}$ by developping $\displaystyle \left(x+\frac{1}{x}\right)\left(x+\frac{1}{x}\righ t)$.

    Then, assuming there exists an integer $\displaystyle n$ such that $\displaystyle x^n+\frac{1}{x^n}$ and $\displaystyle x^{n-1}+\frac{1}{x^{n-1}}$ lie in $\displaystyle \mathbb{N}$, try to develop $\displaystyle \left(x^{n}+\frac{1}{x^{n}}\right)\left(x+\frac{1} {x}\right)$ .
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  3. #3
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    So by assuming that there exists an integer such that and lie in , what we then do is to show that this is also true for n+1 ?
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Yes. I should have written it, sorry.
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  5. #5
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    would it make any difference if we assumed that it's true for n+1, and for n+2, and by using this proving that it's also true for n?
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  6. #6
    Super Member flyingsquirrel's Avatar
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    Yes, it would make a difference : it would be false.

    Let's call $\displaystyle P(n)$ the relation "$\displaystyle x^n+\frac{1}{x^n}$ is an integer"
    If one shows that if $\displaystyle P(n-1)$ and $\displaystyle P(n)$ are true then $\displaystyle P(n+1)$ is true too, as $\displaystyle P(1)$ and $\displaystyle P(2)$ are true, we get that $\displaystyle P(3)$ is true. Then, as $\displaystyle P(2)$ and $\displaystyle P(3)$ are true, we get that $\displaystyle P(4)$ is true... so we'll reach any integer.

    Instead, if you show that $\displaystyle P(n+1)$ and $\displaystyle P(n+2)$ true imply $\displaystyle P(n)$ true, you go decreasing and won't reach all the integers...
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