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Thread: Getting two integers

  1. #1
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    Getting two integers

    Please advise me how to prove the following assertion

    Given two real numbers a , b, there always exists integers m, n such that a < m+n x sqrt(2)< b

    It would be best if we can rely mainly on the basic theorems in number theory, such as the Archimedean Principle,
    Density of Rationals and Density of Irrationals in proving the above assetion
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  2. #2
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    Re: Getting two integers

    Quote Originally Posted by KLHON View Post
    Please advise me how to prove the following assertion

    Given two real numbers a , b, there always exists integers m, n such that a < m+n x sqrt(2)< b

    It would be best if we can rely mainly on the basic theorems in number theory, such as the Archimedean Principle,
    Density of Rationals and Density of Irrationals in proving the above assetion
    This is not true. Let a = 0, b = 1. There are no integers m, n that satisfy the inequality.

    -Dan
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  3. #3
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    Re: Getting two integers

    $\displaystyle 0<3-2\sqrt{2}<1$
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    Re: Getting two integers

    $\displaystyle a<b$

    Find an integer $\displaystyle k\geq 0$

    such that

    $\displaystyle \left(3+2\sqrt{2}\right)^k(b-a)>1 $

    There is an integer $p$ such that

    $\displaystyle \left( 3+2\sqrt{2} \right)^k a< p < \left( 3+2\sqrt{2} \right)^k b $

    multiply by

    $\displaystyle \left(3-2\sqrt{2}\right)^k$

    to get

    $\displaystyle a<p\left(3-2\sqrt{2}\right)^k<b$
    Last edited by Idea; Apr 7th 2018 at 10:27 AM.
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  5. #5
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    Re: Getting two integers

    This is the most interesting question we have had in months.
    Idea's solution is equally interesting. Thank you.
    Last edited by Plato; Apr 7th 2018 at 05:40 PM.
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