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Prove that spec of root 2 contains infinitely many powers of 2.

  1. Oct 5, 2012 #1
    Spec of root 2 is that set of elements floor(k * (root 2)) ; k >= 0 .

    I have no idea of how I can prove the statement in the question.

    Prove that spec of root 2 contains infinitely many powers of 2.
    I need ideas on how to proceed.

    Thank you
     
  2. jcsd
  3. Oct 5, 2012 #2

    Ray Vickson

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    What is meant by the "spec" of root 2? I have never seen that term.

    RGV
     
  4. Oct 5, 2012 #3
    Here, it means Spec√2= {⌊k√2⌋ : k≥0}
     
  5. Oct 5, 2012 #4

    Mark44

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    Spectrum?
     
  6. Oct 5, 2012 #5

    Dick

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    It's little tricky. Here's big hint. Pick k=floor(2^n*sqrt(2)). Think about what binary representation of sqrt(2) looks like. Play around with that for a while.
     
  7. Oct 5, 2012 #6
    I've been thinking about it but I can't seem to move ahead.
    sqrt(2) is irrational. It's binary representation has no pattern.
    It would look like 1.0110101000001001111... and multiplying by 2^n would left-shift it n times.
    Taking floor of this irrational number loses a portion of the number. And inside the Spec definition, we have floor (k * √2) and this floor would lose some number too.
    Finally if √2 * √2 * 2 ^ n = 2 ^(n+1), this attempt leads to an interger smaller than that.
    Where should I be heading?
     
  8. Oct 6, 2012 #7

    Dick

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    Yeah, I'm actually running into the same problem. It's pretty easy to show 2^n*sqrt(2)-floor(2^n*sqrt(2)) is less than 1/2 an infinite number of times and I thought that would cover it, but I keep getting an inequality pointing the wrong way.
     
    Last edited: Oct 6, 2012
  9. Oct 6, 2012 #8

    Dick

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    Ok, I think you can do it. You can either pick 2^n*sqrt(2)-floor(2^n*sqrt(2)) to be in (0,1/2) or (1/2,1) (both happen an infinite number of times). Note it would also work fine if you could show the choice of k=floor(2^n*sqrt(2))+1 works. Just draw some careful diagrams of where all the numbers lie relative to each other.
     
    Last edited: Oct 6, 2012
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