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Square Root Of 2 Irrationality Proof

  1. Jan 24, 2009 #1
    Hi, I'm having trouble understanding some statements in this proof from my textbook:

    "Thus, 2 = s^2/t^2 and 2t^2 = s^2. Since s^2 and t^2 are squares, s^2 contains an even number of 2's as prime factors (This is our Q statement), and t^2 contains an even number of 2's. But then t^2 contains an odd number of 2's as factors. Since 2t^2 = s^2, s^2 has an odd number of 2's. (This is the statement ~Q.) This is a contradiction, because s2 cannot have both an even and an odd number of 2's asfactors. We conclude that sqrt(2) is irrational."

    Why does he assume that t^2 contains an odd number of 2's all of a sudden? ... and even if it did, s^2 would still be even, because it is equal to 2t^2, not t^2. :confused:
     
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  3. Jan 24, 2009 #2

    Hurkyl

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    There is something funny in the phrasing; did you copy it exactly? Anyways, he didn't assume, because having an odd number of factors of 2 is a consequence of the previous observations.
     
  4. Jan 24, 2009 #3
    yes I copied it exactly. ok, let me see if I get it:

    t^2 must have an odd number of 2's because it equals s^2/2, and s^2 has an even number of twos (because any squared number has an even number of twos). but why not stop at that? isn't that a contradiction already? ... and the second statement still makes no sense to me, because if you multiply a number with an odd number of twos as factors (2t) by two, then you have an even number of twos, so why does he say s^2 is odd after 2t^2 = s^2?.

    I assume, as usual, I'm overlooking something very obvious :biggrin:
     
  5. Jan 24, 2009 #4

    Hurkyl

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    Yes it is.
     
  6. Jan 24, 2009 #5
    Well s^2 is an even number right? (since it equals 2t^2 and t is an integer therefore s^2 = 2k, for k \in Z ) but it's also a perfect square which means s is an even number I believe. Well if s is an even number then s^2 must contain an even number of 2s.
     
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