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Which proof technique is it ?

  1. Jun 14, 2014 #1
    Hello,
    Found a book on proofs and went to the exercise section. The proofs are fairly easy. The problem is the question I tried is listed under "direct proofs". I wasn't able to use direct proofs. Here's the question:
    For all x in Z, if 2^(2x) is odd, then 2^(-2x) is odd.

    My thoughts:
    I thought this was a very easy problem because all I had to do was show that 2^(2x)=4^x which is always even and so the proof follows vacuously.
    So I wonder why the author listed this problem under "direct proofs" and not under "vacuous proofs" which also has a section of its own. Is there a direct proof ? Is my proof wrong?

    Thanks
     
    Last edited: Jun 14, 2014
  2. jcsd
  3. Jun 15, 2014 #2

    SteamKing

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    If x > 0, shouldn't 2^(-2x) be equal to 1/[2^(2x)]? How does even/odd work for a rational, non-integer number?
     
  4. Jun 15, 2014 #3

    pwsnafu

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    Perhaps it should be -22x? Are you sure that's what's written?
     
  5. Jun 15, 2014 #4
    Yes I'm sure. Here's a pic of the question:
    2q1s6mv.png

    Why are you insisting on 2^(-2x) if you the premise 2^(2x) is always false ?? I mean the question seems to be similar to the statement: if x^2+1<0, then 1 is even. Since x^2+1<0 is always false, then the statement if x^2+1<0, then 1 is even is always true. I hope I'm explaining myself.
    What I'm asking is can I use the same train of thought to prove the result as shown in the picture. I proved it using a vacuous proof. The question is listed under "direct proofs" meaning that I have to show that for all premises the conclusion has to be true.
     
    Last edited: Jun 15, 2014
  6. Jun 15, 2014 #5

    verty

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    You showed directly that the statement is always true, therefore it is a direct proof. I mean, you didn't say "assume the statement is false, then...".
     
  7. Jun 15, 2014 #6
    the only way i see this working is if x=0, unless im missing something
     
  8. Jun 15, 2014 #7

    verty

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    This is a good point, ##2^{2x}## can be an odd integer. I didn't look at the question closely at all because Medwatt was asking about whether a proof like he suggested would be a direct proof.
     
  9. Jun 16, 2014 #8
    x=0, so it isn't vacuous.
     
  10. Jun 16, 2014 #9
    "For all x in Z..." is NOT what is in the question according to your picture of it. This means something entirely different.
     
  11. Jun 16, 2014 #10

    pwsnafu

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    Basically this.

    The direct proof would start by proving the only integer that makes 22x odd is zero (easy enough), then substituting it into 2-2x to get 1 which is odd.
     
  12. Jun 16, 2014 #11
    Actually, yet again, I'm an idiot!

    "For all x in Z, prove that if..." is equally as valid as "Let x be in Z. If..."
     
  13. Jun 17, 2014 #12

    pwsnafu

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    Huh, you're right, totally missed that.
     
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