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Homework Help: Is this correct(proof)?

  1. Oct 17, 2008 #1
    I figured it out.
     
    Last edited: Oct 17, 2008
  2. jcsd
  3. Oct 17, 2008 #2
    If x is any real number? I'd choose a number more interesting than [tex]x=\pi[/tex]. Remember that the claim employs "either/or", so if both [tex]\pi - x[/tex] and [tex]x+\pi[/tex] are irrational the claim is false.
     
  4. Oct 17, 2008 #3
    So the proof is incorrect because the statement is false? How do you know its an exclusive or?
     
  5. Oct 17, 2008 #4
    Yes. Give a counterexample to disprove the claim. It's an exclusive or because that's the language you use when it's an exclusive or. EITHER x-pi is irrational OR x+pi is irrational is what the claim is for any real x, so if they're both irrational then that isn't "either", that's "both".
     
  6. Oct 17, 2008 #5
    If you changed it to either A... or...B, or both, then the statement and proof would be true?
     
  7. Oct 17, 2008 #6
    well strictly speaking just dropping "either" would work, but nobody ever uses "or" to mean logical or. But yes what you say is right.

    Edit: Yeah Cristo is right I had totally forgotten about any "proof" by the time I wrote this.
     
    Last edited: Oct 17, 2008
  8. Oct 17, 2008 #7

    cristo

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    The proof is incorrect on a more fundamental level: namely, it is not a proof. It gives one example, but it is claimed that this holds for any x, hence a proof must show this. Remember, you cannot prove a general case by one example.
     
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