Is this correct(proof)?

  • Thread starter tronter
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  • #1
186
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I figured it out.
 
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Answers and Replies

  • #2
334
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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.
 
  • #3
186
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So the proof is incorrect because the statement is false? How do you know its an exclusive or?
 
  • #4
334
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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".
 
  • #5
186
1
If you changed it to either A... or...B, or both, then the statement and proof would be true?
 
  • #6
334
0
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.
 
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  • #7
cristo
Staff Emeritus
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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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