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Simple Proof

  1. Jun 22, 2008 #1
    Theorem: Suppose that x is real, then x^2 > or = 0.
    Incorrect Proof: Suppose not. Then x is real and x^2<0. Consider x=3. 9<0, a contradiction. Therefore x^2 >=0.
     
  2. jcsd
  3. Jun 23, 2008 #2
    I'm sorry, but what exactly is the brain teaser at hand here?
     
  4. Jun 23, 2008 #3
    Why the proof is incorrect, even though seemingly it looks like it is.
     
  5. Jun 23, 2008 #4
    I don't see how could this fool anyone... Given the assumption P -> -Q , --Q(3) does not give rise to P -> Q. One needs to show that the assumption P -> -Q gives rise to --Q (a statement for all x being considered, not just 3) and therefore conclude P -> Q by reductio ad absurdum.
     
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