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Contrapositive proof

  1. Apr 13, 2012 #1
    1. The problem statement, all variables and given/known data

    Theorem: Let x,y,ε be ℝ. If x≤ y+ε  for every ε > 0 then x ≤ y.

    Write the above as a logic statement and prove it using contrapositive proof.


    The attempt at a solution

    The contrapositive statement x > y → x > y+ε is only true if ε < 0. Does a contrapositive proof negate the equality of ε?
     
  2. jcsd
  3. Apr 13, 2012 #2

    It should. The negation of x≤ y+ε  for every ε > 0 requires X>y+e for some e<0.
     
  4. Apr 14, 2012 #3

    HallsofIvy

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    I don't know what you mean by "the equality of [itex]\epsilon[/itex]". Are you referring to the in equality "[itex]\epsilon> 0[/itex]"?

    In any case your first statement is incorrect. If x> y then there exist an infinite number of positive [itex]\epsilon[/itex] such that [itex]x> y+ \epsilon[/itex]. x> y implies x-y> 0. Take [itex]\epsilon[/itex] to be any positive number less than x- y.
     
  5. Apr 14, 2012 #4
    HallsofIvy, are you saying that the contrapositive of "for every ε > 0..." is actually "there exists an ε > 0 such that..."?

    I would have thought that the contrapositive should be "there exists an ε < 0 such that...", i.e switch the inequality as well as the universal/existential quantifier
     
  6. Apr 14, 2012 #5
    you can't negate saying that you need an epsilon greater than zero. The negation must be done looking for some nonnegative epsilon. Any will do it, in particular epsilon=y-x.
     
  7. Apr 17, 2012 #6
    A mistake in my previous post. Indeed, to prove ~Q implies ~P you have to show that for some e>0, x > y → x > y+ε, since negating Q means that there is at least one e>0 such that ~Q is true.
     
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