1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook