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Prove Using the Method of Contrapositive

  1. Jan 28, 2016 #1
    Prove both by method of contrapositive.
    1. If a ≤ b + ε, where ε > 0, then b > a.
    2. If 0 ≤ a - b < ε, where ε > 0, then a = b.

    I'll start with problem 1.:
    p: If a ≤ b + ε, where ε > 0
    q: b > a

    neg q: b < a
    neg p: for some ε' > 0 1/2(a - b), a > b + ε'

    define ε' = 1/2(a - b)

    I assume the contrapositive goes something like this: p→q, neg q→neg p,

    After this, do I start with b < a→ a > b + ε', and then substitute the ε' = 1/2( a - b) into neg p and then try to work it into the form b < a ? However, it seem like I should start b > a and work my way to neg p. But how? Am I on the right path? Thanks
     
    Last edited by a moderator: Jan 28, 2016
  2. jcsd
  3. Jan 29, 2016 #2

    Mark44

    Staff: Mentor

    The negation of q is "b is not greater than a" I.e., ##b \le a##
    I don't get what you're doing here. Are there any "for all" qualifiers in the original question?
    Where did the (1/2)(a - b) come from?
    Yes, you have it right.
     
  4. Jan 29, 2016 #3
    yes,,
    Yes, sorry, instead of "where ε > 0" I should have wrote "for all ε > 0".
     
    Last edited by a moderator: Jan 29, 2016
  5. Jan 29, 2016 #4

    Mark44

    Staff: Mentor

    Look up in your textbook or in examples from class how you deal with universal qualifiers in forming the contrapositive.
     
  6. Jan 29, 2016 #5
    The (1/2)(a - b) is a hint the 1st problem gives you. I misplaced it in the problem. The "neg p: for some ε' > 0 1/2(a - b), a > b + ε' ", should read " neg p: There is some ε' > 0 , so that a > b + ε'. And then the hint, define ε' = (1/2)(a - b). Sorry, it was late when I typed the original.

    Also, thank you for confirming that I'm starting in the right direction. I'll work from there, and then post. Thanks much.
     
    Last edited: Jan 29, 2016
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