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Proof by Contradiction

  1. Aug 24, 2016 #1
    Proof by contradiction starts by supposing a statement, and then shows the contradiction.

    1. The problem statement, all variables and given/known data


    Now, there is a statement ##A##.
    Suppose ##A## is false.
    It leads to contradiction.
    So ##A## is true.

    My question:
    There are two statements ##A## and ##B##.
    Suppose ##A## is true.
    Further suppose ##B## is false.
    It leads to contradiction.
    Can I conclude that "if ##A## is true, then ##B## is false."

    Moreover, how do I distinguish between "Proof by Contradiction" and "Proof by Exhaustion"?
     
  2. jcsd
  3. Aug 24, 2016 #2

    andrewkirk

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    No. But you can conclude that if A is true then B is true.
    Proof by exhaustion does not necessarily involve any contradictions. Here's the definition from wiki:
     
    Last edited: Aug 24, 2016
  4. Aug 24, 2016 #3

    Mark44

    Staff: Mentor

    Here's an example of a proof by exhaustion.
    Assume that n is a positive integer. Then n(n + 1)(n + 2) is divisible by 3.

    Case 1: n = 3k, for some integer k
    Then n(n + 1)(n + 2) = 3k(3k + 1)(3k + 2), which clearly has a factor of 3, and so is divisible by 3

    Case 2: n = 3k + 1, for some integer k
    Then n(n + 1)(n + 2) = (3k + 1)(3k + 2)(3k + 3) = (3k + 1)(3k + 2)3(k + 1), which has a factor of 3, so is divisible by 3.

    Case 3: n = 3k + 2, for some integer 8
    Then n(n + 1)(n + 2) = (3k + 2)(3k + 3)(3k + 4) = (3k + 2)3(k + 1)(3k + 4), also has a factor of 3, so is divisible by 3.

    There are no other possible cases. Any integer n falls into one of three equivalence classes: ##n \equiv 0 (\mod 3)##, or ##n \equiv 1 (\mod 3)##, or ##n \equiv 2 (\mod 3)##. IOW, when an integer n is divided by 3, the remainder will be 0, 1, or 2. The three cases above are based on this fact.
     
  5. Aug 24, 2016 #4

    Mark44

    Staff: Mentor

    Reopened...
     
    Last edited: Aug 24, 2016
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