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Disproving an incorrect theorem?

  1. Feb 16, 2016 #1
    Incorrect Theorem:
    Suppose x and y are real numbers and x + y = 10, then x != 3 and y != 8.

    (a) What’s wrong with the following proof of the theorem?

    Proof. Suppose the conclusion of the theorem is false. Then x = 3 and y = 8. But then x + y = 11, which contradicts the given information that x + y = 10. Therefore the conclusion must be true.

    (b) Show that the theorem is incorrect by finding a counterexample.


    So according to the answer it's false because x != 3 can't be proven with x = 3 because that's not the negation, but even so, isn't the theorem true because 3 + 8 = 11 which does contradict the premise... I'm very confused by this....

    For b, the counter example was x = 3 and x = 7 but how does that disprove it? I'm still very confused by counter examples.
    So would it be written as "Suppose x and y are real numbers and x + y = 10, then x = 3 and y = 7" Is that how the counter example would be written?

    Please explain!
     
  2. jcsd
  3. Feb 16, 2016 #2

    micromass

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    That is not the negation of the theorem.
     
  4. Feb 16, 2016 #3
    So the proof is false so therefore the theorem is false or is the theorem already false to begin with, so proof would also be false....
     
  5. Feb 16, 2016 #4
    Never mind. I figured it out. Thank you.
     
  6. Feb 18, 2016 #5

    Mark44

    Staff: Mentor

    In the statement, the conclusion is "then ##x \ne 3 \text{ and } y \ne 8##"

    The negation of the conclusion is ##x = 3 \text{ or } y = 8##. I believe this is what micromass was alluding to.
     
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