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Proving with contrapositive methode instead of contradition

  1. Sep 2, 2013 #1
    Could you tell me how to write a very formal proof of the statment below with the contrapositive methode, if possible.
    (I know how to do it with contradiction)

    Let x be a rational number and y an irrational number, then x times y is irrational.

    Sincerely,
    V. Uljanov
     
  2. jcsd
  3. Sep 2, 2013 #2
    The contrapositive of this is "if xy is rational then x and y are either both rational or both irrational" and this can be proved with similar steps to the proof you already have.
     
  4. Sep 2, 2013 #3
    Thx, but I cant see how to proceed in the same manner. If xy=m/n, then x=m/(n*y), but if y is irrational I am back to the start, and cant say anything about x.

    In contradiction I assumed xy=m/n, and got y=m/(n*x)=ml/nk=m'/n' (x was rational), and this lead to the contradiction of y beeing irrational from the start.
     
  5. Sep 2, 2013 #4
    Exactly as before: assume x is rational, y is irrational and xy is rational, demonstrate the contradiction.

    You see the contrapositive of "A implies B" is "(Not B) implies (not A)". With proof by contradiction you assume (not B) and demonstrate (not A), so proof by contradiction is essentially the same as proof of the contrapositive.

    If this seems a bit circular and tautological it is because this is a bad example. Contrapositive proofs only make sense where there is no obvious proof of "A implies B", but it is possible to prove "(Not B) implies (not A)".

    Here is a better example: "If the angles of a triangle don't add up to 180° the surface is non-Euclidean"; the easiest way to prove this is to prove the contrapositive: if the surface is Euclidean (so we can constuct a line parallel to any side), the angles of a triangle add up to 180°.
     
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