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Clarifications on some types of proofs

  1. Mar 19, 2014 #1
    1. The problem statement, all variables and given/known data

    So I would know how to prove a statement like [itex]\sqrt{2}[/itex] by contradiction, all you have to do is assume to negation. But what about something like

    p → q
    Like if p = (bc mod a != 0), q = (b mod a != 0), how would I prove this, would I negate q or p, or both?
  2. jcsd
  3. Mar 20, 2014 #2


    Staff: Mentor

    First off, ##\sqrt{2}## is not a statement. Since you are using logic to prove or disprove statements, it's important to understand what a statement is. The statement you're referring to is probably "##\sqrt{2}## is an irrational number."

    To prove an implication p ##\Rightarrow## q by contradiction, assume that p is true and that q is false. If you arrive at a contradiction, that means that the assumption itself was faulty, and that the implication must be true. The reason for this is that p ##\Rightarrow## q is equivalent to ~p ##\vee## q. The negation of the implication is equivalent to ~(~p ##\vee## q), which is equivalent to p ##\wedge## ~q. In other words, that p is true and that q is false.

    A proof by contradiction entails assuming that the given statement is false, and then showing that that assumption must itself be false, thereby making the original statement true.
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