Clarifications on some types of proofs

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SUMMARY

This discussion clarifies the method of proving implications using contradiction, specifically in the context of logical statements. To prove an implication of the form p → q, where p is defined as (bc mod a != 0) and q as (b mod a != 0), one must assume p is true and q is false. If this leads to a contradiction, the implication is validated. The discussion emphasizes that the negation of the implication p → q is equivalent to p ∧ ¬q, reinforcing the logical structure necessary for such proofs.

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Homework Statement



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?
 
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Panphobia said:

Homework Statement



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?

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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