Clarifications on some types of proofs

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Homework Statement



So I would know how to prove a statement like \sqrt{2} 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 \sqrt{2} 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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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