Proof by contradiction sqrt(2) is irrational

[JDM]

I'm trying to prove by contradicition that the sqrt(2) is irrational. I followed this example:
http://www.math.utah.edu/~pa/math/q1.html

Which I understand somewhat. What I don't understand is that we are assuming that p and q have no common factors. Then we prove that they do have a common factor greater than 1. That doesn't make sense to me. We added a constraint saying that p and q must have no common factors greater than 1 ( that they must be in simplest form) and then we showed that it fails that constraint. I just don't understand how we can add a constraint to our assumption, prove that the constraint fails, and then say we proved something because it failed some constraint we put on it.

Can anyone help me understand how that works?

 

[Prove It]

I'm trying to prove by contradicition that the sqrt(2) is irrational. I followed this example:
http://www.math.utah.edu/~pa/math/q1.html

Which I understand somewhat. What I don't understand is that we are assuming that p and q have no common factors. Then we prove that they do have a common factor greater than 1. That doesn't make sense to me. We added a constraint saying that p and q must have no common factors greater than 1 ( that they must be in simplest form) and then we showed that it fails that constraint. I just don't understand how we can add a constraint to our assumption, prove that the constraint fails, and then say we proved something because it failed some constraint we put on it.

Can anyone help me understand how that works?

The idea is that a statement can only ever be true or false.

If you make a statement and go through a series of logical steps to reach a contradiction, and each step was valid, then it proves that the original statement is false.

So if you have made a statement that \(\displaystyle \displaystyle \begin{align*} \sqrt{2} \end{align*}\) is rational, and then reach a contradiction (in this case, that there IS a common factor between p and q when we have already stated that there isn't one), then we have shown that the original statement is false.