This is a problem from Discrete Mathematics and its applications
To this proof, I am trying to use proof by contradiction. Here is how the book described the process of proof by contradiction.
I get the idea behind this. A contradiction is basically a compound proposition that no matter what combination the propositions within it take, the compound proposition will always evaluate to false. https://courses.cs.washington.edu/courses/cse311/14au/slides/lecture03-filled.pdf (slide 5) Because the q in this case is false, and the implication we're using, ~p -> q, is true(assuming to be true, like direct proof? Might need clarification on this as well), ~p must be false, which means p is true. So in the end p is proved to be true
Here is my work so far for the problem
What I have so far is a truth table that shows r ^ ~r is a contradiction. I am trying to prove p which is (x*y is irrational. To do proof by contradiction, I know that I am going to have to use ~p, which is x * y is rational. I express ~p with the definition of rationality - exists integers p and q, q!=0, etc... My question is what do I use for r? Just any proposition? Or does it have to relate to the problem? I know that r ^ ~r is a contradiction.