- #1
r0bHadz
- 194
- 17
Homework Statement
If 3n+2 is odd, then n is odd
Homework Equations
The Attempt at a Solution
assume 3n+2 is even, show n is odd
3n+2 = 2f => 3n+3 = 2f + 1 => 3(n+1) = 2f+1
so 3*(n+1) is odd form. Only odd*odd gives you another odd, so n+1 is odd. That means n is even.
We have a contradiction here, as shown, n is even, not odd.
Or does this proof method only work if we take the negation of the conclusion? The reason I think my proof is invalid:
(p) (q) (p->q) (p -> ~q) (~p -> q )
0 ff 1 ff 1 ffffff 1 ffffffffffff 1
0 ff 0 ff 1 ffffff 1 ffffffffffff 0
1 ff 0 ff 0 ffffff 1 ffffffffffff 1
1 ff 1 ff 1 ffffff 0 ffffffffffff 1
If we look at the second row, p = 0, q = 0, and the last column, we will see ~p -> q is 0
There is no contradiction here. ~p is 1, and q is 0. This implication will always be zero. My proof that I did above, it did not prove anything that isn't obvious already, thus there was never a contradiction and it was wrong.
What I think I need to do:
Look at the third row, where p = 1, q = 0, p->q = 0. I need to show that the case p->q = 0 cannot happen. Meaning, that there will be a contradiction. P will have to be 0, or q will have to be 1. This is where the contradictions is. I cannot possibly show this contradiction with ~p->q because ~p is 0 which will always give me a truth of one. I can however show this with p->~q because if I can show that the value of this implication is 1, there will be a contradiction.Does my reasoning seem right folks?