## PhoenixFire Group Title Using the Law of Contrapositive give proof of the theorem: For any integers a,b if ab is odd then either a is odd or b is odd. I tried and got: *give me a second to put it into the first post* 2 years ago 2 years ago

1. PhoenixFire Group Title

p -> q contrapositive is ~q -> ~p So I get the statement: if either a is even or b is even then ab is even Definition of even: x is divisible by 2 Using that definition: $\forall a,b \in \mathbb{Z}: {a \over 2} and {b \over 2} \rightarrow {ab \over 2}$ Re-arranging for a and b for some integer k and s: $a=2k$$b=2s$ Re-arrange the conclusion and substitute a and b in $ab=2t$ $(2k)(2s)=2t$$4ks=2t$$2(2ks)=2t$ Proving the theorem.

2. experimentX Group Title

both must be odd.

3. PhoenixFire Group Title

@experimentX I'm proving the Contrapositive. When you negate the statements it reads "ab is not odd" which is the same as saying "ab is even"

4. experimentX Group Title

"if ab is odd then either a is odd or b is odd." => "if ab is odd then both a and b are odd."

5. experimentX Group Title

using contrapositive : ~p->~q if ab is even, ab must have AT LEAST one even factor.

6. PhoenixFire Group Title

When you negate "a is odd or b is odd" you get "a is even and b is even" p->q contrapositive is ~q->~p p = ab is odd ~p = ab is even q = a is odd or b is odd ~q = a is even and b is even Right?

7. experimentX Group Title

i think the assumption on the Q is wrong.. 2x3 (even x odd) =6 (even) 3x3 (odd x odd) = 9 (odd)

8. PhoenixFire Group Title

I see what you mean. The original theorem states ab is odd then a or b is odd, however if you plug in a as odd and b as even ab is then even. Which would logically mean that it is an invalid theorem.

9. PhoenixFire Group Title

But isn't that why they use different laws such as the Contrapositive? To prove things that don't logically make sense?

10. experimentX Group Title

yep i've already forgotten these types of things ... i think this logic is right though q = a is odd or b is odd => ~q = both a and b are not odd => both a and b are even

11. PhoenixFire Group Title

yeah, it's kind of confusing. I'm only learning these proofs now and even the basic Direct Proof is doing my head in.

12. experimentX Group Title

i hate to read these whole stuff http://en.wikipedia.org/wiki/Proof_by_contrapositive

13. experimentX Group Title

i guess you read the Q wrong ... it says proof by contrapositive is 100% valid (never got logical stuff in my brain though)

14. PhoenixFire Group Title

Well, thanks anyways @experimentX I'll just stick with how I think it should be and ask my lecturer about it later.

15. experimentX Group Title

if you work this way ... from invalid theorem, ... you will reach a valid statement which in not complete.

16. experimentX Group Title

yw