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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*

Mathematics
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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.
both must be odd.
@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"

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Other answers:

"if ab is odd then either a is odd or b is odd." => "if ab is odd then both a and b are odd."
using contrapositive : ~p->~q if ab is even, ab must have AT LEAST one even factor.
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?
i think the assumption on the Q is wrong.. 2x3 (even x odd) =6 (even) 3x3 (odd x odd) = 9 (odd)
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.
But isn't that why they use different laws such as the Contrapositive? To prove things that don't logically make sense?
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
yeah, it's kind of confusing. I'm only learning these proofs now and even the basic Direct Proof is doing my head in.
i hate to read these whole stuff http://en.wikipedia.org/wiki/Proof_by_contrapositive
i guess you read the Q wrong ... it says proof by contrapositive is 100% valid (never got logical stuff in my brain though)
Well, thanks anyways @experimentX I'll just stick with how I think it should be and ask my lecturer about it later.
if you work this way ... from invalid theorem, ... you will reach a valid statement which in not complete.
yw

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