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*

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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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