Ask your own question, for FREE!
Mathematics 22 Online
OpenStudy (phoenixfire):

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*

OpenStudy (phoenixfire):

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.

OpenStudy (experimentx):

both must be odd.

OpenStudy (phoenixfire):

@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"

OpenStudy (experimentx):

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

OpenStudy (experimentx):

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

OpenStudy (phoenixfire):

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?

OpenStudy (experimentx):

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

OpenStudy (phoenixfire):

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.

OpenStudy (phoenixfire):

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

OpenStudy (experimentx):

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

OpenStudy (phoenixfire):

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

OpenStudy (experimentx):

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

OpenStudy (experimentx):

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

OpenStudy (phoenixfire):

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

OpenStudy (experimentx):

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

OpenStudy (experimentx):

yw

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!