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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*
 one year ago
 one year ago
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*
 one year ago
 one year ago

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PhoenixFireBest ResponseYou've already chosen the best response.1
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: \[\forall a,b \in \mathbb{Z}: {a \over 2} and {b \over 2} \rightarrow {ab \over 2}\] Rearranging for a and b for some integer k and s: \[a=2k\]\[b=2s\] Rearrange the conclusion and substitute a and b in \[ab=2t\] \[(2k)(2s)=2t\]\[4ks=2t\]\[2(2ks)=2t\] Proving the theorem.
 one year ago

PhoenixFireBest ResponseYou've already chosen the best response.1
@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"
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
"if ab is odd then either a is odd or b is odd." => "if ab is odd then both a and b are odd."
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
using contrapositive : ~p>~q if ab is even, ab must have AT LEAST one even factor.
 one year ago

PhoenixFireBest ResponseYou've already chosen the best response.1
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?
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
i think the assumption on the Q is wrong.. 2x3 (even x odd) =6 (even) 3x3 (odd x odd) = 9 (odd)
 one year ago

PhoenixFireBest ResponseYou've already chosen the best response.1
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.
 one year ago

PhoenixFireBest ResponseYou've already chosen the best response.1
But isn't that why they use different laws such as the Contrapositive? To prove things that don't logically make sense?
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
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
 one year ago

PhoenixFireBest ResponseYou've already chosen the best response.1
yeah, it's kind of confusing. I'm only learning these proofs now and even the basic Direct Proof is doing my head in.
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
i hate to read these whole stuff http://en.wikipedia.org/wiki/Proof_by_contrapositive
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
i guess you read the Q wrong ... it says proof by contrapositive is 100% valid (never got logical stuff in my brain though)
 one year ago

PhoenixFireBest ResponseYou've already chosen the best response.1
Well, thanks anyways @experimentX I'll just stick with how I think it should be and ask my lecturer about it later.
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
if you work this way ... from invalid theorem, ... you will reach a valid statement which in not complete.
 one year ago
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