PhoenixFire
  • 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*
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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PhoenixFire
  • 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.
experimentX
  • experimentX
both must be odd.
PhoenixFire
  • 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"

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

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