anonymous
  • anonymous
Use the expression for even (2p) and an odd integer (2m+1) to confirm the conjuncture that the product of an even integer and an odd integer is an even integer
Mathematics
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SOLVED
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chestercat
  • chestercat
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thomas5267
  • thomas5267
Just multiply the expression and you should see.
thomas5267
  • thomas5267
And the fact that 2x is always even.
anonymous
  • anonymous
So 2(4) multiplied by (2(7)+1) ?? I'm confused

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anonymous
  • anonymous
Ok the problem comes down to this.... p is an arbitrary integer... so 2p will always be even (i.e. always be divisible by 2 (another way of thinking of something as even)) and 2p+1 will always be odd (could also be 2p-1 depending on your taste) since an odd number always lies 1 to the left or one to the right of an even number. Does this make sense?
anonymous
  • anonymous
So lets multiply those expresions together: \[(2p)*(2p+1) = 4p^2 + 2p = 2(2p^2 + p)\] Now in the definitions above p is an arbitrary integer. No restrictions were place on it, so I am free to defined it however I choose and the definitions of even and odd still hold. Therefore I choose to define: \[ n=2p^2 +p\] As a new arbitrary integer. And thus by the result above I have an integer of the form: \[ 2(2p^2 +p) = 2n\] and thus it must be a positive integer. QED
anonymous
  • anonymous
I still don't understand tbh...
anonymous
  • anonymous
How about this? Since there is no restriction on p in the definitions 2p or 2p+1... I am free to treat the product 2n*(2n+1) = 2(2n^2+n) = 2p (where n is just an integer also) by simply defining p=(2n^2+n)... Since all we know is that p is an integer and the fact that I take an integer n... square it.... multiply it by 2... and then add n to it.... doesnt change the fact that it is still an integer. Thus I can represent this integer (renamed to be p) as 2p which is by definition even. I hope this helps.

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