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anonymous
 one year ago
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
anonymous
 one year ago
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

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thomas5267
 one year ago
Best ResponseYou've already chosen the best response.0Just multiply the expression and you should see.

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.0And the fact that 2x is always even.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So 2(4) multiplied by (2(7)+1) ?? I'm confused

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok 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 2p1 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
 one year ago
Best ResponseYou've already chosen the best response.0So 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
 one year ago
Best ResponseYou've already chosen the best response.0I still don't understand tbh...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How 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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