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anonymous
 one year ago
The product of an odd number and an even number is
anonymous
 one year ago
The product of an odd number and an even number is

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amistre64
 one year ago
Best ResponseYou've already chosen the best response.02n(2n+1) = 2(2n^2+n) = 2k for some integer k

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Idk if that's the answer

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm kinda confused on this question

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0then you should determine what it means to be even and odd ...

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0and maybe what it means to calculate a product

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think it makes an even number like for example it says 3*6=18 5*12=60

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0your assumption is good, 2 (3) = 6 even (odd) even works out in trials

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0in general, i already posted the proofing

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440477601372:dw @amistre64 is there a special reason you used 2n?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0n is a number, even or odd the set of even numbers is: 2n the set of odd numbers is: 2n+1

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0even(odd) 2n (2n+1) 2 (2n^2+n) ^^^^^^ by closure of integers, this is just an integer, let it be k 2k ... is an even number

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0n(n+1) is just a special case and not a generalization 2(7) does not fit

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0i spose i shoulda used i different variable as well ... n_1 and n_2 maybe

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0it didn't make a difference to me n is just a number see above agree result still even

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0if n is even then n + 1 is odd if n is odd then n + 1 is even

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0n(n+1) is a special case of even times odd,

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0how do we prove its good for all cases? 2(7) is not of the form: n(n+1) is it?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0special cases are not general cases ... generality is needed to make a proof of ALL structures of the product of an even and odd.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0let (n,m) be any 2tuple of integers 2n(2m+1) 2(2nm + n) by closure of integers under addition and multiplication; 2mn+n is just some integer, call it k 2k is therefore an even number

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0do you agree odd*odd is always odd and even *even is always even?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0i dont have to agree, i can prove it it general for all cases (2n+1)(2m+1) 4nm+2n+2m+1 2(2nm+n+m)+1 = 2k+1 is an odd number  2n(2m) = 2(2nm) = 2k is an even number

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0is your proof saying yes or no? nothing is defined. I am not questioning what you have. i am just asking to make sure that I still have it correct that odd*odd is odd and even *even is even.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0everything i used has already been defined in the post ... redefining it just seemed like a waste of time :) but yes, its fine
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