The product of an odd number and an even number is

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The product of an odd number and an even number is

Mathematics
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what is 2(3) ?
2n(2n+1) = 2(2n^2+n) = 2k for some integer k
Idk if that's the answer

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what is 2(3) ?
I'm kinda confused on this question
then you should determine what it means to be even and odd ...
and maybe what it means to calculate a product
I think it makes an even number like for example it says 3*6=18 5*12=60
your assumption is good, 2 (3) = 6 even (odd) even works out in trials
in general, i already posted the proofing
Ima go with even
k
|dw:1440477601372:dw| @amistre64 is there a special reason you used 2n?
n is a number, even or odd the set of even numbers is: 2n the set of odd numbers is: 2n+1
even(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
n(n+1) is just a special case and not a generalization 2(7) does not fit
i spose i shoulda used i different variable as well ... n_1 and n_2 maybe
\[2n_1(2n_2+1)=2k\]
it didn't make a difference to me n is just a number see above agree result still even
if n is even then n + 1 is odd if n is odd then n + 1 is even
n(n+1) is a special case of even times odd,
how do we prove its good for all cases? 2(7) is not of the form: n(n+1) is it?
special cases are not general cases ... generality is needed to make a proof of ALL structures of the product of an even and odd.
let (n,m) be any 2-tuple 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
do you agree odd*odd is always odd and even *even is always even?
i 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
is 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.
everything i used has already been defined in the post ... redefining it just seemed like a waste of time :) but yes, its fine
ok thanks
youre welcome

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