The product of an odd number and an even number is

- anonymous

The product of an odd number and an even number is

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- schrodinger

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- amistre64

what is 2(3) ?

- amistre64

2n(2n+1) = 2(2n^2+n) = 2k for some integer k

- anonymous

Idk if that's the answer

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## More answers

- amistre64

what is 2(3) ?

- anonymous

I'm kinda confused on this question

- amistre64

then you should determine what it means to be even and odd ...

- amistre64

and maybe what it means to calculate a product

- anonymous

I think it makes an even number like for example it says 3*6=18
5*12=60

- amistre64

your assumption is good,
2 (3) = 6
even (odd) even
works out in trials

- amistre64

in general, i already posted the proofing

- anonymous

Ima go with even

- amistre64

k

- triciaal

|dw:1440477601372:dw|
@amistre64 is there a special reason you used 2n?

- amistre64

n is a number, even or odd
the set of even numbers is: 2n
the set of odd numbers is: 2n+1

- amistre64

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

- amistre64

n(n+1) is just a special case and not a generalization
2(7) does not fit

- amistre64

i spose i shoulda used i different variable as well ... n_1 and n_2 maybe

- amistre64

\[2n_1(2n_2+1)=2k\]

- triciaal

it didn't make a difference to me n is just a number see above agree result still even

- triciaal

if n is even then n + 1 is odd
if n is odd then n + 1 is even

- amistre64

n(n+1) is a special case of even times odd,

- amistre64

how do we prove its good for all cases?
2(7) is not of the form: n(n+1) is it?

- amistre64

special cases are not general cases ... generality is needed to make a proof of ALL structures of the product of an even and odd.

- amistre64

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

- triciaal

do you agree odd*odd is always odd and even *even is always even?

- amistre64

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

- triciaal

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.

- amistre64

everything i used has already been defined in the post ... redefining it just seemed like a waste of time :)
but yes, its fine

- triciaal

ok thanks

- amistre64

youre welcome

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