anonymous
  • anonymous
The product of an odd number and an even number is
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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amistre64
  • amistre64
what is 2(3) ?
amistre64
  • amistre64
2n(2n+1) = 2(2n^2+n) = 2k for some integer k
anonymous
  • anonymous
Idk if that's the answer

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

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