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anonymous

  • one year ago

The product of an odd number and an even number is

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  1. amistre64
    • one year ago
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    what is 2(3) ?

  2. amistre64
    • one year ago
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    2n(2n+1) = 2(2n^2+n) = 2k for some integer k

  3. anonymous
    • one year ago
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    Idk if that's the answer

  4. amistre64
    • one year ago
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    what is 2(3) ?

  5. anonymous
    • one year ago
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    I'm kinda confused on this question

  6. amistre64
    • one year ago
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    then you should determine what it means to be even and odd ...

  7. amistre64
    • one year ago
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    and maybe what it means to calculate a product

  8. anonymous
    • one year ago
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    I think it makes an even number like for example it says 3*6=18 5*12=60

  9. amistre64
    • one year ago
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    your assumption is good, 2 (3) = 6 even (odd) even works out in trials

  10. amistre64
    • one year ago
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    in general, i already posted the proofing

  11. anonymous
    • one year ago
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    Ima go with even

  12. amistre64
    • one year ago
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    k

  13. triciaal
    • one year ago
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    |dw:1440477601372:dw| @amistre64 is there a special reason you used 2n?

  14. amistre64
    • one year ago
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    n is a number, even or odd the set of even numbers is: 2n the set of odd numbers is: 2n+1

  15. amistre64
    • one year ago
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    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

  16. amistre64
    • one year ago
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    n(n+1) is just a special case and not a generalization 2(7) does not fit

  17. amistre64
    • one year ago
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    i spose i shoulda used i different variable as well ... n_1 and n_2 maybe

  18. amistre64
    • one year ago
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    \[2n_1(2n_2+1)=2k\]

  19. triciaal
    • one year ago
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    it didn't make a difference to me n is just a number see above agree result still even

  20. triciaal
    • one year ago
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    if n is even then n + 1 is odd if n is odd then n + 1 is even

  21. amistre64
    • one year ago
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    n(n+1) is a special case of even times odd,

  22. amistre64
    • one year ago
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    how do we prove its good for all cases? 2(7) is not of the form: n(n+1) is it?

  23. amistre64
    • one year ago
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    special cases are not general cases ... generality is needed to make a proof of ALL structures of the product of an even and odd.

  24. amistre64
    • one year ago
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    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

  25. triciaal
    • one year ago
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    do you agree odd*odd is always odd and even *even is always even?

  26. amistre64
    • one year ago
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    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

  27. triciaal
    • one year ago
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    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.

  28. amistre64
    • one year ago
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    everything i used has already been defined in the post ... redefining it just seemed like a waste of time :) but yes, its fine

  29. triciaal
    • one year ago
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    ok thanks

  30. amistre64
    • one year ago
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    youre welcome

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