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Math Help - New Binary operation

  1. #1
    Newbie
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    New Binary operation

    Hey
    Im new here, and i'm not sure where to put this Q,
    I was studying for the AMO, and I came across a problem in the 2004 AIMO which has me completely stuck ...

    A new binary operation ? defined on all positive x and y, satisfies the following conditions:

    i x?x = x+2
    ii x?y = y?x
    iii x?(x+y) = x+y
    ----x?y------y

    Given the above, find 11?8

    please explain how to do this, i have been to other forums, and unless I ask, they just give a number.

    Thanks
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  2. #2
    MHF Contributor
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    France
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    Hi

    From ii
    11 ? 8 = 8 ? 11

    From iii
    8 ? 11 = 8 ? (8 + 3) = (8 ? 3) 11 / 3

    Now we need to find 8 ? 3
    From ii
    8 ? 3 = 3 ? 8

    From iii
    3 ? 8 = 3 ? (3 + 5) = (3 ? 5) 8 / 5

    Now we need to find 8 ? 3
    From iii
    3 ? 5 = 3 ? (3 + 2) = (3 ? 2) 5 / 2

    Now we need to find 3 ? 2
    From ii
    3 ? 2 = 2 ? 3

    2 ? 3 = 2 ? (2 + 1) = (2 ? 1) 3 / 1

    Now we need to find 2 ? 1
    From ii
    2 ? 1 = 1 ? 2

    1 ? 2 = 1 ? (1 + 1) = (1 ? 1) 2 / 1

    From i
    1 ? 1 = 1 + 2 = 3

    Therefore you can go up the chain to get 11 ? 8
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  3. #3
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    Thanks running-gag
    just one Q, where did you get (8?3)11/3 ???
    then, (3?5)8/5 etc
    all the way down?
    thanks
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  4. #4
    MHF Contributor
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    Rule iii is
    \frac{x?(x+y)}{x?y} = \frac{x+y}{y}

    If you substitute x = 8 and y = 3 you get
    \frac{8?(8+3)}{8?3} = \frac{8+3}{3} = \frac{11}{3}

    Therefore 8?(8+3) = (8?3)\:\frac{11}{3}
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  5. #5
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    yep, thanks, i see,
    just i was keeping x=8, y=11 from the 8?11=8?(8+3), so i got (8?11)11/3
    thanks
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