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Math Help - Integers (587)

  1. #1
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    Integers (587) THANKS CAPTAINBLACK!

    For all numbers x and y, let the operation @ be defined by x @ y=xy-y. If a and b are positive integers, which of the following can be equal to zero?

    I. a @ b
    II. (a+b)@b
    III. a@(a+b)

    a) I only
    b) II only
    c) III only
    d) I and II
    e) I and III

    Can you please explain why? Thanks alot!!
    Last edited by fabxx; September 23rd 2008 at 12:01 AM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by fabxx View Post
    For all numbers x and y, let the operation @ be defined by x @ y=xy-y. If a and b are positive integers, which of the following can be equal to zero?

    I. a @ b
    II. (a+b)@b
    III. a@(a+b)

    a) I only
    b) II only
    c) III only
    d) I and II
    e) I and III

    Can you please explain why? Thanks alot!!
    If a and b are any real numbers we have:

    I. If a@b=0 then ab-b=0, which is the case when b=0 or a=1

    II. If (a+b)@b=0 then (a+b)b-b=0, which is the case when b=0 or (a+b)=1

    III. If a@(a+b)=0 then a(a+b)-(a+b)=0 which is the case when (a+b)=0 or a=1.

    Now if instead a and b are positive integers I. reduces to a=1, II. has no possible solutions in positive integers (explain why), while III. again permits a=1.

    RonL
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    Now if instead a and b are positive integers I. reduces to a=1, II. has no possible solutions in positive integers (explain why), while III. again permits a=1.

    RonL
    Is it because for II. if (a+b)=1 then it either means that a=1-b or b=1-a. Then one of them (a or b) can't be positive because 1-a has to be negative.
    Then you might say that let a=1. If a=1 then b has to equal 0 in order to satisfy (a+b)=1. But 0 is not a positive nor a negative. So (a+b)=1 has no possible solutions in positive integers. (:
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