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Math Help - Range of f

  1. #1
    Member billa's Avatar
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    Range of f

    let f:NxN->N be defined by f(x,y)=(1/2)(x+y)(x+y-1)-(1-y)

    Is f one-to-one? Does f map (NxN) to N or onto N (meaning are all values in N in the range of f or is the range a subset of N)?

    Can I get a hint on how to go about answering either of those questions. I know what each of the questions mean, but I have no idea how to answer them.
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  2. #2
    Senior Member Sampras's Avatar
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    Quote Originally Posted by billa View Post
    let f:NxN->N be defined by f(x,y)=(1/2)(x+y)(x+y-1)-(1-y)

    Is f one-to-one? Does f map (NxN) to N or onto N (meaning are all values in N in the range of f or is the range a subset of N)?

    Can I get a hint on how to go about answering either of those questions. I know what each of the questions mean, but I have no idea how to answer them.
    For the first, does  f(x,y) = f(x',y') \Rightarrow (x,y) = (x',y') ? For the second, put  y=1 . Then  f(x,1) = \frac{1}{2}x(x+1) whose range is....?
    Last edited by Sampras; July 14th 2009 at 08:00 PM.
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  3. #3
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Sampras View Post
    For the first, does  f(x,y) = f(x',y') \Rightarrow (x,y) = (x',y') ?
    It is not 1-1. There are two pairs that map to zero - (-1,1) and (0,1).
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  4. #4
    Member billa's Avatar
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    Quote Originally Posted by Swlabr View Post
    It is not 1-1. There are two pairs that map to zero - (-1,1) and (0,1).
    I think since f:NxN -> N, (-1,1) is not in the domain, so it might still be one-to-one


    For the first, does
    I set f(x,y)=f(x',y') and couldn't show that it implies (x,y)=(x',y')

    <br />
f(x,y)=f(x',y') \implies<br />
x^2+y^2+2xy-x-3y=(x')^2+(y')^2+2x'y'-x'-3y'<br />

    how can i tell that there aren't different pairs such that both of these are equal

    For the second, put . Then whose range is....?
    I am still working on this
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