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Math Help - One to one and onto question

  1. #1
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    One to one and onto question

    Determine whether the given function is one to one and whether it is onto. If the function is both one to one and onto, find the inverse of the function.
    f:R^2---->R^2, f(x,y)=(x+y, y) .

    I know one to one says f(x)=f(y) implies x=y
    Onto means if for every element y in R^2 , there exists an element x in R^2 with f(x)=y.
    I conceptually understand the idea, but don't know how to use these definitions.
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  2. #2
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    Function is one to one...

    Show that if  f(x_1,y_1) = f(x_2,y_2), then  x_1,y_1 = x_2,y_2

    Function is onto...

    For every  (x,y) \in R^2 , show there exists a  x_0, y_0 \in \mathbb{R}^2 such that  f(x_0,y_0) = (x,y)

    Good luck
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  3. #3
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    Ok I think it's the onto that's really driving me crazy.
    f(x0,y0)=(x0+y0,y0). I get there and get stuck.
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  4. #4
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    Quote Originally Posted by kathrynmath View Post
    Determine whether the given function is one to one and whether it is onto. If the function is both one to one and onto, find the inverse of the function.
    f:R^2---->R^2, f(x,y)=(x+y, y) .

    I know one to one says f(x)=f(y) implies x=y
    Onto means if for every element y in R^2 , there exists an element x in R^2 with f(x)=y.
    I conceptually understand the idea, but don't know how to use these definitions.
    I think in this case, rather than grind through the two definitions, just demonstrate that f has an inverse (say for example, something like g: R^2 -> R^2 defined by the rule, (x,y) |-> (x-y, y)?). A reasonable demonstration would then consist of showing that their compositions give the identity map on R^2.
    Last edited by PiperAlpha167; September 30th 2010 at 09:32 PM. Reason: Removed some extraneous stuff.
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  5. #5
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    Quote Originally Posted by kathrynmath View Post
    Ok I think it's the onto that's really driving me crazy.
    f(x0,y0)=(x0+y0,y0). I get there and get stuck.
    Think about what you have. Can you write every single pair (x,y) in terms of (x0 + y0, y0)?

    Remember that x0, y0 can be any number, your job is to show that there will be some numbers x0, y0 that would work for any pair. You can simply define x0 = x - y0 and y0 = y for an arbitary pair (x,y)

    so f(x0,y0) = (x0 + y0, y0) = (x - y0 + y0 , y) = (x , y)
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  6. #6
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    Suppose (a, b) is any pair of real numbers. Can you find (x, y) such that f(x,y)= (x+y, y)= (a, b)? That is the same as solving the two equations x+y= a, y= b for x and y.

    Suppose f(x, y)= f(u, v). Can you show that x= u, y= v? f(x, y)= (x+y, y)= f(u, v)= (u+v, v) so that is the same as the two equations x+y= u+ v, y= v. Does it follow that x= u, y= v?

    In the first part, where you found (x, y) such that f(x,y)= (a, b), you got x= some formula in terms of a and b, y= some formula in terms of a and b. Essentially then, you are saying that (x, y)= f^{-1}(a, b) so you have found your inverse function. (You might want to replace a and b with x and y to have it as f^{-1}(x, y).
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  7. #7
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    x+y=a
    y=b
    Subtracting gives x=a+b
    So atb+y=a
    x=a+b and y=b onto
    I'm going to skip the 1-1 part since I feel like I understand that.
    (a+b,b)=f^{-1}(x,y)
    Something like that?
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