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Homework Help: Verifying that f(x,y) is one to one.

  1. Apr 27, 2012 #1
    1. The problem statement, all variables and given/known data
    How do you show that a function with two variables f(x,y) is one-to-one and onto?
    example f(x,y) = 2x+y

    2. Relevant equations

    Do we have to use linear algebra?
  2. jcsd
  3. Apr 27, 2012 #2


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    f(x,y) represents a 2D surface.

    To have (x,y) -> f(x,y) be 1 to 1 and onto, I think that means that at every height (every value of f) you have a unique point (x,y).

    Unless there is some additional constraint for x and y, so that f(x,y) represents a curve I'm not sure that this is possible.

    In your example f(x,y) = 2x+y is a 2D plane. For each value of f, you get a line 2x +y = const.

    Interesting to think about, though.

  4. Apr 28, 2012 #3


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    Proving that f(x,y) is "one-to-one" and "onto" depends upon the range space!

    Since f(x,y)= 2x+ y is, for numbers x and y, a single number, the "default" assumption would be that f maps R2 to R.

    And to prove, one way or the other, use the definitions!

    A function, f, from set U to set V, is said to be "one to one" if and only if two different values, p and q, in U cannot give the same point in V.

    Here, U is R2 so we can write [itex]p= (x_1, y_1)[/itex] and [itex]q= (x_2, y_2)[/itex]
    . V is R so any point in V is a single number, z. Now, if f= 2x+ y were not one to one, then there would exist [itex]p= (x_1, y_1)[/itex] and [itex]q= (x_2, y_2)[/itex] such that [itex]f(p)= 2x_1+ y_1= 2x_2+ y_2= f(q)[/itex]. That would mean [itex]2(x_1- x_2)= -(y_1- y_2)= 0[/itex] Well, take [itex]x_1= 3[/itex], [itex]x_2= 2[/itex], [itex]y_1= 1[/itex], [itex]y_2= 3[/itex]. Then [itex]x_1- x_2= 3- 2= 1[/itex] so [itex]2(x_1- x_2)= 2[/itex] and [itex]y_1- y_2= -2[/itex] so [itex]-(y_1- y_2)= 2[/itex] also.

    That is, f(3, 1)= 2(3)+ 1= 7 and f(2, 3)= 2(2)+ 3= 7. No, f(x, y)=2x+ y is NOT "one to one".

    A function, f, from set U to set V, is "onto" (the set V) if and only if, for any point q in V, there exist at least one point p, in U so that f(p)= q. To show that f(x,y)= 2x+ y is "onto" R, let z be any real number (any point in R) and look for (x, y) (a point in R2) such that 2x+y= z. In fact (exactly because this function is not "one to one") there are many such points. Take, for example, y to be z- 2 and x to be 1.

    Yes, f(x, y)= 2x+ y is "onto" the set of real numbers.
    Last edited by a moderator: Apr 28, 2012
  5. Apr 30, 2012 #4
    thank you for all your replies would try it out :)
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