1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    jfizzix

    User Avatar
    Science Advisor
    Gold Member

    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.

    -James
     
  4. Apr 28, 2012 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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: Apr 28, 2012
  5. Apr 30, 2012 #4
    thank you for all your replies would try it out :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook