# Verifying that f(x,y) is one to one.

1. Apr 27, 2012

### marble1112

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. Apr 27, 2012

### jfizzix

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.

-James

3. Apr 28, 2012

### HallsofIvy

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 $p= (x_1, y_1)$ and $q= (x_2, y_2)$
. 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 $p= (x_1, y_1)$ and $q= (x_2, y_2)$ such that $f(p)= 2x_1+ y_1= 2x_2+ y_2= f(q)$. That would mean $2(x_1- x_2)= -(y_1- y_2)= 0$ Well, take $x_1= 3$, $x_2= 2$, $y_1= 1$, $y_2= 3$. Then $x_1- x_2= 3- 2= 1$ so $2(x_1- x_2)= 2$ and $y_1- y_2= -2$ so $-(y_1- y_2)= 2$ 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
4. Apr 30, 2012

### marble1112

thank you for all your replies would try it out :)