injective functions R^2->R^2

[xoninhas]

hey everyone, I'm kinda new here so I'm not sure this is the place for this but here it goes:

I have the functions:
f(x,y) = ( x + 2y , x -y )
g(x,y) = ( log(xy) , 1/(x^2 + y^2 )

I want to know if they are injective or/and surjective, but I want to know the method to determinate these things.

I know that for it to be injective: if f(x1,y1) = f(x2,y2) then x1=x2 and y1=y2
for it to be surjective I have more difficulties.

Thanks for any help people!

 

[pka]

Here is the the proof that f is injective.
\(\displaystyle \begin{array}{l} f(a,b) = f(c,d) \\ \left. \begin{array}{l} a + 2b = c + 2d \\ a - b = c - d \\\end{array} \right\} \Rightarrow \quad a = c\,\& \,b = d \\ \end{array}\)

To see if it is surjective if \(\displaystyle \left( {a,b} \right) \in \Re ^2\) can you find an \(\displaystyle \left( {x,y} \right) \in \Re ^2\) such that \(\displaystyle f(x,y) = \left( {x + 2y,x - y} \right) = \left( {a,b} \right)\).

 

[xoninhas]

ok than how do you prove that

f(x,y) = ( log(xy) , 1/(x^2 + y^2) )

is injective using that same technique? I'm interested in the technique to find it out.

Thanks anyway!

 

[stapel]

ok than how do you prove that f(x,y) = ( log(xy) , 1/(x^2 + y^2) ) is injective using that same technique?

Use the same technique (follow the same method and steps) as in the exercise already completed for you.

If you get stuck, kindly reply with a clear listing of all of your work and reasoning so far. Thank you!

Eliz.

 

[xoninhas]

Ooops I just noticed that I never thanked everyone.... in the end I got how to do it, and I realized that each function is a function and you need an eye for the thing!

Anyway thanks for everything!

 

[cadamcross]

By the way, to discuss whether a function is surjective you must explicitly state the codomain.