# Proving 1:1 and onto of 2 variable function

Let $\ f: \mathbb Q^2\to\mathbb Q^2$ be defined by $\ f(x,y) = (x-3y , x + 3y- 1)$ Prove the function is 1:1, onto, find the inverse and $\ f \circ f$.

I was able to prove 1:1 by showing $\ f(x_1,y_1)=f(x_2,y_2) \implies (x_1,y_1)=(x_2,y_2)$, which was relatively simple, but i'm having a hard time using the definition of the onto function with this 2 variable function.

Any push in the right direction would be helpful.

• Hint: Given $(r,s)\in\mathbb{Q}^2$ you must show that there exist $(x,y)\in\mathbb{Q}^2$ such that \begin{align*} x-3y&=r\\ x+3y-1&=s \end{align*} – Ángel Mario Gallegos Apr 12 '16 at 19:41
• So isolating x and y and showing that they belong to Q is proof enough? – Jonathan Apr 12 '16 at 19:54
• Yes, it is enough. – Ángel Mario Gallegos Apr 12 '16 at 19:56
• Start by finding the inverse (which is not difficult). – almagest Apr 12 '16 at 19:56