If I define $f: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R} \times \mathbb{R}$ by $f((x, y)) = (x+3, 4-y)$, how can I prove this function is one-to-one (injective)?
So far I have: $f$ is injective iff. $\forall (x_1, y_1), (x_2, y_2) \in \mathbb{R} \times \mathbb{R}$, if $f((x_1, y_1)) = f((x_2, y_2))$, then $(x_1, y_1) = (x_2, y_2)$.
Suppose $(x_1, y_1)$ and $(x_2, y_2)$ are elements of $\mathbb{R} \times \mathbb{R}$ such that $f((x_1, y_1)) = f((x_2, y_2))$
Then $(x_1+3, 4-y_1)=(x_2+3, 4-y_2)$
This is only possible over the reals if $(x_1, y_1) = (x_2, y_2)$
However my logic is obviously flawed at the final step. How can I prove that $(x_1, y_1) = (x_2, y_2)$?
