# Prove a Function $f: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R} \times \mathbb{R}$ is Injective

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 $(x_1, y_1) = (x_2, y_2)$?

• You may add one line: Then $x_1 + 3 = x_2 + 3$, $4-y_1 = 4-y_2$, then $x_1 = x_2$ and $y_1 = y_2$. Do you feel better now?
– user99914
Oct 5, 2015 at 6:39
• Do you know that the only way $(a,b)$ and $(c,d)$ can be equal is to have $a=c$ and $b=d$? Oct 5, 2015 at 6:39
• Why do you think your logic is obviously flawed in the final step? You come to the right conclusion so it's not that obvious. Oct 5, 2015 at 6:39

Suppose $f(x_1,y_1)=f(x_2,y_2)$. That is, $(x_1 +3,4-y_1)=(x_2+3,4-y_2)$. These row vectors are equal to one another if and only if the corresponding components are equal. That is, $x_1+3 = x_2+3$ and $4-y_1=4-y_2$. Clearly, $x_1=x_2$ and $y_1=y_2$. Hence, $(x_1,y_1)=(x_2,y_2)$ and so $f$ is injective.