Inverse of a multivariable function 3

Question

Consider a function $f : Y_n \rightarrow X_{n^2} $

For any $(a, b) \in Y_n $, let $f((a, b)) = f(a, b) = an + b.$ Prove that $f$ is a bijection and find its inverse $f^{-1}$.

I am struggling to see how I can find an inverse of this multivariable function. I was already able to prove that $f(a,b) = an+b$ is a bijection, I'm just not certain how to find its specific inverse. It would be important to note that $Y_n=X_n \times X_n$ where $X_n = \{0,1,2,...,n-1\}$.

Any help would be appreciated. Thanks!

edit:

I've worked this out a little bit and came up with this, please let me know if this is invalid.

f^-1(f(a,b)) = f^-1(an+b)

let x = an+b

f^-1(x) = f^-1((floor(x/n)*n)+(x % n)) -> (floor(x/n), x%n)

Answer 1

• injectivity: Let $f(a_1,b_1) = f(a_2,b_2)$.
  Then $(a_1-a_2)n = b_1-b_2$.
  If $a_1 \ne a_2$, $|LHS| > n > |RHS|$, which is absurd. Therefore, $LHS = RHS = 0$.
• surjectivity: Let $c \in X_{n^2}$. By Division Algorithm ($n$ is the divisor), there exists unique quotient $a \in \Bbb N \cup\{0\}$ and remainder $b \in X_n$ such that $c = an + b$. Check that $a \in X_n$. It's obvious that $a \ge 0$. If $a \ge n$, $c \ge an \ge n^2$, so $c \notin X_{n^2}$, which is absurd.