Is there any mathematical operation “op”, such that when applied to two integers a, b: a op b = n. We can use that n and reverse the operation, to get n = a op b?
For example, the sum is not reversible to two integers, since 5 + 7 = 12, but we don’t know if 12 = 2 + 10 or 3 + 9 etc.
If $n$ doesn’t have to be an integer, I can define an operator.
For all integers $a$ and $b$, “$\bullet$” defines as follows:
$$a \bullet b = a+b+ai,$$
where $i$ is the square root of $-1$.
Here using $i$, we aim to save one of the elements to turn back. If we define reverse operator as below,
$$\circ(z) = (\operatorname{Im}(z), \operatorname{Re}(z) - \operatorname{Im}(z)),$$
we obtain $(a,b)$ again.
If $n$ has to be an integer, maybe we can define an operator based on “fundamental theorem of arithmetic”.
According to this theorem, all positive integers can be represented as a product of prime powers in a unique way. One can define an operator using this uniqueness.
What about prime number factorisation?
$$f(a,b) = 2^a \cdot 3^b = n$$
Obviously, $n$ can only be a number, which is only divisible by prime factors $2$ and $3$ (like $1$, $2$, $3$, $4$, $6$, $8$, $9$, $12$, $16$, $18$, $24$, ...), but for each of those number, you can easily find $a$ and $b$ back.
For an $m \times n$ matrix, you can always map the pair of indexes $(i,j) \in \{(1,1), \ldots, (1,n), \ldots, (m,1), \ldots, (m,n)\}$ to a single value in $\{1, \ldots, mn\}$: $$(i,j) \to n(i-1)+j.$$ It is a bijection, hence it is reversible: $$r(i,j) = n(i-1)+j \leftrightarrows \begin{cases} i=\lceil r/n\rceil, \\ j=(r-1)\bmod n+1; \end{cases}$$ provided that you know both $m$ and $n$ at all times.
Yes, take any bijection f from ℤ × ℤ into ℤ (there is at least one as the domain and co-domain have the same cardinality; one could also choose f to be surjective). Then, its inverse function "takes you back."
