I attempted the following question:
Show that the number of binary operations on {1,2} having 1 as identity and having 2 as the inverse of 2 is exactly one.
My approach to solve the problem was this:
The question gives a suggestion about a particular operation.
For that operation, the identity is ‘1’.
Also, for that operation, the inverse of 2 is ‘2’.
I was able to understand that, the suggested operation will give the results for all the possible combinations (1,1), (1,2), (2,1) and (2,2).
But I am not able to prove that it is the only possible operation.
Please help me.
Suppose $f$ is the binary operations that satisfy those properties.
We already know that $f(1,x)=x$ and $f(x,1)=x$ for all $x \in \{1,2\}$ since $1$ is the identity.
$f(1,1)=1, f(1,2)=2, f(2,1)=2$.
$f(2,2)=1$ since $2$ is the inverse of $2$, since everything has been fully specified it must be unique.
Suppose not, then we can find $f$ and $g$ both being binary operations that satisfy those properties and some $x, y\in\{1,2\}$ such that $f(x,y) \ne g(x,y)$.
However, we can enumerate all $x$ and $y$ and check that such $x$ and $y$ do not exist. Hence it is a contradiction.
In general, if something has been fully specified, it is unique.