I worked on a problem that reads:
Let $A = \{2, 3, 4, 7\}$ and $B = \{1, 2, 3, ..., 12\}$. Define $a\ S\ b$ if and only if $a | b$. Use the roster method to describe $S$.
My answer also included the ordered pair $(7, 7)$, but that's incorrect. The reason is stated in the definition:
The domain of a relation $R \subseteq A \times B$ is defined as $$dom\ R = \{a \in A\ |\ (a, b) \in R \text{ for some } b \in B\}$$ and the image or range is defined as $$im\ R = \{b \in B\ |\ (a, b) \in R \text{ for some } a \in A\}$$
I have problems understanding this definition.
So in the ordered pair $(a, b) \in R$, the input is the $a$'s from $A$ and the $b$'s from $B$ and the output is the $b$'s from $B$ and the $a$'s from $a$.
But why isn't then the pair $(7, 7)$ a valid solution? Since $7$ is in $A$ and $B$ in the above example?
UPDATE, since some answers pointed out that it's a typo, I think it makes sense to include the reference from where I got this exercise.
This is the original problem (number 7.1.4, page 198/199 in the book):
And this is the solution (page 271 in the book):
The source is A Spiral Workbook for Discrete Mathematics (Harris Kwong) - which is a free textbook and can be downloaded as a PDF here.
What puzzles me a bit is that the author explicity states:
In the last example, $7$ never appears as the first element (in the first coordinate) of any ordered pair. Likewise, $1, 5, 7$, and $11$ never appear as the second element (in the second coordinate) of any ordered pair.
That got me confused as I also expected $(7,7)$ to be part of the solution set $S$.
$7|7$, so $(7,7)\in S$. There isn’t much more to it. The solution you are comparing your solution to must have a typo.
We could discuss the definitions of the domain and image of the relation, but these have nothing to do with whether $(7,7)\in S$.