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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):

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And this is the solution (page 271 in the book):

enter image description here

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 $S$.

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  • $\begingroup$ Yes, it's a typo. As you can see in the answer, $(2,2), (3,3), (4,4) \in S$. Withe the same "logic", also $(7,7) \in S$. $\endgroup$ – Mauro ALLEGRANZA Nov 29 '17 at 8:16
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$7|7$, 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 $(7,7)\in S$.

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  • $\begingroup$ Hm, I see, maybe there is some error in the book. I'm just surprised that the author explicitly added a paragraph why 7 is not a solution (according to him), so that got me confused. $\endgroup$ – Max Nov 29 '17 at 3:30
  • $\begingroup$ Looking into it, it seems to me that the alleged absence of 7 as the first component is a motivation to define the domain and range (and show tha they don't need to correspond to A and B). In other words, those definitions are not part of any proof, they merely follow. Try reading the number 7 as if the author really meant to say 17, and it will make sense (I guess). $\endgroup$ – user491874 Nov 29 '17 at 7:41
  • $\begingroup$ Thanks for looking into it, I think there is not much more to it, indeed. It think I just ignore this exercise and just continue, it is indeed confusing. $\endgroup$ – Max Nov 29 '17 at 8:25
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$(7,7)$ should be included in the subset $S$ of $A \times B$.

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