"We are given:"
"i)"\ \ \quad A \ = \ { 8, 9, 10, 11 }.
"ii)"\ \quad B \ = \ { 2, 3, 4, 5 }.
"iii)" \quad R \ \ "is the relation from" \ A \ "to" \ B, "defined as follows:"
\qquad \qquad \qquad \qquad \qquad \qquad \qquad ( x, y ) \in R \quad hArr \quad y \quad "divides" \quad x.
"We want to find:"
\qquad \qquad "The domain of" \quad R.
"We can proceed as follows."
"1)" \quad R \ \ "can be restated as:"
\qquad \qquad \qquad \qquad \qquad \quad \ ( x, y ) \in R \quad hArr \quad x \quad "is a multiple of" \quad y.
"2)" \quad \ "So, we observe:"
\qquad \qquad x \in "domain of" \ R \quad hArr \quad ( x, y ) \in R, \quad "for some" \ \ y \in B
\qquad \qquad \qquad \qquad \qquad \quad hArr \quad x \quad "is a multiple of" \quad y, \quad "for some" \ y \in B
\qquad \qquad \qquad \qquad \qquad \quad hArr \quad B \ "contains a multiple of" \ x.
\qquad \quad \ "So, from beginning to end here, we conclude:"
\qquad \qquad \quad x \in "domain of" \ R \quad hArr \quad B \ "contains a multiple of" \ x.
"3)" \quad \ "So, to find the domain of" \ R, "we keep those elements of" \ A \ "that are a multiple of something in" \ B. \ \ "This is not hard to do:"
\qquad \qquad \qquad \qquad A \ = \ { 8, 9, 10, 11 } \qquad \qquad B \ = \ { 2, 3, 4, 5 }.
"We see:"
\qquad \qquad \ \ 8 \quad "is a multiple of" \quad 2 \ ( "and" \ 4 ), \qquad 9 \quad "is a multiple of" \quad 3,
10 \quad "is a multiple of" \quad 2, \qquad 11 \quad "is not a multiple of anything in" \ B.
"So, we have now:"
\qquad \qquad \qquad \qquad 8, 9, 10 \quad "are in the domain of" \ R;
\qquad \qquad \qquad \qquad \ \ 11 \quad "is not in the domain of" \ R.
"So, finally, we conclude:"
\qquad \qquad \qquad \qquad \qquad \qquad "domain of" \ R \ = \ { 8, 9, 10 }.
