# "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 \}. #
