## Precalculus (6th Edition)

RECALL: $A \subseteq B$ if all elements of $A$ are also elements of $B$ The elements of $D$ are: $1, 2, 3$ The elements of $B$ are: $2, 4, 6, 8$ Since the elements of $D$ are not in $B$, then $D \not \subseteq B$ . Thus, the given statement is true.