1. The problem asks us to draw the Hasse diagram for the partially ordered set $(\{1, 2, 3, 4, 6, 8, 12\}, \mid)$ where the order relation is divisibility: $a \leq b$ if and only if $a$ divides $b$. 2. We start by listing elements and the divisibility relations: - $1$ divides every element. - $2$ divides $4, 6, 8, 12$. - $3$ divides $6, 12$. - $4$ divides $8, 12$. - $6$ divides $12$. - $8$ divides none greater than itself. - $12$ is the greatest element. 3. To draw the Hasse diagram, we include elements as nodes. Draw an edge (line) from $a$ to $b$ if $a$ divides $b$ and there is no $c$ such that $a$ divides $c$ and $c$ divides $b$ (cover relation). This excludes transitive edges. 4. The cover relations are: - $1 \to 2, 3$ (1 divides 2 and 3, nothing in between) - $2 \to 4, 6, 8$ (2 divides 4, 6, and 8 directly) - $3 \to 6$ - $4 \to 8, 12$ - $6 \to 12$ 5. Arrange nodes vertically so that arrows point upward. $1$ at bottom, $2$ and $3$ above it, then $4,6,8$ above those, and $12$ at the top. 6. The Hasse diagram reflects this partial order without arrows since edges go upward by convention. Final answer: The Hasse diagram nodes and edges are as described with the relations above visualized accordingly.