1. **State the problem:** We are asked to draw the Hasse diagram for the partial order defined by the divisibility relation $a \mid b$ on the set $\{1, 2, 3, 4, 6, 8, 12\}$. 2. **Understand the relation $a \mid b$:** This means $a$ divides $b$ without remainder. For example, $1$ divides all elements, $2$ divides $4, 6, 8, 12$ etc. 3. **Find covering relations:** In a Hasse diagram, we draw edges representing the "covering" relation, meaning $a \mid b$ and there is no $c$ such that $a \mid c \mid b$ with $c \neq a, b$. - From $1$, the elements covered immediately are $2$ and $3$ because they are the smallest multiples in the set. - $2$ covers $4$ and $6$ (since $2 \mid 4$ and $2 \mid 6$, and no intermediate divisor in the set between $2$ and $4$ or $6$). - $3$ covers $6$. - $4$ covers $8$ and $12$. - $6$ covers $12$. - $8$ and $12$ are maximal elements. 4. **Visual structure:** - Bottom level: $1$ - Second level: $2, 3$ - Third level: $4, 6$ - Top level: $8, 12$ 5. **Interpretation:** Edges go upward from smaller divisor to larger multiple without intermediate elements. **Final description:** - $1 \to 2$, $1 \to 3$ - $2 \to 4$, $2 \to 6$ - $3 \to 6$ - $4 \to 8$, $4 \to 12$ - $6 \to 12$ This pictorially arranges nodes so that larger multiples are placed higher, clearly representing the partial order by divisibility. Thus the Hasse diagram corresponds exactly to the described relations.