Let $\Sigma$ be a partial order defined over $\mathbb N^* = \mathbb N \setminus \{0\}$ such that $$n\;\Sigma\;m \Leftrightarrow n=m \text{ or } \text{r}(8,n) \text { is a non-trivial divisor of } \text{r}(8,m)$$
where $\text{r}(x,y)$ is the reminder resulting from the division of $x$ by $y$. Consider the following set $T$ and draw the Hasse diagram, explain why it is a lattice, tell if it is a complemented lattice and deduce $(X, \Sigma)$ isn't a distributive lattice.
$$T = \{1,3,5,6,7,9\}$$
I can demonstrate $S = (X, \Sigma)$ to be a lattice by showing that $\forall (a,b)\in X,\; \sup\{a,b\} \in S$ and $\inf\{a,b\} \in S$.
And I would say it is not complemented because $\sup\{6,5\} = 9 \neq 1$ and $\nexists a \in X : \sup\{6,a\} = 1$. How from that can I deduce it is not distributive? In addition have I correctly calculated that $\sup\{6,5\} = \sup\{3,6\} = \sup\{3,5\} = 9$?
The two prototypical non-distributive lattices are the diamond and the pentagon (see e.g. Wikipedia). To prove that some lattice is not distributive you need to check the distributive law for the middle three elements (i.e. $L -\{\top,\bot\}$) of the diamond or the pentagon sublattice.
I hope this helps ;-)