A bit confused on how to approach this problem since I have only worked with sets for a similar question. Any help would be great!
Given two integers $a$ and $b$, we say that $a$ divides $b$ is there is some integer $k$ such that $a \cdot k = b$. Prove that this relation is reflective and transitive but not symmetric.
Given $a, b \in \mathbb{Z}$, we denote $a \sim b$ if $a$ divides $b$.
Now to check that the relation $\sim$ is reflexive, we want to check $$\forall a \in \mathbb{Z}, \,a \sim a.$$ This is immediate, since $$\forall a \in \mathbb{Z}, a \cdot 1 = a \, \text{and} \, 1 \in \mathbb{Z}$$, and hence $$\forall a \in \mathbb{Z}, \,a \, \text{divides} \, a.$$
To check that $\sim$ is transitive, we want to show that $$\forall a,b,c \in \mathbb{Z}\, \text{with} \, a \sim b \, \text{and} \, b \sim c, \, a \sim c.$$ Now, since $a$ divides $b$, $$\exists k \in \mathbb{Z} \,\text{s.t.}\, ak = b.$$ Also, since $b$ divides $c$, $$\exists l \in \mathbb{Z} \, \text{s.t.}\, bl = c.$$ Hence $$a(kl) = (ak)l = bl = c.$$ Since $k, l \in \mathbb{Z}, kl \in \mathbb{Z}.$ It follows that $a$ divides $c$. It follows that $\sim$ is transitive. To show that $\sim$ is not symmetric, we only need one example, and @drhab has already provided one.