How to show that the following relation is transitive (What is the idea?)

Question

Suppose we consider the following set $$ X = \big\{ (p,q) : p \in \mathbb{Z} \ \text{and} \ q \in \mathbb{N} \big\}$$

Define the following relation $$(p_1,q_1) \sim (p_2,q_2) \iff p_1q_2 = p_2q_1$$

We can easily verify that this relation is both symmetric and reflexive. I am having trouble many with transitive. The issue that I am having is we can't really "divide". So what we assume is the following:

Suppose that $(p_1,q_1) \sim (p_2,q_2)$ and $(p_2,q_2) \sim (p_3,q_3)$. This means that the following equations are satisfied:

$$p_1q_2 = p_2q_1$$ $$p_2q_3 = p_3q_2$$

We would like to show that the following equation is satisfied: $$p_1q_3 = p_3q_1$$

I am kinda lost in how to do that?

Answer 1

From $p_1q_2=p_2q_1$ and $p_2q_3=p_3q_2$ we obtain $$ (p_2q_2)(p_1q_3) = (p_2q_2)(p_3q_1), $$ which is equivalent to $$ (p_2q_2)(p_1q_3-p_3q_1) = 0. $$ Hence either $p_2q_2=0$ or $p_1q_3=p_3q_1$. Since $p_2q_2=0$ only if $p_2=0$, and in this case we must also have $p_3=p_1=0$ by hypothesis, we are still left with the equality $p_1q_3=p_3q_1$.