(related to Proposition: Order Relation for Rational Numbers is Strict Total)

In order to avoid complicated distinctions of cases, we can without loss of generality assume the integers $a,b,c,d,e,f\in Z$ to be all positive integers. We first show the trichotomy of the order relation "$<$" for rational numbers \(\frac ab,\frac cd, \in\mathbb Q\):

- By the definition of the order relation for rational numbers,
- $\frac ab < \frac cd$ is equivalent to $\frac ab - \frac cd < 0$,
- $\frac ab > \frac cd$ is equivalent to $\frac ab - \frac cd > 0$, and
- $\frac ab = \frac cd$ is equivalent to $\frac ab - \frac cd = 0$.

- Therefore, the order relation of rational numbers "$<$" can be reduced to the order relation for integers "$<$".
- Now, the trichotomy of the order relation for rational numbers follows from the trichotomy of the order relation for integers.
- Therefore, only one of the cases $\frac ab = \frac cd$, or $\frac ab < \frac cd$, or $\frac ab > \frac cd$ can occur at once.

We now show the transitivity.

- Let $\frac ab < \frac cd$ and $\frac cd < \frac ef$.
- It follows $\frac ab - \frac cd < 0$ and $\frac cd - \frac ef < 0$.
- Adding both sides of the inequations in the set of rational numbers results in $\frac ab- \frac cd + \frac cd- \frac ef < 0$.
- The existence of inverse rational numbers with respect to additions results in $\frac ab- \frac ef < 0$.
- This means that $\frac ab < \frac ef$.∎