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Proof

(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 $\frac ab < \frac ef$.
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