Rational Numbers form Totally Ordered Field

Theorem

The set of rational numbers $\Q$ forms a totally ordered field under addition and multiplication: $\struct {\Q, +, \times, \le}$.

Proof

Recall that by Integers form Ordered Integral Domain, $\struct {\Z, +, \times, \le}$ is an ordered integral domain

By Rational Numbers form Field, $\struct {\Q, +, \times}$ is a field.

In the formal definition of rational numbers, $\struct {\Q, +, \times}$ is the field of quotients of $\struct {\Z, +, \times, \le}$

By Total Ordering on Field of Quotients is Unique, it follows that $\struct {\Q, +, \times}$ has a unique total ordering on it that is compatible with its ring structure.

This article, or a section of it, needs explaining. In particular: Review the ordering / total ordering question You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code.

Thus $\struct {\Q, +, \times, \le}$ is a totally ordered field.

$\blacksquare$

Sources

• 1964: Walter Rudin: Principles of Mathematical Analysis (2nd ed.) ... (previous) ... (next): Chapter $1$: The Real and Complex Number Systems: Introduction
• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $23$. The Field of Rational Numbers