# Rational Numbers form Totally Ordered Field

Jump to navigation
Jump to search

## 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 quotient field of $\struct {\Z, +, \times, \le}$

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

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): $\S 23$