Set of Rational Cuts forms Ordered Field
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Theorem
Let $\RR$ denote the set of rational cuts.
Let $\struct {\RR, +, \times, \le}$ denote the ordered structure formed from $\RR$ and:
- the operation $+$ of addition of cuts
- the operation $\times$ of multiplication of cuts
- the ordering $\le$ of cuts.
Then $\struct {\RR, + \times, \le}$ is an ordered field.
Proof
We demonstrate that $\struct {\RR, +, \times}$ is a field by showing it is a subfield of the structure $\struct {\CC, +, \times}$, where $\CC$ denotes the set of all cuts.
We do this by establishing that all $4$ criteria of the Subfield Test are satisfied.
We note that $0^* \in \RR$, where $0^*$ is the rational cut associated with the (rational) number $0$:
- $0^* = \set {r \in \Q: r < 0}$
So $\RR \ne \O$.
Thus criterion $(1)$ is satisfied.
$\Box$
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$\blacksquare$
Sources
- 1964: Walter Rudin: Principles of Mathematical Analysis (2nd ed.) ... (previous) ... (next): Chapter $1$: The Real and Complex Number Systems: Real Numbers