# The reals (constructed as Dedekind cuts) form a field

The real num­bers, when con­structed as Dedekind cuts over the ra­tio­nals, form a field.

We shall of­ten write the one-sided Dedekind cut $$(A, B)$$ noteRe­call: “one-sided” means that $$A$$ has no great­est el­e­ment. as sim­ply $$\mathbf{A}$$ (us­ing bold face for Dedekind cuts); we can do this be­cause if we already know $$A$$ then $$B$$ is com­pletely de­ter­mined. This will make our no­ta­tion less messy.

The field struc­ture, to­gether with the to­tal or­der­ing on it, is as fol­lows (where we write $$\mathbf{0}$$ for the Dedekind cut $$(\{ r \in \mathbb{Q} \mid r < 0\}, \{ r \in \mathbb{Q} \mid r \geq 0 \})$$):

• $$(A, B) + (C, D) = (A+C, B+D)$$

• $$\mathbf{A} \leq \mathbf{C}$$ if and only if ev­ery­thing in $$A$$ lies in $$C$$.

• Mul­ti­pli­ca­tion is some­what com­pli­cated.

• If $$\mathbf{0} \leq \mathbf{A}$$, then $$\mathbf{A} \times \mathbf{C} = \{ a c \mid a \in A, a > 0, c \in C \}$$. we’ve missed out the com­ple­ment in this no­ta­tion, and can’t put set-builder sets in boldface

• If $$\mathbf{A} < \mathbf{0}$$ and $$\mathbf{0} \leq \mathbf{C}$$, then $$\mathbf{A} \times \mathbf{C} = \{ a c \mid a \in A, c \in C, c > 0 \}$$.

• If $$\mathbf{A} < \mathbf{0}$$ and $$\mathbf{C} < \mathbf{0}$$, then $$\mathbf{A} \times \mathbf{C} = \{\}$$ write down the form of the set

where $$(A, B)$$ is a one-sided Dedekind cut (so that $$A$$ has no great­est el­e­ment).

(Here, the “set sum” $$A+C$$ is defined as “ev­ery­thing that can be made by adding one thing from $$A$$ to one thing from $$C$$“: namely, $$\{ a+c \mid a \in A, c \in C \}$$ in set builder no­ta­tion; and $$A \times C$$ is similarly $$\{ a \times c \mid a \in A, c \in C \}$$.)

# Proof

## Well-definedness

We need to show firstly that these op­er­a­tions do in fact pro­duce Dedekind cuts.

Firstly, we need ev­ery­thing in $$A+C$$ to be less than ev­ery­thing in $$B+D$$. This is true: if $$a+c \in A+C$$, and $$b+d \in B+D$$, then since $$a < b$$ and $$c < d$$, we have $$a+c < b+d$$.

Next, we need $$A+C$$ and $$B+D$$ to­gether to con­tain all the ra­tio­nals. This is true: this, it’s quite boring

Fi­nally, we need $$(A+C, B+D)$$ to be one-sided: that is, $$A+C$$ needs to have no top el­e­ment, or equiv­a­lently, if $$a+c \in A+C$$ then we can find a big­ger $$a' + c'$$ in $$A+C$$. This is also true: if $$a+c$$ is an el­e­ment of $$A+C$$, then we can find an el­e­ment $$a'$$ of $$A$$ which is big­ger than $$a$$, and an el­e­ment $$c'$$ of $$C$$ which is big­ger than $$C$$ (since both $$A$$ and $$C$$ have no top el­e­ments, be­cause the re­spec­tive Dedekind cuts are one-sided); then $$a' + c'$$ is in $$A+C$$ and is big­ger than $$a+c$$.

this section

### Ordering

this section

iden­tity, as­so­ci­a­tivity, in­verse, commutativity

## Ring structure

mul­ti­plica­tive iden­tity, as­so­ci­a­tivity, distributivity

inverses

## Order­ing on the field

a ⇐ b im­plies a+c ⇐ b+c, and 0 ⇐ a, 0 ⇐ b im­plies 0 ⇐ ab

Parents: