# 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

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