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

We shall often write the one-sided [dedekind_cut Dedekind cut] $(A, B)$ %%note:Recall: "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_order 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 $\mathbf{A} \times \mathbf{C} = \{ a c \mid a \in A, a > 0, c \in C \}$. [todo: 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 $\mathbf{A} \times \mathbf{C} = \{\}$ [todo: write down the form of the set]

where $(A, B)$ is a one-sided [dedekind_cut 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_cut Dedekind cuts].

Addition

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: [todo: 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 $a+c$.

Multiplication

[todo: this section]

Ordering

[todo: this section]

Additive commutative group structure

[todo: identity, associativity, inverse, commutativity]

Ring structure

[todo: multiplicative identity, associativity, distributivity]

Field structure

[todo: inverses]

Ordering on the field

[todo: a <= b implies a+c <= b+c, and 0 <= a, 0 <= b implies 0 <= ab]