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

We shall often write the one-sided Dedekind cut ^[1] as simply $A$ (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 $0$ for the Dedekind cut $({r \in Q ∣ r < 0}, {r \in Q ∣ r \geq 0})$ ):

$(A, B) + (C, D) = (A + C, B + D)$

$A \leq C$ if and only if everything in $A$ lies in $C$ .

Multiplication is somewhat complicated.
- If $0 \leq A$ , then $A \times C = {a c ∣ a \in A, a > 0, c \in C}$ .
  we've missed out the complement in this notation, and can't put set-builder sets in boldface
- If $A < 0$ and $0 \leq C$ , then $A \times C = {a c ∣ a \in A, c \in C, c > 0}$ .
- If $A < 0$ and $C < 0$ , then $A \times C = {}$
  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, ${a + c ∣ a \in A, c \in C}$ in set builder notation; and $A \times C$ is similarly ${a \times c ∣ a \in A, c \in C}$ .)

Proof

Well-definedness

We need to show firstly that these operations do in fact produce 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:

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

this section

Ordering

this section

Additive commutative group structure

identity, associativity, inverse, commutativity

Ring structure

multiplicative identity, associativity, distributivity

Field structure

inverses

Ordering on the field

a <= b implies a+c <= b+c, and 0 <= a, 0 <= b implies 0 <= ab

^{^︎}
Recall: "one-sided" means that $A$ has no greatest element.