# Homotopy Type Theory an axiomatization of the real numbers > history (Rev #6, changes)

Showing changes from revision #5 to #6: Added | Removed | Changed

## Idea

I am going to define this in terms of Archimedean ordered Q-algebras…

## Definition

### Commutative $\mathbb{Q}$-algebras

A commutative ring $R$ is a commutative $\mathbb{Q}$-algebra if there is a commutative ring homomorphism $h:\mathbb{Q} \to R$.

### Totally ordered commutative rings

A commutative ring $R$ is a totally ordered commutative ring if it comes with a function $\max:R \times R \to R$ such that

• for all elements $a:R$, $\max(a, a) = a$

• for all elements $a:R$ and $b:R$, $\max(a, b) = \max(b, a)$

• for all elements $a:R$, $b:R$, and $c:R$, $\max(a, \max(b, c)) = \max(\max(a, b), c)$

• for all elements $a:R$ and $b:R$, $\max(a, b) = b$ implies that for all elements $c:R$, $\max(a + c, b + c) = b + c$

• for all elements $a:R$ and $b:R$, $\max(a, 0) = a$ and $\max(b, 0) = b$ implies $\max(a \cdot b, 0) = a \cdot b$

• for all elements $a:R$ and $b:R$, $\max(a, b) = a$ or $\max(a, b) = b$

### Totally ordered commutative $\mathbb{Q}$-algebras

Given totally ordered commutative rings $R$ and $S$, a commutative ring homomorphism $h:R \to S$ is monotonic if for all $a:R$ and $b:R$, $\max(h(a), h(b)) = h(\max(a, b))$.

A totally ordered commutative ring $R$ is a totally ordered commutative $\mathbb{Q}$-algebra if there is a monotonic commutative ring homomorphism $h:\mathbb{Q} \to R$.

### Strictly ordered integral ring

A totally ordered commutative ring $R$ is a strictly ordered integral ring if it comes with a strict type order family$\lt$ such that

• $0 \lt 1$for all elements $a:R$ and $b:R$, $a \lt b$ is a proposition
• for all elements $a:R$ , and b:R a \lt a , if is false$0 \lt a$ and $0 \lt b$, then $0 \lt a + b$
• for all elements $a:R$ , and$b:R$ , if and 0 c:R \lt a , and if 0 a \lt b c, then  0 \lt a \cdot \lt b or $b \lt c$
• for all elements $a:R$ and $b:R$, if  0 a \lt \max(a, b -a) is false and 0 b \lt \max(b, a -b) , is false, then 0 a \lt = \max(a b \cdot b, -a \cdot b)
• for all elements $a:R$ and $b:R$, if $a \lt b$, then $b \lt a$ is false.
• $0 \lt 1$
• for all elements $a:R$ and $b:R$, if $0 \lt a$ and $0 \lt b$, then $0 \lt a + b$
• for all elements $a:R$ and $b:R$, if $0 \lt a$ and $0 \lt b$, then $0 \lt a \cdot b$
• for all elements $a:R$ and $b:R$, if $0 \lt \max(a, -a)$ and $0 \lt \max(b, -b)$, then $0 \lt \max(a \cdot b, -a \cdot b)$

### Strictly ordered integral $\mathbb{Q}$-algebras

Given strictly ordered integral rings $R$ and $S$, a monotonic commutative ring homomorphism $h:R \to S$ is strictly monotonic if for all $a:R$ and $b:R$, $a \lt b$ implies $h(a) \lt h(b)$.

A strictly ordered integral ring $R$ is a strictly ordered integral $\mathbb{Q}$-algebra if there is a strictly monotonic commutative ring homomorphism $h:\mathbb{Q} \to R$.

### Archimedean ordered integral $\mathbb{Q}$-algebras

A strictly ordered integral $\mathbb{Q}$-algebra $A$ is an Archimedean ordered integral $\mathbb{Q}$-algebra if for all elements $a:A$ and $b:A$, if $a \lt b$, then there merely exists a rational number $q:\mathbb{Q}$ such that $a \lt h(q)$ and $h(q) \lt b$.

### Sequentially Cauchy complete Archimedean ordered integral $\mathbb{Q}$-algebra

Let $A$ be an Archimedean ordered integral $\mathbb{Q}$-algebra and let

$A_{+} \coloneqq \sum_{a:A} 0 \lt a$

be the positive elements in $A$. $A$ is sequentially Cauchy complete if every Cauchy sequence in $A$ converges:

$isCauchy(x) \coloneqq \forall \epsilon \in A_{+}. \exists N \in I. \forall i \in I. \forall j \in I. (i \geq N) \wedge (j \geq N) \wedge (\vert x_i - x_j \vert \lt \epsilon)$
$isLimit(x, l) \coloneqq \forall \epsilon \in A_{+}. \exists N \in I. \forall i \in I. (i \geq N) \to (\vert x_i - l \vert \lt \epsilon)$
$\forall x: \mathbb{N} \to A. isCauchy(x) \wedge \exists l \in A. isLimit(x, l)$