# NOTE: ordered field

Field is a nonempty set $$\mathbb{F}$$ with two binary operations

1. addition: $$+$$
2. multiplication: $$\cdot$$

satisfying the following five axioms.

1. Commutative Law: If $$a, b \in \mathbb{F}$$, then $$a + b = b + a$$ and $$a \cdot b = b \cdot a$$
2. Distributive Law: If $$a, b, c \in \mathbb{F}$$, then $$a \cdot (b + c) = a \cdot b + a \cdot c$$
3. Associative Law: If $$a, b, c \in \mathbb{F}$$, then $$(a + b) + c = a + (b + c)$$ and $$(a \cdot b) \cdot c = a \cdot (b \cdot c)$$
4. Identity Law: There are special elements $$0, 1 \in \mathbb{F}$$, where $$a + 0 = a$$ and $$a \cdot 1 = a$$ for all $$a \in \mathbb{F}$$
5. Inverse Law: For each $$a \in \mathbb{F}$$, there is an element $$-a \in \mathbb{F}$$ such that $$a + -a = 0$$. If $$a \neq 0$$, then there is also an element $$a^{-1} \in \mathbb{F}$$ such that $$a \cdot a^{-1} = 1$$.

### 0.1 examples

• Natural numbers $$\mathbb{N}$$ didn't form a field, it didn't satisfy $$a + (- a) = 0$$ since there has no $$-a \in \mathbb{N}$$ for any $$a \in \mathbb{N} \gt 0$$
• Integers $$\mathbb{Z}$$ is not a field, it didn't satisfy $$a \cdot a^{-1} = 1$$, for example, where $$a = 2$$, then $$a^{-1} = \frac{1}{2}$$, but $$\frac{1}{2} \notin \mathbb{Z}$$
• $$\mathbb{Q}$$ form a field

## 1 positive set

With an additional axiom, you will get an ordered field $$\mathbb{F}$$

#### 1.0.1 Order Axiom

There is a nonempty subset $$P \subseteq \mathbb{F}$$, called the positive elements, such that

1. If $$a, b \in P$$, then $$a + b \in P$$ and $$a \cdot b \in P$$
2. If $$a \in \mathbb{F}$$ and $$a \neq 0$$, then $$a \in P$$ or $$-a \in P$$

## 2 Metric properties

Since positive elements are all you need to define the absolute value, it give metric properties(distance function). You can define inequality formally now.

#### 2.0.1 Inequality in Ordered Field

If $$\mathbb{F}$$ is an ordered field and $$a, b \in \mathbb{F}$$, then we say $$a < b$$ if $$b - a \in P$$, where $$a \leq b$$ is $$a = b$$ or $$a < b$$

## 3 Standard properties of inequalities

1. If $$a < b$$, then $$a + c < b + c$$
2. Transitivity: If $$a < b$$ and $$b < c$$, then $$a < c$$
3. If $$a < b$$, then $$ac < bc$$ if $$c > 0$$, else $$ac > bc$$
4. If $$a \neq 0$$, then $$a^2 > 0$$

### 3.1 Proofs

#### 3.1.1 First: If $$a < b$$, then $$a + c < b + c$$

By definition of inequality, we know the following conversion

$a + c < b + c \\ \to (b+c) - (a+c) \in P \\ \to b + c - a - c = b - a$

And we already know $$b - a \in P$$

Q.E.D.

#### 3.1.2 Second: If $$a < b$$ and $$b < c$$, then $$a < c$$

By definition of inequality, $$a < c$$ mean $$c - a \in P$$

$c - a \\ \to (c - b) + (b - a)$

We know $$c - b \in P$$ and $$b - a \in P$$, by first rule of Order Axiom, we know

If $$a, b \in P$$, then $$a + b \in P$$

So $$c - a = (c - b) + (b - a) \in P$$

Q.E.D.

#### 3.1.3 Third: If $$a < b$$, then $$ac < bc$$ if $$c > 0$$, else $$ac > bc$$

By definition of inequality, we know

1. If $$c > 0$$

$ac < bc = bc - ac \in P$

In this case, $$bc - ac = c (b - a)$$, we know

1. $$c \in P$$
2. $$b - a \in P$$ since $$a < b$$

Thus, by first rule of Order Axiom, $$c (b - a) \in P = ac < bc$$

2. If $$c < 0$$

$ac > bc = ac - bc \in P$

In this case, $$ac - bc = c (a - b)$$, we know

1. $$c \notin P$$, so $$-c \in P$$, by second rule of Order Axiom
2. $$a - b \notin P$$ since $$a < b$$, so $$-(a - b) \in P$$

Thus, by first rule of Order Axiom, $$-c \cdot -(a - b) \in P$$

And $$-c \cdot -(a - b) \in P = c (a - b) \in P = ac - bc \in P = ac > bc$$

Q.E.D.

#### 3.1.4 Fourth: If $$a \neq 0$$, then $$a^2 > 0$$

1. If $$a \in P$$, by first rule of Order Axiom, $$a \cdot a \in P$$
2. If $$a \notin P$$, by first rule of Order Axiom, $$-a \cdot -a \in P$$, and $$-a \cdot -a = a \cdot a$$

Q.E.D.

Date: 2022-03-04 Fri 00:00