NOTE: ordered field

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

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

satisfying the following five axioms.

1. Commutative Law: If a,b∈Fa, b \in \mathbb{F}a,b∈F, then a+b=b+aa + b = b + aa+b=b+a and a⋅b=b⋅aa \cdot b = b \cdot aa⋅b=b⋅a
2. Distributive Law: If a,b,c∈Fa, b, c \in \mathbb{F}a,b,c∈F, then a⋅(b+c)=a⋅b+a⋅ca \cdot (b + c) = a \cdot b + a \cdot ca⋅(b+c)=a⋅b+a⋅c
3. Associative Law: If a,b,c∈Fa, b, c \in \mathbb{F}a,b,c∈F, then (a+b)+c=a+(b+c)(a + b) + c = a + (b + c)(a+b)+c=a+(b+c) and (a⋅b)⋅c=a⋅(b⋅c)(a \cdot b) \cdot c = a \cdot (b \cdot c)(a⋅b)⋅c=a⋅(b⋅c)
4. Identity Law: There are special elements 0,1∈F0, 1 \in \mathbb{F}0,1∈F, where a+0=aa + 0 = aa+0=a and a⋅1=aa \cdot 1 = aa⋅1=a for all a∈Fa \in \mathbb{F}a∈F
5. Inverse Law: For each a∈Fa \in \mathbb{F}a∈F, there is an element −a∈F-a \in \mathbb{F}−a∈F such that a+−a=0a + -a = 0a+−a=0. If a≠0a \neq 0a=0, then there is also an element a−1∈Fa^{-1} \in \mathbb{F}a−1∈F such that a⋅a−1=1a \cdot a^{-1} = 1a⋅a−1=1.

examples

• Natural numbers N\mathbb{N}N didn't form a field, it didn't satisfy a+(−a)=0a + (- a) = 0a+(−a)=0 since there has no −a∈N-a \in \mathbb{N}−a∈N for any a∈N>0a \in \mathbb{N} \gt 0a∈N>0
• Integers Z\mathbb{Z}Z is not a field, it didn't satisfy a⋅a−1=1a \cdot a^{-1} = 1a⋅a−1=1, for example, where a=2a = 2a=2, then a−1=12a^{-1} = \frac{1}{2}a−1=21​, but 12∉Z\frac{1}{2} \notin \mathbb{Z}21​∈/Z
• Q\mathbb{Q}Q form a field

positive set

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

Order Axiom

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

1. If a,b∈Pa, b \in Pa,b∈P, then a+b∈Pa + b \in Pa+b∈P and a⋅b∈Pa \cdot b \in Pa⋅b∈P
2. If a∈Fa \in \mathbb{F}a∈F and a≠0a \neq 0a=0, then a∈Pa \in Pa∈P or −a∈P-a \in P−a∈P

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.

Inequality in Ordered Field

If F\mathbb{F}F is an ordered field and a,b∈Fa, b \in \mathbb{F}a,b∈F, then we say a<ba < ba<b if b−a∈Pb - a \in Pb−a∈P, where a≤ba \leq ba≤b is a=ba = ba=b or a<ba < ba<b

Standard properties of inequalities

1. If a<ba < ba<b, then a+c<b+ca + c < b + ca+c<b+c
2. Transitivity: If a<ba < ba<b and b<cb < cb<c, then a<ca < ca<c
3. If a<ba < ba<b, then ac<bcac < bcac<bc if c>0c > 0c>0, else ac>bcac > bcac>bc
4. If a≠0a \neq 0a=0, then a2>0a^2 > 0a2>0

Proofs

First: If a<ba < ba<b, then a+c<b+ca + c < b + ca+c<b+c

By definition of inequality, we know the following conversion

And we already know b−a∈Pb - a \in Pb−a∈P

Q.E.D.

Second: If a<ba < ba<b and b<cb < cb<c, then a<ca < ca<c

By definition of inequality, a<ca < ca<c mean c−a∈Pc - a \in Pc−a∈P

We know c−b∈Pc - b \in Pc−b∈P and b−a∈Pb - a \in Pb−a∈P, by first rule of Order Axiom, we know

If a,b∈Pa, b \in Pa,b∈P, then a+b∈Pa + b \in Pa+b∈P

So c−a=(c−b)+(b−a)∈Pc - a = (c - b) + (b - a) \in Pc−a=(c−b)+(b−a)∈P

Q.E.D.

Third: If a<ba < ba<b, then ac<bcac < bcac<bc if c>0c > 0c>0, else ac>bcac > bcac>bc

By definition of inequality, we know

1. If c>0c > 0c>0

   ac<bc=bc−ac∈Pac < bc = bc - ac \in Pac<bc=bc−ac∈P

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

   1. c∈Pc \in Pc∈P
   2. b−a∈Pb - a \in Pb−a∈P since a<ba < ba<b

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

2. If c<0c < 0c<0

   ac>bc=ac−bc∈Pac > bc = ac - bc \in Pac>bc=ac−bc∈P

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

   1. c∉Pc \notin Pc∈/P, so −c∈P-c \in P−c∈P, by second rule of Order Axiom
   2. a−b∉Pa - b \notin Pa−b∈/P since a<ba < ba<b, so −(a−b)∈P-(a - b) \in P−(a−b)∈P

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

   And −c⋅−(a−b)∈P=c(a−b)∈P=ac−bc∈P=ac>bc-c \cdot -(a - b) \in P = c (a - b) \in P = ac - bc \in P = ac > bc−c⋅−(a−b)∈P=c(a−b)∈P=ac−bc∈P=ac>bc

Q.E.D.

Fourth: If a≠0a \neq 0a=0, then a2>0a^2 > 0a2>0

1. If a∈Pa \in Pa∈P, by first rule of Order Axiom, a⋅a∈Pa \cdot a \in Pa⋅a∈P
2. If a∉Pa \notin Pa∈/P, by first rule of Order Axiom, −a⋅−a∈P-a \cdot -a \in P−a⋅−a∈P, and −a⋅−a=a⋅a-a \cdot -a = a \cdot a−a⋅−a=a⋅a

Q.E.D.