Understanding Ordered Fields: A Beginner's Guide

[zli034]

Could anyone explain me the concept of Ordered Field? I have googled it, all came up are definitions I don't know how to handle. Numbers and calculations with statistical means I can understand fairly simple; but pure math has never worked for me.

Can anyone make the ordered field with simple understandable meanings?

 

[xxxx0xxxx]

I wouldn't sweat this to much, it means the elements of the field can be put in order according to some relation.

 

[micromass]

Hi zli034,

May I ask why you need this? Maybe my explanation will be better if I know this.

Basically, we want to generalize number system such as \mathbb{R} and \mathbb{Q}. In particular, there are some results on these systems that can be explained by means of a minimal number of axioms. Why do we do this? First, to get a better understanding of \mathbb{R} and \mathbb{Q}. Second, because there are a lot of other systems out there which also share the same properties and the theory of ordered fields will unify these systems.

What do we have on \mathbb{R} and \mathbb{Q}? Well, we have an addition. This addition + satisfies following properties:

• Associativity: a+(b+c)=(a+b)+c
• Neutral element: a+0=a=0+a
• Inverse element: a+(-a)=0=(-a)+a
• Commutativity: a+b=b+a

Anything else which shares the same properties is called an abelian group. The theory of abelian groups is extremely useful and it arises everywhere!

Now, what else do we have on \mathbb{R} and \mathbb{Q}? A multiplication of course? This satisfies:

• Associativity: a(bc)=(ab)c
• Neutral element: a1=a=1a
• Inverse element: aa^{-1}=1=a^{-1}a for all nonzero a.
• Commutativity: ab=ba
• Distributivity: a(b+c)=ab+ac

This is called a field. Other fields include the complex numbers. But there is something on \mathbb{R} and \mathbb{Q} that the complex numbers don't have: an order!

Basically, we have a relation \leq that satisfies

• Reflexivity: a\leq a
• Anti-symmetry: a\leq b~\text{and}~b\leq a implies a=b
• Transitivity: a\leq b~\text{and}~b\leq c implies a\leq c

This is what we call a partial order. But there order on \mathbb{R} and \mathbb{Q} satisfies some additional properties:

• Every two elements are comparable: for all a,b we have either a\leq b or b\leq a (or both).
• If a\leq b, then a+c\leq b+c
• If 0\leq a and 0\leq b, then 0\leq ab

A structure which satisfies all these axioms is called an ordered field...

 

[xxxx0xxxx]

Hi zli034,

May I ask why you need this? Maybe my explanation will be better if I know this.

Basically, we want to generalize number system such as \mathbb{R} and \mathbb{Q}. In particular, there are some results on these systems that can be explained by means of a minimal number of axioms. Why do we do this? First, to get a better understanding of \mathbb{R} and \mathbb{Q}. Second, because there are a lot of other systems out there which also share the same properties and the theory of ordered fields will unify these systems.

What do we have on \mathbb{R} and \mathbb{Q}? Well, we have an addition. This addition + satisfies following properties:

• Associativity: a+(b+c)=(a+b)+c
• Neutral element: a+0=a=0+a
• Inverse element: a+(-a)=0=(-a)+a
• Commutativity: a+b=b+a

Anything else which shares the same properties is called an abelian group. The theory of abelian groups is extremely useful and it arises everywhere!

Now, what else do we have on \mathbb{R} and \mathbb{Q}? A multiplication of course? This satisfies:

• Associativity: a(bc)=(ab)c
• Neutral element: a1=a=1a
• Inverse element: aa^{-1}=1=a^{-1}a for all nonzero a.
• Commutativity: ab=ba
• Distributivity: a(b+c)=ab+ac

This is called a field. Other fields include the complex numbers. But there is something on \mathbb{R} and \mathbb{Q} that the complex numbers don't have: an order!

Basically, we have a relation \leq that satisfies

• Reflexivity: a\leq a
• Anti-symmetry: a\leq b~\text{and}~b\leq a implies a=b
• Transitivity: a\leq b~\text{and}~b\leq c implies a\leq c

This is what we call a partial order. But there order on \mathbb{R} and \mathbb{Q} satisfies some additional properties:

• Every two elements are comparable: for all a,b we have either a\leq b or b\leq a (or both).
• If a\leq b, then a+c\leq b+c
• If 0\leq a and 0\leq b, then 0\leq ab

A structure which satisfies all these axioms is called an ordered field...

I believe that sqrt[a+bi*a-bi] forms a partial order on C with all complex numbers having the same magnitude parceled into and ordered set of equivalence classes.

It may not have a total order as do the reals.

 

[micromass]

Well, of course the complex numbers have an order. Every set has an order
The complex numbers also have a total order, as does every set.

But it doesn't have an order which makes it into an ordered field. I should have been more clear on that...

 

[zli034]

How cool is that!

I counted there are 4 addition properties, 5 multiplication properties.

Let's call the order, 10th property.

 

[zli034]

I would like to continue this Question & Answer because the new confusions.

From the book I'm reading that set Q is an Archimedean Ordered Field. However set R of real number will obey all the axioms for Archimedean Ordered Field together with one more axiom, called the Completeness Axiom, which is not satisfied by Q.

Q is a subset of R, why is it the thing assumed in R is not satisfied by Q?

 

[Hurkyl]

I believe that sqrt[a+bi*a-bi] forms a partial order on C with all complex numbers having the same magnitude parceled into and ordered set of equivalence classes.

It may not have a total order as do the reals.

Aside: after fixing your parentheses... that isn't a partial ordering; that is a (total) pre-ordering.

 

[JThompson]

I would like to continue this Question & Answer because the new confusions.

From the book I'm reading that set Q is an Archimedean Ordered Field. However set R of real number will obey all the axioms for Archimedean Ordered Field together with one more axiom, called the Completeness Axiom, which is not satisfied by Q.

Q is a subset of R, why is it the thing assumed in R is not satisfied by Q?

Firstly, a subset does not necessarily have all of the same properties as any of its supersets, so it's perfectly reasonable that a subset of R would not satisfy the Completeness Axiom.

Consider \{x\in \mathbb{Q}: x\leq \sqrt{2}\}. Is there a least upperbound for this set in Q? No, because Q is dense in R.