Show that the field of rational functions is not a complete ordered field

[Theorem.]

Homework Statement

Show that R(x) cannot be made into a complete ordered field, where R(x) is the field of rational functions.

Homework Equations

Definition of a complete ordered field: An ordered field O is called complete if supS exists for every non empty subset S of O that is bounded above

The Attempt at a Solution

$$R(x)= \frac{P(X)}{Q(X)}$$

where P(X) and Q(X) are polynomials and Q(x) is not equal to 0.

I have proven in the past that Q (the rational numbers) is not a complete ordered field using the subset {y$$\in$$R: y$$\geq$$0 and $$y^{2} \leq$$2}. Yet in the case of the rational functions as a field I am finding it very difficult.

I am fairly new to proofs (I switched from a non-proof based calc program to a honors proofed-base program) and I just don't know where to start. I understand the definition of a complete ordered field, a supremum and an infimum. I know that I have to prove in some way that a set of the rational functions is both bounded from above but can't have a supremum I just don't have a clue where I can start.

Please help me get started!

 

[Office_Shredder]

It might help to notice that the rational numbers are a subfield of the rational functions.

What kind of coefficients are your polynomials allowed to have here?

 

[Dick]

You haven't defined the ordering you are using on the rational functions. That might help. Hint: is 1<x?

 

[Dick]

It might help to notice that the rational numbers are a subfield of the rational functions.

What kind of coefficients are your polynomials allowed to have here?

You'll want to take them real. Otherwise, sure, having a subset Q is going to make it not complete. There is a 'usual' field order on the real rational functions, which I'd like Theorem. to define.

 

[Theorem.]

It might help to notice that the rational numbers are a subfield of the rational functions.

What kind of coefficients are your polynomials allowed to have here?

I thought about your first point but I am having difficulty using this to my advantage. I know that, say for example, $$\sqrt{2} is not \in Q$$
but $$f(x)=\sqrt{2}$$ is still a rational function.

As for the coefficients, nothing has been specified in the question

 

[Office_Shredder]

Ok yeah obviously the coefficients can be real numbers. For some reason I had a brain fart and thought R(x) was supposed to stand for rational functions over x (R meaning rational) but it's just notation for extending the real numbers by an element x.

You haven't defined the ordering you are using on the rational functions

I don't think you get to choose. The claim is that there is NO ordering that makes it a complete ordering.

The key is fairly similar to the rational number situation. Inside of R(x) is an element which does not have a square root

 

[Theorem.]

Ok yeah obviously the coefficients can be real numbers. For some reason I had a brain fart and thought R(x) was supposed to stand for rational functions over x (R meaning rational) but it's just notation for extending the real numbers by an element x.

R(x) is the field of rational functions in x

I don't think you get to choose. The claim is that there is NO ordering that makes it a complete ordering.

Yes I think you are correct here- I am trying to show that the field of rational functions in x is always not a complete ordered field.

The key is fairly similar to the rational number situation. Inside of R(x) is an element which does not have a square root

you mean which doesn't have a rational square root?
Then how would this proof be any different they proving that Q can not be completely ordered? The proof for Q was rather easy because it wasn't too hard to define a set in which the most obvious upper bound could not be the supremum- I just don't see how I could do that here : (

 

[Dick]

I don't think you get to choose. The claim is that there is NO ordering that makes it a complete ordering.

Perhaps you are right. That is what it says. I was just thinking that for someone who just "switched from a non-proof based calc program to a honors proofed-base program" that seems a little involved. So how many field orderings on the rational functions are there? I don't think the square root thing is going work though. Your ordering is defined only on rational functions. How can you say a rational function is less than a 'square root' if the square root isn't a rational function?

 

[Theorem.]

Perhaps you are right. That is what it says. I was just thinking that for someone who just "switched from a non-proof based calc program to a honors proofed-base program" that seems a little involved. So how many field orderings on the rational functions are there? I don't think the square root thing is going work though.

what do you mean by how many field orderings there are? I am a little confused by this.

Your ordering is defined only on rational functions. How can you say a rational function is less than a 'square root' if the square root isn't a rational function?

I also don't see how the square root method can work here. I really don't have any Ideas though- it's really hard for me because I am completely new to the proof style of thinking. I have spent a lot of time trying to understand the various methods but it doesn't come quickly.

 

[Office_Shredder]

We can prove that Q is not complete because the square root of 2 is not a rational number.

Similarly, look at all functions p(x)/q(x) such that $$\left( \frac{p(x)}{q(x)} \right)^2<x$$

If x is negative (i.e. smaller than 0 in the ordering) substitute -x in place for that. The key is that the square root of x (which would be the obvious supremum) is not a rational function.

I haven't worked out the full details but this sounds like a really good place to start

ASIDE: I'm actually not certain how many field orderings there are for R(x). There's only one really for the rationals, since the ordering for the integers is fixed immediately, and the ordering for all the elements between 0 and 1 is fixed by noting that you can put them over a common denominator, and then clear the denominator. There might not be a lot of choices for R(x) as well; but I don't know enough to state something like that off the top of my head

 

[Dick]

We can prove that Q is not complete because the square root of 2 is not a rational number.

Similarly, look at all functions p(x)/q(x) such that $$\left( \frac{p(x)}{q(x)} \right)^2<x$$

If x is negative (i.e. smaller than 0 in the ordering) substitute -x in place for that. The key is that the square root of x (which would be the obvious supremum) is not a rational function.

I haven't worked out the full details but this sounds like a really good place to start

ASIDE: I'm actually not certain how many field orderings there are for R(x). There's only one really for the rationals, since the ordering for the integers is fixed immediately, and the ordering for all the elements between 0 and 1 is fixed by noting that you can put them over a common denominator, and then clear the denominator. There might not be a lot of choices for R(x) as well; but I don't know enough to state something like that off the top of my head

Doesn't look so good to me. I guess I don't see how that tells me sup (p(x)/q(x))^2 doesn't exist just because it doesn't happen to be 'x'. Tough to say since we don't know what the ordering is. Here's the field ordering I know for the rational functions. Define a rational function (a*x^n+...)/(x^m+...) where a is nonzero to be 'positive' if a is positive. It's nonarchimedean.

 

[Theorem.]

We can prove that Q is not complete because the square root of 2 is not a rational number.

Similarly, look at all functions p(x)/q(x) such that $$\left( \frac{p(x)}{q(x)} \right)^2<x$$

If x is negative (i.e. smaller than 0 in the ordering) substitute -x in place for that. The key is that the square root of x (which would be the obvious supremum) is not a rational function.

I haven't worked out the full details but this sounds like a really good place to start

This makes sense to me.
could I define a set as follows:
$$S:={\frac{p(x)}{q(x)} \in R(x) : 0 \leq \frac{p(x)}{q(x)} and ( \frac{p(x)}{q(x)})^{2} \leq x }$$

 

[Dick]

This makes sense to me.
could I define a set as follows:
$$S:={\frac{p(x)}{q(x)} \in R(x) : 0 \leq \frac{p(x)}{q(x)} and ( \frac{p(x)}{q(x)})^{2} \leq x }$$

It would be really tough to define that if you don't know what $$\leq$$ means.

 

[Theorem.]

It would be really tough to define that if you don't know what $$\leq$$ means.

Oh you mean we'd have to first look at the properties of R(x) according to the order axioms?

 

[Dick]

Oh you mean we'd have to first look at the properties of R(x) according to the order axioms?

In some sense. I mean we had better figure out what '<=' means before we start using it in set definitions.

 

[Office_Shredder]

Define a rational function (a*x^n+...)/(x^m+...) where a is nonzero to be 'positive' if a is positive. It's nonarchimedean.

Actually I'm thinking this might be close to the only field ordering. I'm going to assume for a second that the real numbers must have their ordinary ordering induced by the rational function ordering (this is not obvious, and I have no proof as of yet)

Since the real numbers are complete, we can figure out that assuming x>0 (otherwise look at -x):

Either x is larger than every positive real number or x is smaller than every positive real number. EDIT: Actually, this is a lie, see end of post

Observe: A polynomial $$ax^n+...$$ has its sign determined only by a. We can do this by induction: if n=0 we're done., If it holds through n, for n+1:
The question is whether $$ax^{n+1}+bx^n+...$$ is positive or negative. If a and b have the same sign, by induction we know we're adding a positive $$ax^{n+1}$$ and positive $$bx^n+...$$ or two negatives, so we have the right sign. If the signs are opposite, WLOG let a>0 and b<0. I'll be swapping b for -b to make it clear where the negative term is.

By induction we know $$x^n$$ is necessarily larger than all of the smaller terms in the ... so $$ax^{n+1}-bx^n+...>ax^{n+1}-bx^n-x^n=ax^{n+1}-(b+1)x^n$$

If this is negative, then $$ax^{n+1}<(b+1)x^n\Rightarrow x<\frac{b+1}{a}$$ which is a contradiction.

It immediately follows that the only ordering is the one that you propose.

If x is smaller than every positive real number, 1/x is larger than all of them and we can re-write every rational function as a rational function of 1/x's by dividing the numerator and denominator by a large enough power of x. Then we get the same ordering only looking at rational functions of 1/x's.

All that's left to do then is show that the real numbers have their original ordering induced (this might not be true but my spidey senses are tingling).

To deal with the lie about the size of x: {m real number| m<x} is either bounded or unbounded. If it's unbounded, then since we have the normal ordering on the reals (again, assuming that for the moment) we have x is larger than all real numbers. If it's bounded, it has a supremum we will call y. Then either y-x or x-y is positive and smaller than every real positive number. Let's suppose x-y is positive for a minute. Then a polynomial f(x) in x can be turned into a polynomial in x-y by letting g(x)=f(x+y), then expanding g(x) into a polynomial in x, again, and then looking at g(x-y) (which equals f(x) and is a polynomial in x-y). then apply the analysis for x smaller than all positives only using polynomials in x-y

 

[Dick]

It's not hard to show the real numbers have their usual ordering. Every positive number is a square. That makes it 'positive' in the field ordering sense as well. In fact, looking at this in the clear light of morning you may not need to know the ordering in any terrible detail. Think about the set R (all real numbers). If you knew that was bounded you would be (almost) done. See if you can figure out Office_Shredder's proof of that. Or make your own, of course!

 

[Theorem.]

Thank you both for your help. I found out exactly how I could complete the proof- although it was getting there that still proved a task for me. I just need more practice writing proofs for that part though.
Thanks again