4

I'm trying to prove that an Archimedian field is a subfield of the real numbers, my plan is to use the fact that the rationals are dense within the field and their Dedekind completion is the real numbers.

However, I seem to run into some trouble writing a neat proof for the fact that the two properties of Archimedian fields are equivalent:

Let $F$ be an ordered field, the following are equivalent:
1. For every $x\in F$ there exists a natural number $n$ such that $x < n$
2. The rationals are dense in $F$

(In fact, I only need $(1)\implies (2)$ to be proved)

I have the essence of the proof, assume by contradiction some $a<b$ are two numbers that have no rational numbers in between, wlog we can assume $a,b\in (0,1)$, take some $n$ which is large enough and show some $\frac{k}{n}$ has to be between $a$ and $b$ otherwise there is some $x$ which is an upper bound of $\mathbb{N}$ in the field. However it seems that I can't write it elegantly.

Asaf Karagila
  • 393,674

3 Answers3

4

There is a positive integer $n$ greater than $(b-a)^{-1}$, and there is a least positive integer $k$ greater than $n\cdot a$. Thus $a\lt\frac{k}{n}$, and because $\frac{k-1}{n}\leq a$, $\frac{k}{n}=\frac{k-1}{n}+\frac{1}{n}\lt a +(b-a) = b$.

Jonas Meyer
  • 53,602
  • Beat me to it, slightly more elegantly... – Arturo Magidin Feb 10 '11 at 21:42
  • Much obliged :-) – Asaf Karagila Feb 10 '11 at 21:45
  • 1
    You can also use the ceiling function to get a fraction between $x-\epsilon$ and $x+\epsilon$. See Proposition 6 of a previous answer of mine – kahen Feb 10 '11 at 21:56
  • This is simpler viewed geometrically: by AP we can scale the interval $\rm\ I = (a,b)\ $ by an integer factor $\rm:n:$ to obtain $\rm:n:I = (na,nb)$ of length $> 1:.\ $ Thus by AP there is an integer $\rm:k\in n:I:$ hence a rational $\rm:k/n\in I:.\ $ QED $\ $ Essentially we exploit the fact that the problem is invariant under rational scaling symmetries to reduce it to a floor/ceiling computation, which exists the (generalized) division algorithm. See answer for further elaboration. – Bill Dubuque Feb 11 '11 at 03:18
  • @Bill: I agree that thinking of it geometrically lends more insight, and indeed the intuition for taking $n>(b-a)^{-1}$ in my answer is simply that the mesh size $\frac{1}{n}$ becomes smaller than the distance from $a$ to $b$, so that at least one of the "ruler marks" measured in $\frac{1}{n}^\text{th}$s must lie between $a$ and $b$. – Jonas Meyer Feb 11 '11 at 03:31
  • 1
    @Bill: Thank you for your time and answer, I find geometrical interpretations unhelpful in most cases because I don't have a strong geometrical thinking. Jonas' proof fit perfectly into my thinking process. – Asaf Karagila Feb 11 '11 at 06:59
  • @Asaf: As I said, the key to solving this problem simply is recognizing the invariance under rational scaling symmetries. Searching for key innate symmetries is one of the most ubiquitous mathematical problem solving techniques. It is at the heart of Galois theory, Lie symmetries for PDE's etc. It is crucial to understand these ideas in elementary contexts such as here else you will never see the forest for the trees in more complex scenarios. – Bill Dubuque Feb 11 '11 at 07:05
  • @Bill: As I said in my original question. I essentially knew the solution but just under unfortunate circumstances could not write it cleanly. As per the other concepts you mentioned such as PDE and Galois theory, I don't have much interest in those at the time being. – Asaf Karagila Feb 11 '11 at 07:17
  • @Asaf: Those were just a couple of uncountably many examples. As I said, symmetry-based problem solving is ubiquitous in mathematics. You would be robbing yourself of one of the most powerful (and beautiful) problem-solving techniques in all of mathematics if you purposely choose to ignore it. – Bill Dubuque Feb 11 '11 at 07:20
  • @Asaf: Since you essentially asked how to present the proof "elegantly", you shouldn't be surprised by replies like mine that try to point out the heart of the matter (considered from various viewpoints). – Bill Dubuque Feb 11 '11 at 07:27
  • @Bill: I have no qualms against new viewpoints, and more answers. – Asaf Karagila Feb 11 '11 at 07:32
4

For my take on this, see the proof of Proposition 108 in these notes.

Community
  • 1
Pete L. Clark
  • 97,892
4

This is quite trivial if viewed correctly. In a nutshell it boils down to the observation that we can employ rational-based rulers, i.e. we can change our unit of measurement on the number line from $1$ to any rational $\rm\:s\:$. More precisely, suppose we wish to find a rational in the real interval $\rm\ (a,b),\ a,b\in \mathbb R\:$. Simply choose some rational $\rm\:r < a\ $ and a positive rational step size $\rm\ s < b-a\:.\ $ Now, starting from $\rm\:r\:,\ $ keep taking steps of size $\rm\:s\:.\ $ By the Archimedean Property (AP) eventually you'll surpass $\rm\:b\:$, and taking a step back from the first such point necessarily lands you at a rational point in $\rm\ (a,b)\:,\ $ since the step size is smaller than the interval. Since the proof uses only the Archimedean property it works for any Archimedean field. Essentially it's simply employing the (Euclidean) division algorithm for rationals. Equivalently, instead of choosing the unit $1$ for our number line we instead choose the unit to be the rational number $\rm\:s\:$ that is our step size. In fact these ideas go all the way back to Euclid, who used it to (effectively) compute the gcd of rationals (greatest common measure of line segments).

It proves illuminating to examine the proof in JM's answer from this geometric viewpoint: by AP we can scale the interval $\rm\ I = (a,b)\ $ by an integer factor $\rm\:n\:$ to obtain $\rm\:n\:I = (na,nb)$ of length $> 1\:.\ $ Thus by AP there is an integer $\rm\:k\in n\:I\:$ hence a rational $\rm\:k/n\in I\:.\ $ QED $\ $ Notice how much more intuitive the proof is when presented in this geometric form. The key to its success is that the problem is invariant under rational scaling symmetries. Thus we can scale it to reduce it to what amounts to a trivial floor or ceiling computation (these functions exist by AP and induction).

Bill Dubuque
  • 272,048