# I don't understand why this axiom makes the difference between a completely ordered field and an ordered field

I've been reading about ordered fields and completely ordered fields, and I'm stuck on the difference between the rational and real numbers that makes the rational numbers an ordered field and the real numbers a completely ordered field. I see that the reals are complete, and the rationals aren't, but how does that correlate with one being a completely ordered field and the other just an ordered field?

My book, "A Friendly Introduction to Analysis" by Witold A. J. Kosmala, gives the following so-called 'completeness axiom' that separates a completely ordered field from an ordered one:

I understand what the axiom says, but I have a hard time grasping what it really means for the field that it applies to, and what it really means for the difference between a completely ordered and an ordered field. Furthermore, why do the rationals not satisfy this axiom? My thoughts were that a subset of the rationals also contain a supremum, namely the greatest rational number in that subset. What part don't I understand?

All help is appreciated.

• the set of rationals $\{x \in \mathbb{Q} | x^2<2\}$ has no supremum. There is no greatest rational number in this set, nor outside the set neither. – ziggurism Nov 14 '18 at 19:50
• When you say "I see that the reals are complete, and the rationals aren't", what is your understanding of "complete"? Is there some different definition you have? Or do you just mean "complete" in some vague intuitive sense? – Eric Wofsey Nov 14 '18 at 20:04
• @EricWofsey I understand it now, but just for clarification: yes, I meant that in the most intuitive way you could think of - probably not the best idea, but I had a hard time explaining my question properly. I meant it as it's written down in the book I'm using: "Intuitively, if we were to line up the rational numbers, there would be plenty of "holes" in them. Real numbers, on the other hand, would not have that property." – Benjamin Caris Nov 14 '18 at 20:28
• I suggest writing "complete ordered field", rather than "completely ordered field" (just a small remark). – Malkoun Dec 16 '18 at 17:08

In the real numbers, if $$(a_n)$$ is a bounded, monotone increasing sequence, i.e. $$a_1 \le a_2 \le a_3 \le \dots$$ then

$$\lim_{n} a_n = \sup\{a_n : n \in \mathbb N\} \tag{*}$$

Thus completeness of an ordered field implies that every bounded, monotone sequence has a limit.

Monotone sequences are important to the real numbers because if you take your favourite real number and look at it's decimal expansions. E.g.

$$3, 3.1, 3.14, 3.141, 3.1415, 3.14159, \dots$$

you get a monotone increasing sequence of rational numbers. So having every infinite decimal expansion exist (in this case $$\pi$$) follows from having limits for monotone sequences.

For a non-example in the rationals, consider the sequence

$$\frac{1}{1}, \frac{3}{2}, \frac{8}{5}, \frac{21}{13} \dots, \frac{F_{2n}}{F_{2n-1}},\dots$$

where the numerator and denominator are Fibbonacci numbers. This is monotone increasing and converges to the golden ratio, $$\frac{1 + \sqrt 5}{2}$$.

The proof of this fact is not important for this discussion so either accept that it's true or imagine some other increasing sequence of rational numbers that converge to an irrational number.

Anyways, with $$a_n = F_{2n}/F_{2n-1}$$ as above, the set $$\{a_1,a_2,a_3,\dots\}$$ has no maximum, because $$a_n < a_{n + 1}$$ so if we said the maximum was $$a_n$$ we'd have a problem because $$a_{n + 1}$$ is bigger. By $$(*)$$ (which you should try to prove) we have

$$\sup\{a_1,a_2,a_3,\dots\} = \frac{1 + \sqrt 5}{2}$$

So the set has no rational supremum but it does have a supremum in the reals.