Why don't rational numbers fulfill the completeness axiom?

1. Aug 18, 2012

johnqwertyful

1. The problem statement, all variables and given/known data

Show that Q does not fulfill the completeness axiom.

2. Relevant equations

"Every non empty set of rational numbers that contains an upper bound contains a least upper bound" (show this is false)

3. The attempt at a solution

I've sat on this question for a few days, but I can't think of ANYTHING. I tried thinking of trying to show the opposite, to see where it would lead me. I couldn't start there. Any help JUST getting started?

I understand what the completeness axiom is. I understand what rational numbers are. I have no clue how to start.

2. Aug 18, 2012

SteveL27

Consider the set of rationals whose square is less than 2. Does that set have an LUB?

3. Aug 18, 2012

Curious3141

All you need is a counterexample. As Steve said, you can consider the set of rationals that are strictly less than the square root of 2; this set is clearly bounded by a rational number, e.g. 2, but you cannot construct an LUB.

If you prefer a constructive counterexample, consider the set of rationals constructed from the partial sums for the Taylor's expansion for e (the base of natural logs). Can you see why the set is rational and bounded, but has no LUB?

Last edited: Aug 18, 2012
4. Aug 18, 2012

johnqwertyful

I can't believe it was that simple, thanks so much. I've been thinking about it for a few days. It's so obvious now, I can't believe I didn't get it before.

5. Oct 23, 2013

dr.vj

Actually Steve didn't say that, he said "square less than 2", which is completely different from square root of 2, because the set of rationals doesn't contain square root of 2 and therefore you cannot use that to test completeness. But of course the equivalent square less than 2. Which doesn't contain irrationals per se.

6. Oct 23, 2013

R136a1

But $\sqrt{2}$ is a real number, and you can compare rational numbers with real numbers. So if you take

$$\{x\in \mathbb{Q}~\vert~x<\sqrt{2}\}$$

then that's a perfectly valid set in $\mathbb{Q}$. So Curious was right.

7. Oct 23, 2013

dr.vj

Yes, your'e right. sqrt of 2 belongs to real numbers and that is the exact reason, why we cannot use it for defining incompleteness of rational numbers. You can use, as Steve said "set of rationals whose square is less than 2". Because 2 belongs to rationals, but sqrt 2 doesn't.
It is this small difference that lead to the proof of completeness and hence we say real numbers are complete. Whereas rationals have infinite holes more than the infinite amount of rationals itself.

8. Oct 23, 2013

dr.vj

I think it is a big mistake to say that x less than sqrt(2), earlier you said x belongs to rationals, but sqrt(2) is not rational, it is real. You cannot just take some thing from real number system for defining something in rationals.

9. Oct 23, 2013

R136a1

The only thing you want is to find a subset of $\mathbb{Q}$ that has no supremum in $\mathbb{Q}$. And

$$\{x\in \mathbb{Q}~\vert~x<\sqrt{2}\}$$

is a perfectly valid subset of $\mathbb{Q}$. There is nobody that says that you can't use real numbers to define sets of rational numbers (unless the OP didn't define the real numbers yet).

And $\sqrt{2}$ is not in the set anyway, so you don't include any irrationals.

10. Oct 23, 2013

dr.vj

I have a video of Stanford mathematician Keith Devlin talking about this particular aspect of why you can't define in terms of real numbers for proving something about rational numbers.
He in particular worked out this proof without going anywhere near real numbers.

sqrt(2) is not in any subset of rationals. Absolutely not true!!

We have to use sets of rationals and the argument is about rationals and not reals.
Hence I conclude.

11. Oct 23, 2013

R136a1

Sure, I agree. I never said that, did I?

$$\{x\in \mathbb{Q}~\vert~x<\sqrt{2}\}$$

does not define a valid set of rational numbers.

12. Oct 23, 2013

Ray Vickson

I am suspicious of your statement of the completeness axiom; the problem occurs with your word 'contains'---as in "... contains an upper bound...". This seems to say that the upper bound itself is in the set we are discussing, and so is rational. (That would automatically eliminate such counterexamples as have already been shown to you, because those sets would not actually 'contain' an upper bound that is rational---so they do not contradict your statement). Statements of the completeness axiom that I have seen all say more-or-less the same thing: "every nonempty upper-bounded set of real numbers contains a least upper bound", or maybe "every bounded set of real numbers has a least upper bound". (Note that these do NOT claim that the original upper bound is in the set itself!) You want to show that if you substitute the word "rational" for "real" you no longer have a true statement. That is NOT quite you stated!

In fact, as stated the result is false: if we have a set $S \subset \mathbb{Q}$ which contains an upper bound $u \in \mathbb{Q}$, then $u$ IS a least upper-bound of $S$, because $u$ is an upper bound and no number smaller than $u$ is an upper bound (as that would exclude $u$).

Last edited: Oct 23, 2013
13. Oct 24, 2013

dr.vj

Hi
Since I am not a real expert and I am just starting to self-teach maths in my part-time, I asked another expert and his answer was:

The problem with using √2 in your definition of the set is due to the fact that you are working in the rationals, defining a set of rational numbers and, as we all know by now, √2 is not a rational so doesn't exist while we are in the rationals. If we were in the reals then √2 would be there, but in the rationals it is a hole in the number line. How can you use something that doesn't exist to define your set?
It is a bit like trying to imagine a four dimensional object in our three dimensional world, you are pretty sure it exists but know nothing about it, so would not use it to describe anything else - you stick to what you know in your world.

In my opinion, it is like defining Earthians by describing a property that Martians hold.

Vijay.

14. Oct 24, 2013

Office_Shredder

Staff Emeritus
The set is well-defined as long as you can use the existence of the real numbers. You have to be careful if you are in a real analysis course where you are going to construct the real numbers from the rationals, because in that context you cannot assume that sqrt(2) is an irrational number (or that it's a number at all!)

15. Oct 24, 2013

R136a1

As long as you defined the real numbers, the set is perfectly well-defined and a subset of the rationals. If you did not yet define the real numbers, then I agree there is a problem.

The condition $x<\sqrt{2}$ is perfectly well-defined for rational numbers $x$ because inequalities between real numbers are well-defined and because every rational number is real.

Likewise, the set

$$\{x\in \mathbb{Z}~\vert~x= 1/2\}$$

Is a perfectly well-defined subset of $\mathbb{Z}$ that happens to be the empty set.

16. Oct 24, 2013

dr.vj

I think you gave the perfect reasoning for this. As I have stated before about the Stanford mathematician, he was actually teaching real analysis, where we don't have prior knowledge of these numbers. Thanks.