How to understand countability of rational vs irrational numbers

[The question was edited 2024-01-09. Changes are marked with these brackets, mostly.]

[Edited again 2024-01-10 adding "final remarks".]

I'm struggling with getting my head around some aspects of transfinite numbers and sets in general. This question is about my struggle with the properties and countability of rational vs irrational numbers.

I think these are common facts, please correct me if I got anything wrong:

1. The rational as well as irrational numbers are "dense" on the real number line, meaning roughly that the resolution is unlimited, and there are no gaps of size >0.

2. The rational as well as irrational numbers are an "incomplete" total order, meaning roughly that there are indeed gaps (of size 0), and ... (yada yada existence of supremums).

3. The (set of) rationals is countable while the irrationals aren't. (I have not seen a direct proof of the latter, only that it follows from reals being proved uncountable while rationals are proved countable.) [Update 2024-01-09: the remark has been answered in a comment.]

• (a.) As a consequence of density; between every pair of distinct irrational numbers there is at least one (and actually infinitely many) rational numbers, no matter how small the distance as long as >0.

• (b.) Symmetrically, exactly the same for irrationals between rationals.

1. The Dedekind cut is one way (of several) to construct all the reals from the rationals. Each real number $$x$$ is uniquely associated with the set of all rational numbers that are strictly smaller than $$x$$.

"5b." [Probably irrelevant but included for symmetry] Except that it may be without purpose, is there anything to stop me from defining "Hallström cuts" in exactly the same way as Dedekind cuts, but instead using the irrational numbers as a base for modeling all the reals.

1. [Probably not relevant here] The rational numbers are closed under basic arithmetic ops (add and mul). The irrationals aren't.
• [Sidenote] I slightly wonder if there exists subsets of the irr:s that are closed under such ops. [Update 2024-01-09: this sidenote has been answered in a comment.]

The question

My main headache: given (4) above, how can it be possible that one type has larger cardinality than the other (acc to (3) above)? They have such striking symmetry with each other. Intuitively and informally, they should be sort of perfectly interleaved with each other. (I know intuition is not necessarily a reliable guide here. I also realise that assuming perfect interleaving is not a consistent model.)

Put another way: where could all the additional irrationals be hiding from being forced into 1-1 correspondence with the rationals? If there were a hideout anywhere with several (distinguishable) irrs in a row without any rats in between them, it would be in breach of (4) and would also invalidate the D. cut.

Another argument: The D. cut is not using the power set of the set of rats, only an interval. The only difference between different cuts is the upper bound of the interval. ([Edit 2024-01-09] The upper bound is a real number here.) When using only the domain of the rational numbers, how can there exist more such distinct intervals (uncountable) than there are rats (countable)??

Here is a new attempt to quantify my mental issue with the countable rational numbers, by focusing on the Dedekind cuts.

Construct the Dedekind cuts for all rational numbers, and only those to begin with. Now for every rat num $$r$$ there is a uniquely associated set of rat nums, containing the open interval $$(-\infty, r)$$. For two rat nums $$r_2>r_1$$, $$D(r_2)=D(r_1)\cup[r_1, r_2)$$, so the sets only differ "on the top", so to say.

Now, at least intuitively, in constructing all these intervals $$D(r)$$, all available rats have been consumed already. How could there exist any more intervals of that same kind, containing rat nums exclusively.

If there was the largest rat num in the open interval (I know there isn't), then every rat num $$r$$ would clearly be busy being the top member of exactly one $$D(s)$$.

If there was such a thing as "the next rat num" (I know there isn't), it would be clear that the set of all $$D(r)$$ so far equals the set of all possible $$D(x)$$ at all (for any real number $$x$$).

(The outcome of this: I'm trying to claim that the set of all possible D cuts, mentally, cannot exceed the set of all possible rat nums $$r$$.)

Given the argument in the prev paragraphs, how can now unique D. cuts be added for the irrational numbers when they were already exhausted by the rational mumbers? There would be no "free" rat nums left that can be used to uniquely create a new cut. It seems to require that there are more rat nums than there are rat nums. (Sic.)

Yet another angle: in how many different ways can one create an interval containing rational numbers only? (Please note that the interval can be bounded by real numbers but contains only rationals.) Arguably only countably many. Imagine adding one rat at a time to the top of the interval, ignoring the fact that the rats cannot really be enumerated in order like that. Any enumeration of rats can only go on for a countable number of times.

[2024-01-10 "Final remarks" by the author]

Thank you for helpful comments and answers. It may look like it but I have not been trying to prove that something is wrong, because I can't. I have tried to reason about my main source of "headache" in this subject, and was seeking some guidance about it.

For the record, I already know that for the time being I must accept (reluctantly) such parts of math that I am intuitively at odds with (at least until I develop ability to prove inconsistencies).

• There is no such thing as 'the upper bound of the interval', not in the rational numbers, anyway. The least upper bound can be one of those 'gaps of size zero' you were talking about. Commented Jan 5 at 7:44
• The countability is not so easy to grasp intuitively. The algebraic irrational number still form a countable set ! Countable means that there is a method to list all elements in such a way that it is guaranteed that every element will eventually occur. That this is not necessarily the case , even in the countable case : Consider the method to list first the even numbers , then the odd ones. What is wrong with this ? Correct, $1$ will never be reached. Commented Jan 5 at 7:44
• A similar question: math.stackexchange.com/questions/4743188
– Karl
Commented Jan 5 at 8:10
• Answer to your Sidenote: The set of all expressions of the form $\sum_{i=1}^n a_i \pi^i$ with $n\geq 1$, $a_1, a_2,\ldots, a_n$ integer $\geq0$, and $a_n>0$ consists of irrational numbers only. And it is closed unter addition and multiplication. Commented Jan 5 at 10:14
• I don't know exactly what you mean by a "direct proof", but the standard Cantor diagonalization argument used to show that the reals are uncountable work: given a countable list of irrational numbers, you may construct an irrational number different from every number of the list by a diagonal argument (modify the original argument to make sure you are constructing a non-repeating decimal. E.g. use digits 1 or 2 for the first digit, use digits 8 and 9 for the following two digits, then 1/2 again for the next three digits, then 8/9 again for the next four digits, and so on...) Commented Jan 6 at 3:08

The simple short answer is that infinity is weird and does not fit well with intuition which was developed in the real world of finite things.

An early challenge to intuition is whether or not there is a single infinity. Is the set of integers greater than or equal to zero bigger than the set of integers strictly greater than zero because it has one more element? Is the set of all integers twice as big because it has the negatives as well?

We can only start to answer this after we have defined what we mean for two, possibly infinite sets, to have the same size. The most common definition is that two sets have are the same size if a bijection (one-to-one and onto map) can be found between them. We quickly run into another challenge. With familiar finite sets, if we find a map which is one-to-one but not onto then we would deduce that one is smaller than the other. So, this suggests that the examples which I just gave are different sizes. However, we find that the existence of a one-to-one but not onto map does not preclude a bijection. A bijection can be found between all of those sets so they are the same size. We find that further intuitively bigger sets are also the same size e.g. the rational numbers and the algebraic numbers. The algebraic numbers are those which are the roots of polynomials with integer (or rational) coefficients. They include some irrationals, e.g. $$\sqrt 2$$ but not others, e.g. $$\pi$$.

We usually say "cardinality" rather than "size" which mainly emphasizes that we are using this interpretation. I will stick to "size" as we are talking about intuition.

We might start to develop the intuition that all infinite sets are the same size but now we find that the real numbers are strictly bigger. A bijection between the natural numbers and the real numbers is not possible.

How much bigger? Is it the next biggest set or is there something with an intermediate size? Now it gets really weird. Not only do we not know but we can prove that we cannot prove that there is or is not a set with an intermediate size. Look up the Continuum Hypothesis if you want to know more.

We know that the reals are not the biggest set. For any set, we can construct a bigger one so there is no biggest one.

Now back to some of your points. It is indeed counter-intuitive that all of these are true at once:

1. Between any rationals there is an irrational
2. Between any irrationals there is a rational
3. There are more irrationals than rationals

1 and 2 suggest that they alternate along the real line and hence there are equally many but 3 contradicts this. Along with the Continuum Hypothesis, I think that you just have to work on understanding the proofs and then learn to accept it. I still find it weird but not as weird as the Continuum Hypothesis.

The problem with Hallström cuts is that you don't have the irrationals until you have the reals. The irrationals are just the reals which are not rational. So, you would have to obtain the irrationals somehow else before you could start work on Hallström cuts. We know how to construct the rationals without the reals so we can use them as a starting point. You could perform cuts on the algebraic numbers and that would work because we could construct them without the reals. However, you would just get something isomorphic to the usual reals.

It is often said, without good evidence, that contemplating infinity drove Cantor mad. Even if not true, the existence of the story shows that struggling to understand infinity is common.

• +1 for the first and last paragraphs, which I borrowed in my answer to stress what I think the OP is struggling with. Commented Jan 10 at 16:06
• +1 for pretty much all of it. It was a long answer but who am I to say that. I (OP) especially appreciate the commentary in the middle about the counter-intuitive facts. I reckon this is the best answer that I'm gonna get, and I'm going to accept it. For the record, I already know that I for the time being must accept (reluctantly) such parts of math that I am intuitively at odds with. Commented Jan 10 at 21:44
• Actually, I considered saying more. There is a great book: The Road to Reality by Roger Penrose. He looks at many weird areas of maths, e.g. complex numbers, and shows uses for them in physics. He seems surprised when it comes to infinite cardinals that they have not (yet) proved useful in physics. Also search for "unreasonable effectiveness of mathematics", this will lead you to an article by Eugene Wigner expressing surprise at how useful maths is in physics. Together. they are saying that maths is very useful but not a perfect description of reality. Commented Jan 10 at 22:46

The essential answer to your question is that you can't trust your intuition about infinity. All you can do is follow the formal definitions to their logical conclusions and then train your intuition to bring it into accord with those conclusions. Until you do that, none of the carefully written answers here is likely to satisfy you. The first and last paragraphs from @badjohn say it well:

The simple short answer is that infinity is weird and does not fit well with intuition which was developed in the real world of finite things. ...

It is often said, without good evidence, that contemplating infinity drove Cantor mad. Even if not true, the existence of the story shows that struggling to understand infinity is common.

Given the argument in the prev paragraphs, how can now unique D. cuts be added for the irrational numbers when they were already exhausted by the rational mumbers? It seems to require that there are more rat nums than there are rat nums.

There are Dedekind cuts which are not of the form $$D(r)$$ for any rational number $$r$$. For example, the Dedekind cut corresponding to the interval $$(-\infty, \sqrt{2})$$ (i.e., the set of all rational numbers less than $$\sqrt{2}$$) is not equal to $$D(r)$$ for any rational $$r$$. You can replace $$\sqrt{2}$$ by any irrational number.

• Thanks, but this is exactly the argument I'm feebly trying to make. It is "not reasonable" that $D(\sqrt{2})\not =D(r)$ for all rational $r$ because there would be no free rat num left to put in its set to make it unique. Between any $r_1$ and $D(\sqrt{2})$ there are always many $r_2$ that already have their cut, and the argument is that $D(\sqrt{2})$ cannot possibly be different from all of those. This all informal, I know. But I think there is something deeply fishy about the D cuts. It is always dodged by saying "there is no rat $r$ closest to irr $i$". Commented Jan 9 at 21:36
• You've given the correct reason "there is no rat $r$ closest to irr $i$" so why do you still think it is fishy?
– Ted
Commented Jan 9 at 21:59
• To me it is like a fallacy by denying access to the evidence, much like the Heisenberg uncertainty or the Schrödinger cat. We can never go deep enough to really check whether the Dedekind claims hold at the smallest scale (because there is no smallest scale). Commented Jan 9 at 22:14
• Please note: I have no problem with the D cuts as long as we limit it to comparing two arbitrary reals. That always works. My issue is with the idea that one could uniquely construct all those cuts at once for every real (and implicitly collect them in a set). Commented Jan 9 at 22:18
• You keep making arguments relying on things that you know are wrong, like "if we ever allowed a rat to exist immediately above sqrt(2)", "ignoring the fact that the rats cannot really be enumerated in order like that", etc. Why do you do that?
– Ted
Commented Jan 10 at 0:02

To be honest this post is giving me headache so I'm only gonna address the new additions today (9th Jan 2023) in the end. I'll use $$r$$ to consistently stand for some rational number.

Unfortunately, all arguments you've made seem to me to be begging the question: you start by assuming all Dedekind cuts are $$D(r)$$, and conclude there can be no more other Dedekind cuts. I think you need to let go of this premise, and start thinking "maybe there are Dedekind cuts that are not $$D(r)$$". Until then, we are going nowhere.

You can also take a look at this post by Eric Wofsey. It gives a proof that a countable dense totally ordered set cannot be Dedekind complete. In it, the author assumes $$\mathbb{R}$$ is countable and arrives at a contradiction using only the dense total order on $$\mathbb{R}$$. Hopefully, since this proof relies purely on countability and dense total order, with no arithmetics, algebra, or calculus involved, it can convince you better. If it doesn't, I really doubt if you will ever be convinced.

• Thank you for your link. However it appears to be another proof of the uncountability of the reals, which was not quite the issue. I have read and (reluctantly) accepted a couple of such proofs. Unfortunately I fail to see the connection to my rambling about D cuts, and I can only guess the meaning of "D. complete", but nevermind. Besides I think you missed the point of said rambling, but you're not alone and who could blame you. I do understand the general point you are making that I need to relax and accept some things as they are (until I develop ability to prove otherwise). Commented Jan 10 at 22:04
• @AndersH The intent of the proof I linked is, as you said, different, but it would be helpful to this issue if you replace $\mathbb{R}$ that is assumed countable with the actual countable $\mathbb{Q}$. A total order being Dedekind complete just means every Dedekind cut is $D(x)$ for some $x$ in it. Thus, the proof basically says because $\mathbb{Q}$ is countable, there will always be a Dedekind cut (in fact, uncountably many) that is not $D(r)$, which I was hoping can show you the very premise you repeatedly rely on, that every Dedekind cut must be $D(r)$, is false. Commented Jan 10 at 22:31