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I understand that irrational numbers are uncountable. I've seen the proof and it makes perfect sense. However, I came up with this (most likely false) proof that says that they're countable. Chances are, I made a mistake and the proof doesn't mean anything, but I just want to be sure. Here is the proof:

You can't find a solid chunk (range of numbers) that does not contain a rational number. Which means that irrational numbers are just points on the number line, not lines. Which means you just need to name all the points to count the irrationals.

What is the mistake here?

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    $\begingroup$ How do you propose to name them? You'll run out of names long before you run out of numbers. That's the essence of uncountability. (note that what I said is quite imprecise, but it should make you think). $\endgroup$
    – Deepak
    Dec 31, 2018 at 3:39
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    $\begingroup$ There are uncountable sets which do not contain any continuous segment of the number line. Another famous example is the Cantor set: en.wikipedia.org/wiki/Cantor_set $\endgroup$
    – Noble Mushtak
    Dec 31, 2018 at 3:40
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    $\begingroup$ Geometric intuition does not give much help when thinking about cardinality. Both the irrationals and the rationals "look like" sprinkled dust, but somehow the irrationals have "more dust" than the rationals. What you need to do is start thinking precisely about functions and their behavior. How exactly do you propose to build a surjection from the naturals to the irrationals, just by using the fact that the irrationals are totally disconnected? Along the samelines, a seemingly-promising observation is that the rationals are dense, so they should be "equivalent to" the irrationals somehow; $\endgroup$
    – Noah Schweber
    Dec 31, 2018 at 3:46
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    $\begingroup$ but that's also wrong. It's totally plausible at first, but all the density of the rationals tells you is that the set of irrationals can be approximated by (in some sense) a countable set, and you can't actually close the gap and get true countability. $\endgroup$
    – Noah Schweber
    Dec 31, 2018 at 3:47
  • $\begingroup$ It is a common misconception that every point of a totally disconnected space is an isolated point. Had this been true, then your proof can actually be made rigorous. However, counterexamples are abundant. $\endgroup$
    – shalop
    Dec 31, 2018 at 22:23

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In mathematical language, the claim that you can't find a solid chunk of the line (meaning an interval) that does not contain a rational is correct, and it is usually stated that the rationals are dense in the line.

I think your intuition is then that there is one irrational between "neighboring" rationals, which is the sense of saying that the irrationals are points on the line instead of lines. This is the problem. Between any two rationals there are an uncountable infinite number of irrationals and a countable infinite number of rationals. When you think of dense sets you need to banish the thought of "neighboring elements" from your mind. They do not exist because between any two there is another.

The difference between the rationals and the reals as a field is that the reals are complete. If you think of the interval $[0,1]$ you can make a binary expansion of a number. The expansion will converge, but without the completeness axiom you don't know that it will converge to a number in your field. A classic example is $\sqrt 2 -1$. If you make its binary expansion the approximation by increasing numbers of bits is a Cauchy sequence. The reals have an axiom that says it converges to a real, while the rationals do not. Now the classic Cantor proof applies to say the reals are not countable.

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Let's go through this sentence-by-sentence.

You can't find a solid chunk (range of numbers) that does not contain a rational number.

This is true: The rational numbers are dense in the reals, so every segment $[a, b]$ in the real numbers contains some rational number.

Which means that irrational numbers are just points on the number line, not lines.

This is half-true: Because of the first statement, there is no segment $[a, b]$ which is a subset of the irrationals, since all such segments contain at least one rational number. However, on the other hand, these irrational numbers are not just isolated points, meaning that in any segment $[a, b]$ which contains an irrational number $q$, there will always be another irrational number $p\neq q$ such that $p$ is also in the segment $[a, b]$.

Thus, it is wrong to think about the irrational numbers as just isolated points on the number line, because they aren't: Even though there is no segment consisting of entirely irrational numbers, there are still irrational numbers which are arbitrarily close to each other.

Now, you could also say the same thing about the rationals: Even though there is no segment consisting of entirely rational numbers, there are still rational numbers which are arbitrarily close to each other. Thus, whether or not a set can be thought of as a bunch of isolated points or not doesn't really have anything to do with if the set is countable or not. This gets us to the third statement:

Which means you just need to name all the points to count the irrationals.

This statement is false, and does not follow from the above two statements because of the argument I have presented above. Just because there is no continuous segment which is a subset of the irrationals does not automatically mean you can make a list all of the irrational numbers by just listing all of the points. This sentence here is pretty vague and does not highlight any way to actually create a bijection between the set of natural numbers and the set of irrational numbers, which is what it really means to be a "countable set." Thus, this proof is invalid because of this vague third statement, which does not actually construct a clear way to make a list of all the irrational numbers.

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This isn't really an answer, but it might help...

You seem to be thinking of the rationals and irrationals interleaved 1-to-1, so we could "name" the irrationals using the rational to their left (for irrational $j$, the rational just less than $j$). What happens when we do this for $\sqrt{2} = 1.414\dots$? With which rational do we name it?

We might try to sneak up on the name through $1$, $\frac{14}{10}$, $\frac{141}{100}$, $\frac{1414}{1000}$, $\dots$, but this never settles down to a single fraction. (That it does not settle down to a single fraction is what it means when we say $\sqrt{2}$ is irrational.) Between any of these rational numbers and $\sqrt{2}$, there are infinitely many rationals and infinitely many irrationals, and there is no way to pick the rational with which to "name" $\sqrt{2}$.

One way to see where you are misled is the phrase "Between any of these rational numbers and $\sqrt{2}$, there are infinitely many rationals and infinitely many irrationals". This says there are infinitely many rationals and infinitely many irrationals in every interval (chunk of the real numbers), but it cannot possibly tell you if there are more (with respect to cardinality) of one than the other -- "infinitely many" is not precise enough to do so.

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"You can't find a solid chunk (range of numbers) that does not contain a rational number." - True, this is just the statements that the rationals are dense in the reals (and also irrationals).

"Which means that irrational numbers are just points on the number line, not lines." - True

"Which means you just need to name all the points to count the irrationals." - Basically saying "we need to give a name to everything in this bucket". As @Deepak noted, this is the hard part. Note that whatever (finite!) "name" you come up with, it can be mapped to the naturals via eg dictionary sorting. This is basically the statement that the computable irrationals are still countable - whether "sqrt of pi" or "8th root of polynomial: ___". Beyond these are an uncountable sea of irrationals that cannot be computed/described/etc by any finite means. How do you plan to name these?

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Q: "What is the mistake here?" A: I believe the mistake is assuming that there is a countable number of segments.

In your 'proof', you say each segment (which contained only rational numbers) is terminated at an irrational number. Since there are an uncountable number of segments, there are an uncountable number of irrational numbers (each one being at the end of its 'companion' segment).

[EDIT: Removed trailing garbage from earlier edit]

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