This one always breaks my brain, or at least my intuition. I understand that the reals infinitely outnumber the rationals, and the diagonal proof makes perfect sense to me. But the "between two reals there is always a rational" just beats my intuition like there's candy inside it.

Between any two reals, there’s a whole copy of R.

The original comment assumes its conclusion: it reduces to this smaller copy — then just jumps to the conclusion. It never actually tells us how to measure the ratio.

There's a simple constructive proof using high-school level thinking. ... Two different reals have different decimal expansions. Go out far enough that they differ. Since this is about intuition, let's just assume the larger one is positive and irrational, and thus has an infinitely long expansion. Since the truncation has a finitely long decimal expansion, it's rational. And it's between the two original reals. Q.E.D. ... A full proof for all cases can be built similarly.

It doesn't break my brain that there's a rational between any two reals; it breaks my brain that this doesn't imply equivalent sizes between the reals and the rationals.

> So there must be an equal number of them.

(Pretending to dig out my imaginary high school/freshman hat.)

If you said between every two reals there's one rational (and between every two rational there's one real), I'd assume you meant exactly one (i.e. in the conversational sense), and then there's an equal number of them. But you said at least one, so it could be two or three, so now I'm not sure. And if there are infinitely many of them... but infinity is not even a number, so who knows?!

“So there must be an equal number of them” is not correct.

Q is indeed dense in R, but firstly it’s very clear that there isn’t an equal number of them because rational numbers are a subset of the real numbers and there exists at least one irrational number (I pick “e”) that is in R but not in Q. So R must be at least bigger than Q.

Additionally you can’t say that between any two rationals there must be a real number because all rational numbers are also real numbers. You can say that Q is dense in R, but if you try to say R is dense in Q what you’re trying to say is “the bits of R that are not in Q are dense in the rest of R” which ends up with a bit of set logic to just be the same statement as the first one.

> Q is indeed dense in R, but firstly it’s very clear that there isn’t an equal number of them because rational numbers are a subset of the real numbers and there exists at least one irrational number (I pick “e”) that is in R but not in Q. So R must be at least bigger than Q.

This isn't a correct explanation, because I can use the same explanation to show that there are more integers than that there are even integers.

"it’s very clear that there isn’t an equal number of them because even numbers (let's call it E) are a subset of the natural numbers (let's call that N) and there exists at least one odd number (I pick 1) that is in N but not in E. So N must be at least bigger than E."

But as I explain in another thread, that doesn’t apply because E and N are in 1-to-1 correspondence which is not the case with Q and R

That’s what makes this statement incorrect:

> firstly it’s very clear that there isn’t an equal number of them because rational numbers are a subset of the real numbers and there exists at least one irrational number (I pick “e”) that is in R but not in Q

There are N not in E, but E and N have the same cardinality.

You have a second technical mistake as well:

> Additionally you can’t say that between any two rationals there must be a real number because all rational numbers are also real numbers.

They’re obviously referring to Q as a subset of R, and for any two elements of subset Q there is indeed a member of R not in Q.

Why does this work? Just because there is 1 number which is in R but not in Q doesn’t prove anything, does it?

If it did, why are Even N and (Even + Odd N) the same size?

Even N and (even + odd) N have the same size because there is a 1 to 1 correspondence between the sets. For every even number x in N I can pair it with x/2 which is also in N and that gives me all of N so N must be at least as large as even N. Likewise for x in N I can pair x with 2x and that gives me all the even N, so even N must be at least as large as N. So N and even N must have the same size.

However, while I can pair every rational number up with a real number, if I try to go the other way I find there is no rational number to pair with some (actually infinitely many) real numbers. I picked one, e, to show this. So the real numbers must be strictly larger than the rationals. Because the rationals are a strict subset of the reals it’s simpler than the Even/odd natural numbers because you don’t need the correspondence- every x in Q is in R but there exists at least one x in R not in Q, and no amount of correspondence shenanigans can get around that so |R|>|Q|.

> every x in Q is in R but there exists at least one x in R not in Q, and no amount of correspondence shenanigans can get around that so |R|>|Q|.

This also applies to evens and naturals: evens are a strict subset of naturals and there exists naturals which are not evens.

A counter example in the reals is (0,1) and R have the same cardinality, despite there being real numbers not in the interval (0,1).

You have the fact incorrect: for infinite sets, all B being a strict subset of A shows is that |A| >= |B|.

> So there must be an equal number of them

There definitely are not. Rationals are countable, reals are not.

It's tricky to prove things about the sizes of infinite sets, and "there's infinitely many of them interleaved with each other" doesn't do it, at least to my knowledge. I don't know of a way to prove equipotence except by creating a 1-1 relationship between the sets in question.

> I don't know of a way to prove equipotence except by creating a 1-1 relationship between the sets in question.

The usual definition of two sets having the same size (cardinality) is that they are in bijection. So ultimately, any proof that |A| = |B| will reduce to a "creating a 1-1 relationship between the sets in question", or at least showing that one exists.

Ok cool, I didn't remember that was the actual definition.

> I don't know of a way to prove equipotence except by creating a 1-1 relationship between the sets in question.

That relies on first defining it as "there is 1-to-1 mapping". But for sequences of numbers you could instead compare growth rates. Then the natural numbers and the even natural numbers wouldn't be the same size. Which is nice, because the latter sequence is a proper subset of the former. The former grow twice as fast, so you could say they are twice as large. But it only works for ordered sequences. Anyway, it all depends on definitions.

You can get infinitely many of such perplexing puzzles when you bring in infinity into our logic framework. The existence of these nonsensical puzzles prove that infinite doesn't fit into our logical framework, and one can get any number of astounding and amusing results they want, when you mix common mathematical logic and infinite.

You'd think so, but there aren't equal numbers of them.