[1]: https://math.stackexchange.com/questions/2602418/proof-there...
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.
(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?!
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.
If it did, why are Even N and (Even + Odd N) 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|.
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|.
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."
> 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.
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.
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.
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.