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If the rationals are dense in the rationals, meaning there is a rational between every 2 irrationals, then there is essentially a rational number for every irrational.

If there is aleph irrationals, why isnt there aleph rationals if there exists a unique rational for every irrational number?

David Behar
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  • Aleph what? There are $\aleph_0$ rationals, but it famously independent of ZFC whether or not there are $\aleph_1,\aleph_2,\cdots$ many irrationals – FShrike Sep 15 '23 at 20:48
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    It's not like there's a unique rational between every two irrationals. Indeed there are infinitely many. And there are infinitely many irrationals between any two rationals. Hard to use density results like this to check cardinality. – lulu Sep 15 '23 at 20:48
  • The word "essentially" hides a lot. If there were a consistent way to pick a distinct rational number for every irrational number, this would indeed be a problem - but if we try, we find that we can't use the idea that there is a rational between every two irrationals to do that. – Misha Lavrov Sep 15 '23 at 20:50
  • For $A,B$ dense in totally ordered $C$, Density only allow you to conclude $|B|\leq|2^A|$, not $|B|\leq|A|$. – user10354138 Sep 15 '23 at 20:51

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meaning there is a rational between every 2 irrationals, then there is essentially a rational number for every irrational.

Perhaps, but this neglects that we only posit the existence of one rational, and certainly nothing about uniqueness. For instance, between $-\sqrt 2$ and $\sqrt 2$ you have the rational $0$. And between $-\sqrt 5$ and $\sqrt 5$ as well. Or between $-x$ and $x$, for any positive irrational $x$. (In fact, the density of $\mathbb{Q}$ ensures that there are infinitely many rationals between any two reals, and the irrationals satisfy this too, making matters a fair bit more complicated.)

This is certainly not enough to argue anything about cardinality.

Moreover, if the irrationals $\mathbb{R} \setminus \mathbb{Q}$ are indeed countable, then $$ \mathbb{R} = \mathbb{Q} \cup (\mathbb{R} \setminus \mathbb{Q}) $$ giving the reals as a finite union of disjoint countable sets, hence being countable. However, it is easy to prove that $\mathbb{R}$ is uncountable and $\mathbb{Q}$ is countable, so that means that $\mathbb{R} \setminus \mathbb{Q}$ must be uncountable.

PrincessEev
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Cantor proved that $\mathbb{Q}$ is countable and $\mathbb{R}$ is uncountable.
There are two important facts which I think you are aware of:

  1. Between any two distinct rationals, there are infinitely many irrationals.
  2. Between any two distinct irrationals, there are infinitely many rationals.


However the two infinities I mentioned above are not the same (I will omit further discussion about this, it is easily available online). One of them has a larger cardinality than the other.
So your argument "If the rationals are dense in the rationals, meaning there is a rational between every 2 irrationals, then there is essentially a rational number for every irrational" is flawed.
Also words like "aleph irrationals", "aleph rationals" doesn't make any sense.

Ajin Shaji Jose
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Elaborating on @lulu 's comment in hopes of clearing up your misunderstanding.

Suppose you have two different numbers of any kind. Then the whole open interval between them is just a scaled down copy of the interval $(0,1)$ so it contains lots of rationals (countably many) and even more irrationals.

Ethan Bolker
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