Why is the cardinality of irrationals greater than of rationals?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.