So I was walking on some subset of a spherical metric, when I came across an idea that I haven't yet been able to disprove or shake. It goes as follows:

Let the rational numbers be countable an lie dense in the real numbers. Then the irrational numbers must also lay dense in the reals.

**Proof:** Suppose $x,y$ real numbers and $x\neq y$, assume without loss of generality that $x<y$. Then there exists a rational number $q$ such that $x+\pi<q<y+\pi$, so $x<q-\pi<y$, which is irrational. $\blacksquare$

- Take any arbitrary 2 elements $x,y$ from the irrational numbers. Then because the denseness of the rationals, there lies a rational $c$ inbetween.
- Take any arbitrary 2 elements $p,q$ from the rational numbers. Then because of the denseness of the irrationals, there lies a rational $c$ inbetween.

On a first glance, this would suggest that the cardinality of the irrationals is equal to the cardinality of the rationals. This is obviously not true. The question is, why not? For this I have been unable to find a rigorous argument that does not use the fact that the reals are uncountable.

PS: I'm aware of: Intuitive explanation for how could there be "more" irrational numbers than rational? . This question does however not have any satisfactory answers because they are all either built on intuition (This question specifically looks at *rigorous* ways to prove this.) or use the uncountability of the reals (I already know this proof. I'm looking for an alternative.)