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I know the real numbers and the rational numbers have different cardinalities and I understand the proofs that the rational numbers are countable and the real numbers are uncountable.

I also know that the rational numbers are dense in the real numbers. However I am having trouble understanding how/why the density of the rationals does not contradict the difference in cardinality between the rationals and the reals.

I can't seem to wrap my mind around this so any motivation would be extremely helpful!

orojack
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    I'm finding it hard to write a good answer without knowing more about why you are inclined to think there is a contradiction. – spaceisdarkgreen Apr 01 '23 at 03:15
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    The reason it doesn't supply a contradiction is that "I don't understand how this could be" does not qualify as a contradiction. There are many things in mathematics that boggle the mind Things for which you can study for decades and not really understand how they can be, but a contradiction requires being able to actually prove that a statement is true, and also prove that the same statement is false. Until you demonstrate those two proofs, it isn't a contradiction. – Paul Sinclair Apr 02 '23 at 02:02
  • Not sure if this will help, but it might help to get away from the "weakly defined" notion of dense as the property that between any two points of the ambient space (e.g. the reals) there is an element of the set (e.g. the rationals, the irrationals, the reals, etc.) asserted to be dense in that space. Instead, consider a more precisely defined notion of density in which we consider how many points in the set are between two given points in the ambient space. In this sense, the density of the rationals in the reals is different from the density of the reals in the reals. (continued) – Dave L. Renfro Apr 06 '23 at 18:04
  • The rationals (as a subset of the reals) have the property that between any two real numbers there exists a total of countably infinitely many rational numbers -- one sometimes says that the rationals are $\aleph_0$-dense in the reals. On the other hand, the reals (as a subset of the reals) have the property that between any two real numbers there exists a total of continnum many real numbers -- one sometimes says that the reals are $c$-dense in the reals. Also, the irrationals are $c$-dense in the reals, if you want an example where the set and space are different. (continued) – Dave L. Renfro Apr 06 '23 at 18:04
  • This idea of more precisely defining density notions can be extended to other ways (besides cardinality) of measuring how many points are between two given points, such as those based on Lebesgue measure and Hausdorff (or fractal) dimension, among other possibilities. For example, in this answer I point out that the set of discontinuities of the derivative of a function can be $c$-dense in the reals, (continued) – Dave L. Renfro Apr 06 '23 at 18:12
  • and even more than this -- there exist derivatives such that between any two points the set of discontinuities of the derivative has Hausdorff dimension $1$ (the maximum possible Hausdorff dimension for a subset of the reals). – Dave L. Renfro Apr 06 '23 at 18:12

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Salaam!
The set of rational numbers $\mathbb{Q}$ is countable because it has the same cardinality as the natural numbers $\mathbb{N}$.
The set of rational numbers $\mathbb{Q}$ is dense in the set of real numbers $\mathbb{R}$.
However, the cardinality of the rationals $\mathbb{Q}$ is not equal to the cardinality of the reals $\mathbb{R}$.
I can see why you find it difficult to make sense of it, but I would remind you that the set of irrational numbers is dense in the set of real numbers $\mathbb{R}$, too! Thus, casually speaking, it kind of makes the "size" of the reals $\mathbb{R}$ much larger than that of the rationals $\mathbb{Q}$, and so the cardinality of the reals $\mathbb{R}$ is much bigger than that of the rationals $\mathbb{Q}$.
I hope you find this make sense!

Nada Effat Ali
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