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Suppose we impose the relation on the set of all real numbers that if the difference of two numbers is a rational number then they related. Then we are forming an equivalence relation on the set of all real numbers and each of the equivalence classes will be countably infinite. So I can say that the cardinality of the set of all equivalence classes is at least countable. But I cannot figure out its exact cardinality. I am not able to prove that it cannot be uncountable or neither can I get a bijection with $\mathbb{N}$. Let me know what you think about its cardinality. Any help will be truly appreciated.

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Working in set theory with the axiom of choice, your equivalence relation partitions the real line into some number $\kappa$ of countably infinite sets. It follows that the cardinality of the continuum is the product $\kappa\cdot\aleph_0$. But (using AC again), the product of two infinite cardinal numbers is just the larger of the two. So the cardinality of the continuum equals $\kappa$.

In the absence of the axiom of choice, the situation is more complicated. It is still true that the number of equivalence classes is at least the cardinal of the continuum, but that inequality could be strict. (If you think it's unimaginable that there could be strictly more than continuum pieces in a partition of the continuum, then you should either accept the axiom of choice or stretch your imagination.)

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It should be clear that each rational number has a corresponding equivalence class under this relation.

Given a fixed rational number $r$, the set of points $(x,y)$ in the plane satisfying $y = x - r$ is equal to the set of all points in the plane satisfying $x-y=r$ which is the equivalence class corresponding to $r$. Each equivalence class is now just a line in the plane, which is in bijection with the real numbers.

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