# Cardinality of “$x-y\in\Bbb Q$”-equivalence class of $1/\sqrt2$

For $x,y\in I:=[0,1]$ define the relation on $I$ as $x-y\in \Bbb Q$.

How big (using cardinal number) is the cardinality of the equivalence class $[1/\sqrt2]$?

I have tried to solve it by finding the equivalence class but I'm not sure about the method that I can solve the question.

Please, can you help me to solve it.

Hint: Every member of that equivalence class will be of the form $\frac1{2^{1/2}}+q$ for some rational $q$. (Why?) How many numbers of such form are there? This gives you an upper bound on the cardinality of the equivalence class (that is, there are at most that many elements in the equivalence class). You should be able to show that there are at least that many elements in the equivalence class, as well, by giving an explicit set of that many numbers of the appropriate form.
There are two ways of interpreting your question. One is, given $x\in I$, how large is $[x]$. This set is countably infinite, that is, of cardinality $\aleph_0=|\mathbb N|$. This is simply because $\mathbb Q$ is countable, so if $x-y\in\mathbb Q$, then $y$ can only be one of countably many numbers. And also, $\mathbb Q$ is dense, so there are rationals arbitrarily close to $0$ (both positive and negative), which means that there are reals within rational distance of $x$, both from above and from below, and as close to $x$ as desired. This shows that $[x]$ is infinite, but we already showed it is countable. In particular, $|[2^{-1/2}]|=\aleph_0$.
The other interpretation is what is the size of the collection of all equivalence classes. This is a trickier question. The collection has the same size as the reals, $\mathfrak c=|\mathbb R|$, but any proof of this fact uses some non-trivial fragment of the axiom of choice. See here, and the slides I link to in the answer there.