TITLE: Theorem 2.22 from RCA Rudin
QUESTION [6 upvotes]: I read this interesting result from Rudin's book but I would like to clarify some confusing moments.
As I understood $(\mathbb{R}^1, +)$ is group and $(\mathbb{Q}, +)$ is subgroup. 
He considers cosets $E_r=r+\mathbb{Q}$ for $r\in \mathbb{R}$. Since $E_r\neq \varnothing$ then by axiom of choice we can costruct set $E$ which contains only one element from each $E_r$.
Question 1: I am not sure that $(E+r)\cap (E+s)=\varnothing$ for $r,s\in \mathbb{Q}$, $r\neq s$. Let's consider cosets $\sqrt{2}+\mathbb{Q}$ and $1+\sqrt{2}+\mathbb{Q}$ and let $E$ contains $\sqrt{2}$ and $\sqrt{2}+1$ from the first and second cosets, respectively. But $E+1$ also contains $\sqrt{2}+1$. So we see that $E\cap (E+1)\neq \varnothing$. 
Question 2: Suppose that I have mistake in my previous question. Let $y$ and $z$ lie in the same coset of $Q$? Where's the contradiction?
Question 3: Why such $y$ exists? I am about point $(b)$.
Would be very thankful for help!

REPLY [4 votes]: Question 1: The cosets $\sqrt{2}+\mathbb{Q}$ and $1+\sqrt{2}+\mathbb{Q}$ are actually the same.  If $x=\sqrt{2}+q\in\sqrt{2}+\mathbb{Q}$, then $q-1$ is also rational, so $x=1+\sqrt{2}+(q-1)$ is also in $1+\sqrt{2}+\mathbb{Q}$, and similarly conversely.
Question 2: By assumption, $E$ contains exactly one point from each coset.  So if $y,z\in E$ are two distinct points which are in the same coset, this is a contradiction.
Question 3: Again, by assumption, $E$ contains exactly one point from each coset.  In particular, it contains exactly one point from the coset $x+\mathbb{Q}$, and this is the point we call $y$.