Sequence $(x_n)$ of irrationals in $[0,1]$ such that sets $\{y + x_n: y \in \mathbb{Q} \cap [0,1]\}$ are disjoint.

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Let $E = \mathbb{Q} \cap [0,1]$. Show there exists a sequence $(x_n)_{n \geq 1}$ in $[0,1]$ such that the sets $E + x_n = \{y + x_n: y \in E\}$ are disjoint.

The sequence must consist of irrationals with irrational differences between terms, but I am struggling to generalise this. Any help would be much appreciated.

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For an explicit construction you could use $x_n = q_n \sqrt{p_n}$ where $p_n$ denotes the $n$-th prime number and $q_n\in\Bbb Q$ such that $x_n\in [0,1]$ (you scale $\sqrt p_n$ down).

Now $x_n-x_m$ is indeed irrational for $m\neq n$ as otherwise $$ (x_n-x_m)^2 = q_n^2p_n - 2q_nq_m\sqrt{p_np_m}+p_mq_m^2 $$ would show that $\sqrt{p_np_m}$ is rational, what is not true.