for every irrational number $i_1 \in [0, 1]$, there is only one irrational number $i_2 \in [0, 1]$ such that $i_1 - i_2 \in \Bbb Q$

Question

I'm currently working on a proof where the truth of the following statement would help me a lot:

Given the interval $[0, 1] \subset \Bbb R$, for every irrational number $i_1 \in [0, 1]$, there is only one irrational number $i_2 \in [0, 1]$ such that $i_1 - i_2 \in \Bbb Q$, and that is $i_2 = i_1$, so $i_1 - i_2$ would be identical with $0$ (which is rational, of course).

The background is that we defined an equivalence relation on $[0, 1]$, namely $x \sim y$ iff $x - y \in \Bbb Q$. Now I would like to show that every equivalence class that contains an irrational number only contains that one irrational number. Now, is this plausible? If so, is there a short way to prove it? If not, I guess that this wouldn't be the right approach they seek us to take, I guess.

Answer 1

No, the statement you are trying to prove is false. If $i_1$ is irrational, then $i_1-(i_1+q)$ is rational for any rational $q$, while $i_1+q$ is irrational.

Now for any $i_1\in[0, 1]$, we may find infinitely many $q\in\mathbb{Q}$ such that $i_1+q$ is also in $[0, 1]$ (exercise); so in fact there are infinitely many $i_2$s for every $i_1$.

Indeed, of the uncountably many $\sim$-equivalence classes, most (= all but countably many) contain only irrationals.

Answer 2

Counterexample:

• $i_1=\pi-3$
• $i_2=\pi-3.1$

Since $i_1 \neq i_2$ yet $i_1-i_2=0.1\in\mathbb{Q}$.