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I must explicitly construct a choice set on $\Bbb Q$.

The rational equivalence relation is as follows "Two numbers in a set are rationally equivalent provided their difference is rational".

I don't think this is possible, since any two rational numbers have rational difference, and so they cannot satisfy the requirement that any two points in the choice set have irrational difference.

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  • $\begingroup$ You're right that any two rational numbers have rational difference, but I'm pretty sure by "two numbers" it means two real numbers. $\endgroup$ – Dan Simon Feb 29 '16 at 6:24
  • $\begingroup$ Indeed that is the definition of the rational equivalence, but I am asked to construct it on $\Bbb Q$, the rationals. $\endgroup$ – Dr. John A Zoidberg Feb 29 '16 at 6:54
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    $\begingroup$ Perhaps you're overthinking it: unlike on $\Bbb R$, it's really easy to specify a choice set on $\Bbb Q$ :) $\endgroup$ – BrianO Feb 29 '16 at 9:41
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You're right that it's not possible to have two points in the choice set, since they would have to have irrational difference. But that doesn't mean the choice set can't exist: it just means it must have fewer than two points! Indeed, if $x\in\mathbb{Q}$ is any element, then the set $S=\{x\}$ is a choice set, since for each element $y\in\mathbb{Q}$, there exists a unique element of $S$ which is equivalent to $y$ (namely, $x$).

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