0
$\begingroup$

$X$ is a subset of the reals such that for each distinct pair $x, y \in X$, their difference is irrational. Determine the elements of $X$.

My thinking is as follows:

If $X$ is empty or a singleton, this condition cannot be satisfied, so consider $X$ with at least two elements, $x$ and $y$.

For $x-y$ to be irrational, we have two cases: (i) $x$ and $y$ are irrational, (ii) only one of $x$ and $y$ is irrational (WLOG, assume $x$ is irrational).

If $X$ has three elements, we must have either (i) all three elements are distinct irrationals, or (ii) two elements are distinct rationals and the other is rational.

This is the also case for any cardinality of $X$, so in general we have: "$X$ with $n$ elements, we have $n-1$ distinct irrationals and only one rational.

So at its "largest", $X$ is all of the irrationals, plus one rational?

Can I then express $X$ as some disjoint union of sets?

Thanks.

$\endgroup$
8
  • $\begingroup$ $X$ is a subset of $\mathbb R$. There's no way to determine the elements of $X$ as there are uncountably infinitely possible sets $X$ could be. $X =\emptyset$ works. So does $X= \{3\}$ or $X= \{7,\pi, \sqrt 2\}$ etc....."If X is empty or a singleton, this condition cannot be satisfied" The condition will be vacuously satisfied. $\endgroup$
    – fleablood
    Oct 23, 2020 at 18:39
  • $\begingroup$ Yes, of course we can't explicitly state the possible elements of X, I just meant to verify that my general final statement about the elements of X is correct. Do you agree with it? $\endgroup$
    – Natasha
    Oct 23, 2020 at 18:41
  • 2
    $\begingroup$ @Natasha: No, it’s not correct: $\sqrt2$ and $1+\sqrt2$ are both irrational, and their difference is rational. $\endgroup$ Oct 23, 2020 at 18:42
  • 1
    $\begingroup$ "f X has three elements, we must have either (i) all three elements are distinct irrationals" If the set is $\{\sqrt 3, \sqrt 3+3, \sqrt 3+ 5\}$ this fails. "so at its "largest", X is all of the irrationals, plus one rational?" No. irrational $x$ and $r - x$ for rational $r$ fails. So I'm finding this a bit of an unanswerable question. $\endgroup$
    – fleablood
    Oct 23, 2020 at 18:43
  • 1
    $\begingroup$ You are correct that there is at most one rational. It shouldn't be too hard to show the cardinality can (but doesn't have to) be uncountable.... then again it's not a trivial proof. $\endgroup$
    – fleablood
    Oct 23, 2020 at 18:45

1 Answer 1

1
$\begingroup$

There are many such sets. The problem is, to produce a very large such set.

Define on ${\mathbb R}$ the equivalence relation $$x\sim y\qquad:\Leftrightarrow\qquad x-y\in{\mathbb Q}\ .$$ It is easy to see that there are uncountably many equivalence classes. Using the axiom of choice, choose a single element $x$ from each class, and form the set $X$ of all these $x$. (You then have a representant system of the quotient set ${\mathbb R}/{\mathbb Q}$.)

This set $X$ is large, and has the property you desire.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.