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Please help me find an injection, $F$, from the set of real numbers into itself such that $F(x) - F(y)$ is an irrational number for any two distinct real numbers $x$ and $y$.

Thank you.

  • 5
    This is impossible: If $x = y$, you have $F(x) - F(y) = F(x) - F(x) = 0$, which is rational. Are there any additonal conditions?2012-05-16
  • 0
    Do you mean that $F(x)-F(y) \notin \mathbb{Q}$ for all distinct pairs of real numbers $x$ and $y$?2012-05-16
  • 5
    Hint: use the axiom of choice to select one element from each set of the form $x+\mathbb Q$. You'll have to find support for the claim that this gives you a set of cardinality continuum.2012-05-16
  • 2
    If the range of $F$ is *measurable* with positive measure, then the set of differences contains an interval about zero, so contains some rationals. So for your example you need either something non-measurable (as Leonid did) or measure zero. For the latter, think of some things like Cantor sets.2012-05-16
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    What do you mean by "find"? You can prove that such an $F$ must exist, in ZFC, but the word "find" might mean you want a definition. Given that it ultimately involves the axiom of choice, there is probably no "constructive" definition.2012-05-16
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    Yes I meant F(x) - F(y) is not rational....2012-05-22

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