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I was wondering if you could help me with this question?

"Consider the following relation on $\mathbb{R}$, the set of real numbers:

aRb $\iff_{def}$ a - b $\in \mathbb{Z}$

(a) Prove that this is an equivalence relation (done)

(b) Prove that the set $[\pi]_R$ = {x $\in \mathbb{R}$ | xR$\pi$} is in bijective correspondence with $\mathbb{Z}$"

I guess I have to find a function going from $\mathbb{Z}$ to $[\pi]_R$ and prove that it's bijective, but I have no idea of what this function would look like !

Thank you

Progranma
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1 Answers1

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What about $f:[\pi]_R \to \mathbb Z$ such that $x \mapsto x-\pi$?

Check that $f(x) \in \mathbb Z$ (it follows from the definition of $R$),
that $f$ is surjective (hint: given $k \in \mathbb Z$, what is $f(k+\pi)$?),
and that, for $x,y \in \mathbb R$ such that $xRy$, if $x-\pi = y -\pi$, then that $x=y$.

amrsa
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  • So I think it gives "Take x $\in [\pi]_R$ , then $x - \pi \in \mathbb{Z}$, and so $f(x) \in \mathbb{Z}.$ For the surjectivity : Take $y \in \mathbb{Z}$. Pose $x= y + \pi$. x is in$ [\pi]_R$ (since $y + \pi - \pi = y \in \mathbb{Z}$). Moreover $f(y+\pi) = y + \pi - \pi = y$ so there exist x so that $ y = f(x)$"? Thank you very much for the help ! – Progranma Dec 16 '19 at 00:50
  • Also, I always have trouble to understand what I have to check for the function to work (in more than the injectivity and the surjectivity) : is checking that f(x) is in the co-domain the only thing to check? Do we need also to make sure that the function we defined is functional (single-valued and total)? – Progranma Dec 16 '19 at 00:56
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    @Progranma Yes, you must verify all that, although we sometimes hand-wave those parts you mentioned since they follow, in this case, from the fact that the correspondence is defined in terms of know functions (here, subtraction; often, other functions, not necessary from arithmetic). But it is perhaps safer to do all those verifications. – amrsa Dec 16 '19 at 09:48