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Let $aRb $ iff $b - a$ is an integer.

$5 - 0$ is an integer, so $5 \in [0].$ In fact, $[0] = \mathbb Z$. Does it mean $\mathbb Z \in \mathbb R/R$?

$5.14159265359 - \pi$ is an integer, so $5.14159265359 \in [\pi]$ and $[\pi] \in \mathbb R/R$? Is that right?

Can we list all/most/main elements of $\mathbb R/R$ in this manner?

Vladimir Putin
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  • $5.14159265359 - \pi$ is not an integer. You're referring to $5.14159265359\ldots = \pi + 2$. – anomaly Sep 30 '15 at 14:07
  • Yes, $\Bbb Z \in \Bbb R / R$. In fact, we can identify $\Bbb R / R$ with $\Bbb S^1$, by showing that the quotient map $\Bbb R \to \Bbb R / R$ and the exponential map $\Bbb R \to \Bbb S^1$ defined by $x \mapsto e^{2 \pi i x}$ make the same identifications of elements of $\Bbb R$. – Travis Willse Sep 30 '15 at 14:08

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Yes, $\mathbb{Z} \in \mathbb{R} / R$, and $[\pi] \in \mathbb{R}/R$. Regarding the last question, $$\mathbb{R}/R = \{[x] \mid 0 \leq x < 1 \},$$ and for $0 \leq x,y < 1$ you have $[x] = [y]$ if and only if $x = y$, so this gives you a way to think about all the elements of $\mathbb{R}/R$.

fuglede
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  • It's perhaps worth saying, too, that the equivalence class $[x]$ is just $x + \Bbb Z$, where $x$ is any representative of the class. – Travis Willse Sep 30 '15 at 14:09
  • And that in case that notation is new to OP, this means $x + \mathbb{Z} = { x + n \mid n \in \mathbb{Z} }$. – fuglede Sep 30 '15 at 14:11