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I know the quotient set which is the integers modulo n.

The set of all congruence classes of the integers for a modulus n is called the ring of integers modulo n and they exist.

My question is whether the set of all congruence classes for the reals for a modulus m aka the ring of reals modulo m exists?

Can we do modulo arithmetic on the reals? Can we have a set of all congruence classes for the reals (aka a quotient set) for a modulus m and form a ring of reals modulo m?

  • $m\Bbb Z$ is a subgroup, but not an ideal of $\Bbb R$: $\Bbb R/m\Bbb Z$ is a group under addition, but not a ring. – Angina Seng Jul 01 '17 at 15:56
  • can you give a dumbed down non-algebraic theory answer assuming I don't know algebra? –  Jul 01 '17 at 16:02
  • are strictly irrationals and rationals modulo m existent? –  Jul 01 '17 at 20:14

3 Answers3

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Problem is, the reals form a field, and so the only ideals are the reals themselves and the trivial one.

Addendum

Expanding a bit, one can show that the equivalence classes of an equivalence relation $R$ on a ring form a ring with respect to the natural definitions (analogues of those for the integers) if and only if the equivalence relation is a congruence modulo an ideal, that is, $x R y$ iff $x - y \in I$, for some ideal $I$.

Andreas Caranti
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Consider the ring homomorphism from the real line to the unit circle $$f \colon \mathbb{R} \to S^1, \quad x \mapsto e^{2\pi i x/m}.$$ It is clearly surjective and its kernel is $m\mathbb{Z}$. So, for any integer $m$, the quotient $\mathbb{R}/ m\mathbb{Z}$ is an infinite, abelian group isomorphic to $S^1$.

On the other hand, $\mathbb{R}/m \mathbb{Z}$ is not a ring, because $m\mathbb{Z}$ is not an ideal of $\mathbb{R}$. If you want instead to consider the ideal generated by $m\mathbb{Z}$ this is the whole $\mathbb{R}$ (since a field has no proper ideals), so the corresponding quotient is the zero ring.

Francesco Polizzi
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Suppose $m$ is a nonzero real number (an integer, if you prefer).

You can talk about the ring of reals modulo $m$; all of the construction goes exactly the same. The problem is that the identity $$ x \equiv y \pmod m $$ holds for every pair of real numbers. For example, $x-y$ is always divisible by $m$: the quotient is $(x-y)/m$.

Consequently, there is only one congruence class, and so the reals modulo $m$ is the zero ring.

One could instead define an equivalence relation by insisting that $(x-y)/m$ is an integer. For rings, this equivalence relation is not a congruence relation, so it doesn't make sense to speak of the ring of reals modulo this relation.

However, it is a congruence relation for the additive abelian group of reals, so it does make sense to speak of the additive group of reals modulo relation, which could also reasonably go by the term "reals modulo $m$". This is much less trivial, and is more often what is meant should you ever hear the phrase. It is most commonly used with $m=1$ or $m=2\pi$.

  • $\mathbb{R}/\mathbb{Z}$ is isomorphic to $S^1$, and the same is true for $\mathbb{R}/m \mathbb{Z}$ for any integer $m$. – Francesco Polizzi Jul 01 '17 at 16:11
  • @FrancesncoPoizzi: Yes. While the sentence you write is true, $\mathbb{Z}$ is not the ideal generated by $m$, and thus has nothing to do with the ring of reals modulo $m$. –  Jul 01 '17 at 16:13
  • I would say that it makes no sense to talk of the "quotient ring" by something that is not an ideal. In this case, the "quotient group" is the only possible construction. – Francesco Polizzi Jul 01 '17 at 16:15
  • @FrancescoPolizzi: That's demonstrably false, because it is standard terminology to speak of a ring modulo an element of the ring. Such as in phrases like "the integers modulo $m$". –  Jul 01 '17 at 16:16
  • The standard terminology is the ring modulo the ideal generated by an element, but whatever. – Francesco Polizzi Jul 01 '17 at 16:17
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    @FrancescoPolizzi: Both are standard. (and if I wanted to be an extreme purist, I would point out that we should be talking about the quotient by congruence relations, not by ideals or subgroups or similar) –  Jul 01 '17 at 16:21
  • what rationals, p-adics, irrationals modulo m existent? –  Jul 01 '17 at 20:14