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I'm working with a strange and perhaps (mathematically) nonsensical realm.

I'd like to know if we can work with $\mathbb{R}/ \mathbb{Z}$.

For example, if we take $(\pi \bmod 3)$ we get $(\pi - 3)$.

I'm wondering if I always end up with the same equivalence class modulo 3 if I truncate the reals in this way.

In general, I'd like to know if I can work this way modulo any integer. I want to know if this will work for addition, subtraction and multiplication.

Matt Groff
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  • This might be of interest:

    http://crazyproject.wordpress.com/2010/01/04/the-reals-mod-the-integers-are-a-group/

     – user47805 Apr 08 '13 at 16:35
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    Addition and subtraction of reals will be well defined modulo integers. Multiplication will not be (in the following sense): $1.4$ and $0.4$ are in the same coset of the integers, but if you square them, the results won't be: $$1.4^2=1.96,\qquad 0.4^2=0.16$$ The same thing happens with other pairs of reals. When multiplication is out, so will division.

    In the language of abstract algebra: $\mathbb{Z}$ is a (normal) subgroup of the additive group of reals, but it is not an ideal of the ring of reals.

     – Jyrki Lahtonen Apr 08 '13 at 16:41
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    Another example, $\frac{1}{2} \sim \frac{3}{2}$, but $\frac{1}{2}\frac{1}{4} = \frac{1}{8} \not\sim \frac{3}{2}\frac{1}{4} = \frac{3}{8}$. – copper.hat Apr 08 '13 at 16:45
  • Thanks for the counterexamples. I will still try to find a way to get this to work... – Matt Groff Apr 08 '13 at 16:49
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    Note that working in $\mathbb{R}/\mathbb{Z}$ does not allow you to do "mod $3$". To do this, you should look at $\mathbb{R}/3\mathbb{Z}$ (which happens to be isomorphic to $\mathbb{R}/\mathbb{Z}$ as abelian groups). – Tobias Kildetoft Apr 08 '13 at 16:54
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    $\Bbb R/\Bbb Z$ is indeed a group under addition, isomorphic to the cirlce group $\Bbb T$. In particular, this means that it is closed under addition and subtraction. – A.P. Apr 08 '13 at 16:55
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    If it were closed under all ring operations (i.e. a ring congruence) then multiplying $\rm: 1\equiv 0:$ times $\rm: r\equiv r:$ yields $\rm:r \equiv 0.:$ Or, equivalently the ideal $\rm:I = { r,:, r\equiv 0}:$ contains $,1,$ so $\rm:I = \Bbb R.$ Thus, as a ring quotient $\rm:\Bbb R/\Bbb Z \cong 0.:$ But, as groups, the quotient is nontrivial. It is known as the circle or torus group. – Math Gems Apr 08 '13 at 17:04

1 Answers1

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$\mathbb Z$ is a normal subgroup of $\mathbb R$ (as a group under addition), so the quotient group $\mathbb R/\mathbb Z$ makes sense in the usual way in group theory.

$\mathbb Z$ is not an ideal of $\mathbb R$ (as a ring with addition and multiplication), so $\mathbb R/\mathbb Z$ does not make sense in the usual way in ring theory.

GEdgar
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    It does make sense as a ring, since the intended denotation of $\rm,\Bbb Z:$ is the ideal generated by $\rm,\Bbb Z.:$ Therefore $\rm:R/\Bbb Z:$ denotes $\rm:R/\Bbb ZR \cong R/R \cong 0,:$ just as $\rm:R/2:$ denotes $\rm:R/2R.:$ Generally, if one want to force some equalities to be true in an equational algebra (variety) such as groups, rings, etc. then one works in the algebra modulo the congruence generated by those equalities. – Math Gems Apr 08 '13 at 17:38