# The differences between $\mathbb{R}/ \mathbb{Z}$ and $\mathbb{R}$

The cosets of $\mathbb{Z}$ in $\mathbb{R}$ are all sets of the form $a+\mathbb{Z}$, with $0 ≤ a < 1$ a real number. Adding such cosets is done by adding the corresponding real numbers, and subtracting 1 if the result is greater than or equal to 1. -- Examples of Quotient Group, Wiki

I cannot figure out the differences between $\mathbb{R}/ \mathbb{Z}$ and $\mathbb{R}$. Besides, "subtracting 1 if the result is greater than or equal to 1", what does "the result" mean here? Why do we need to subtract 1? I was wondering what is the background of $\mathbb{R}/ \mathbb{Z}$.

• There is a non-zero element in the quotient which is its own inverse. – Thomas Andrews Feb 2 '12 at 4:51
• So many high quality answers I don't feel the need to post another one!! – user38268 Feb 2 '12 at 11:01
• A useful property is the following: If $G$ is a finite subgroup of $\mathbb{R}/\mathbb{Z}$ then $G$ is cyclic. The easiest way to see this is through the correspondence provided by Dylan below. – AnonymousCoward Feb 2 '12 at 21:02
• Now $\mathbb{R}/\mathbb{Z}$ has non-trivial finite subgroups, but how many does $\mathbb{R}$ have? – AnonymousCoward Feb 2 '12 at 21:05
• $\mathbb{R}^n/\mathbb{Z}^n$ is an $n-$torus! – Maxime May 15 '12 at 21:57

The elements of $\mathbb{R}$ are real numbers. The elements of $\mathbb{R}/\mathbb{Z}$ are sets of real numbers, pairwise disjoint, where each set contains reals that differ from each other by integer distances, and the set contains all reals that differ from the reals in the set by integer distances.

So one of the elements of $\mathbb{R}/\mathbb{Z}$ is $$\{ \ldots, -2, -1, 0, 1, 2, 3, \ldots\}$$ Another element of $\mathbb{R}/\mathbb{Z}$ is $$\{\ldots, \pi-3, \pi-2, \pi-1, \pi, \pi+1, \pi+2,\ldots\}$$ and so on. Every real number is in one and only of the elements of $\mathbb{R}/\mathbb{Z}$ (which, remember, are sets of real numbers).

(The elements of $\mathbb{R}/\mathbb{Z}$ are the equivalence classes of real numbers under the equivalence relation "$x\sim y$ if and only if $x-y\in\mathbb{Z}$").

Now, the set $\mathbb{R}/\mathbb{Z}$ can be made into a group as well; that is, we can define an "addition of classes". One way to define this "addition of classes" is to first give a "name" to each element of $\mathbb{R}/\mathbb{Z}$. Since every one of the sets contains one, and only one, real number in $[0,1)$, we will represent the set that contains the real number $r\in[0,1)$ by writing $[r]$. So the first set I wrote above is called $$, the second set I wrote above is called $[\pi-3]$, and so on.

(Added. Given a real number, how can you tell what element of $\mathbb{R}/\mathbb{Z}$ it is in? Remember that $\lfloor x\rfloor$ is defined to be the largest integer $n$ such that $n\leq x$. The "fractional part of $x$" is defined to be $x-\lfloor x\rfloor$. It is not hard to check that $x\in[x-\lfloor x\rfloor]$.)

Now that every set has a name, we define addition of classes. The way we are going to define addition (which I will write $\oplus$ to distinguish it form the addition of real numbers) is as follows: if $[r]$ and $[s]$ are two classes, then $[r]\oplus [s]$ is: $$[r]\oplus [s] = \left\{\begin{array}{ll} \ [r+s] &\text{if }0\leq r+s\lt 1\\ \ [r+s-1] &\text{if }1\leq r+s. \end{array}\right.$$ Because $0\leq r,s\lt 1$, then $0\leq r+s\lt 2$, so one and only one of those situations will happen, and the result will always be a "proper name" for an element of $\mathbb{R}/\mathbb{Z}$ (that is, the answer is of the form $[a]$ with $0\leq a\lt 1$).

Then one can show that this makes $\mathbb{R}/\mathbb{Z}$ into a group.

What the text you are quoting is doing is that it is writing "$r+\mathbb{Z}$" where I wrote "$[r]$" above. The reason it does this is that this is the standard notation when performing the kind of construction that is being discussed. This is covered in any book on abstract algebra that discusses groups.

• "under the equivalence relation "$x\sim y$ if and only if $x-y\in\mathbb{Z}$" Is it right? If $x=1.1, y=1.2$, then $x-y=0.1$, which is not in $\mathbb{Z}$. But it seems that $x$ and $y$ are in the same equivalence class here.(?) – Kou Feb 2 '12 at 15:40
• @Kou: No, $1.1$ and $1.2$ are not in the same equivalence class here (precisely because $x-y\notin\mathbb{Z}$). What gives you the impression that they are? (Honest question; if there is confusion, I need to clear it up) – Arturo Magidin Feb 2 '12 at 15:58
• Oh, so $1.1, 2.1, 3.1$ are in the same equivalence class? Since they share the same decimal part? – Kou Feb 2 '12 at 16:12
• @Kou: They are in the same equivalence class because their difference is an integer, yes. (Better to think in terms of the difference, which is independent of representation, than of their decimal representation which, well, depends on their representation). – Arturo Magidin Feb 2 '12 at 17:16
• Thanks for your great patience to me, a high school student. – Kou Feb 2 '12 at 20:47

I wrote some words about the algebraic structure on $\mathbf R/\mathbf Z$, but I think that's been treated very well in the other answers. By the tags I feel that we should also talk about the topological and analytic aspects of this group, so here's a start in that direction. I certainly wish that I knew more about this.

$\mathbf R/\mathbf Z$ is often called the circle group, because if we view $S^1$ as lying inside of the complex plane then the map $F\colon\mathbf R \to S^1$, $F(x) = e^{2\pi ix}$ induces a bijection $\mathbf R/\mathbf Z \to S^1$ which is both an isomorphism of groups and a homeomorphism of topological spaces (if we place the quotient topology on $\mathbf R/\mathbf Z$); in particular, $\mathbf R/\mathbf Z$ is naturally a compact space.

$\mathbf R/\mathbf Z$ is a nice home for periodic functions, since if $f\colon \mathbf R \to \mathbf R$ is such that $f(x + n) = f(x)$ for all $x \in \mathbf R$ and $n \in \mathbf Z$, then $f$ descends to a map $\mathbf R/\mathbf Z \to \mathbf R$. Moreover, Pontryagin duality says that Fourier analysis on the circle is related to functions on a certain "dual group", and the dual group of the circle is $\mathbf Z$, which is discrete! On the other hand, $\mathbf R$ is its own dual.

As for differential operators, that's a large subject and I can only sketch the situation and give Wiki references. One can view $\partial/\partial x$ as a vector field on $\mathbf R$. Each $a \in \mathbf R$ induces a translation diffeomorphism $L_a\colon \mathbf R \to \mathbf R$, $L_a(x) = x + a$, and the field $\partial/\partial x$ is called left-invariant if it is unchanged under the corresponding pushforwards: $L_{a*}(\partial/\partial x) = \partial/\partial x$.

In general, given a smooth map of manifolds $F\colon N \to M$ a vector field on $N$ may not correspond to a vector field on $M$. But you can check that in this case we can define a vector field on $S^1$, which people usually call $\partial/\partial\theta$, by $(\partial/\partial\theta)_{F(P)} = F_*(\partial/\partial x)_P$.

• Hm. fourier-analysis has been taken off, so this looks a little out of place now. I was always worried about the level of it (and my own ignorance of abstract analysis), but I'll leave it up for the time being. – Dylan Moreland Feb 2 '12 at 5:43
• There seemed to be no indication of how it was connected to Fourier analysis (the only context given was Wikipedia), which is why I took it off. If the query arose from Fourier analysis, this was definitely hidden from our view... – Arturo Magidin Feb 2 '12 at 5:53
• +1: Fourier-analysis aside, I imagine that this geometric viewpoint (circle as opposed to the numberline) may help the OP to grasp the meaning of the quotient group - at least this particular quotient group. – Jyrki Lahtonen Feb 2 '12 at 6:13
• Actually I met this problem because I am reading something on Fourier analysis... Could you explain more about "Since $\mathbf R/\mathbf Z$ is like the real line looped around itself, it seems like a nice home for periodic functions." I couldn't understand the words like "$S^1$ inherits the translation-invariant differential operator $d/dx$ from the real line $\mathbf R$". – Kou Feb 2 '12 at 15:08
• @ArturoMagidin Could you also help me with the problem I mentioned above? Thanks. – Kou Feb 2 '12 at 20:49

One difference between ${\bf R}/{\bf Z}$ and $\bf R$ is that $${1\over2}+{1\over2}=0$$ in the former but not in the latter.

EDIT: Perhaps it would be more precise to write $$\left({1\over2}+{\bf Z}\right)+\left({1\over2}+{\bf Z}\right)=0+{\bf Z}$$ but $${1\over2}+{1\over2}\ne0$$

Perhaps even better simply to say that if $x+x$ is the identity element in $\bf Z$ then $x$ is itself the identity element of $\bf Z$, but in ${\bf R}/{\bf Z}$ there is an $x$ other than the identity such that $x+x$ is the identity.

Every real number has two parts: an integer part and a decimal part. In other words, we can write every $x \in \mathbb{R}$ as a sum of elements $n_{x} \in \mathbb{Z}$ and $u_{x} \in [0,1]$. Let us write $x = n_{x} + u_{x}$. Now, we can use this to define an equivalence relation on $\mathbb{R}$: we say that $x \equiv y$ if $u_{x} = u_{y}$, so that their decimal parts are the same. For a given element $x \in \mathbb{R}$, the equivalence class $[x]$ is just $u_{x} + \mathbb{Z}$.

The quotient group $\mathbb{R}/\mathbb{Z}$ arises by "collapsing" together these equivalence classes. We are thus left with something akin to the unit interval $[0,1]$, as the rest of $\mathbb{R}$ is made of shifts of the unit interval by $\mathbb{Z}$. However, since $0$ and $1$ have the same decimal expansion, they are equal to each other. What we get, topologically, is a line with the ends tied together: a circle. Thus $\mathbb{R}/\mathbb{Z} \cong S^{1}$, and hence it is compact.

Moreover, we need to be careful in adding equivalence classes. Clearly, $[0.25] + [0.25] = [0.5]$. However, $[0.75] + [0.75] = [1.5]$. Since we working with the unit interval, we subtract one, to get $[0.5]$. The equivalence class is the same, only now we are representing it by an element in $[0,1)$.

• I think you want $[0,1)$ rather than $[0,1]$. – Jonas Meyer Feb 2 '12 at 5:27
• I think the standard term is fractional part; after all, it's independent of decimal notation. (Of course, it's not always a fraction, either, but you can't win 'em all...) – Rahul Feb 2 '12 at 5:41

$(a+\mathbb Z)+(b+\mathbb Z)$ is found by adding $a$ and $b$, the result of which is $a+b$. If $a+b<1$, then $(a+\mathbb Z)+(b+\mathbb Z)=(a+b)+\mathbb Z$. If $a+b\geq 1$, then $(a+\mathbb Z)+(b+\mathbb Z)=(a+b-1)+\mathbb Z$.

But this is only if you follow the stated convention of only listing representatives from $[0,1)$. The fact is, $(a+b)+\mathbb Z$ and $(a+b-1)+\mathbb Z$ are different names for the exact same set, so you don't really need to subtract $1$.