For the set $ \Bbb R$, define two elements in $ \Bbb R$ to be equivalent if their difference belongs to $ \Bbb Q$.

I can prove that this defines an equivalence relation. (see Verify an equivalence relation.)

However, I would like to be able to define the class and the partition. Is it possible to place an element into a class without comparing it to other members of the class?

A similar problem, where the difference belongs to $ \Bbb Z$, any element r $\in \Bbb R$ can be classified by setting up class Am such that m = r - g(r) where g(r) returns the nearest integer less than r. As a result, any r $\in \Bbb R$ can be put into some class Am, $0<=m<1$. There may be other indexing schemes.

But I can't come up with anything for the difference belonging to $ \Bbb Q$.

This problem is from Pinter's A Book of Abstract Algebra.


  • 1
    $\begingroup$ "I would like to be able to define the class and the partition" -- what do you mean by "the class"? $\endgroup$ – BrianO Oct 14 '15 at 20:28
  • $\begingroup$ equivalence classes I presume $\endgroup$ – BrianO Oct 14 '15 at 20:38
  • $\begingroup$ Pinter is the only author I'm familiar at all with - he states, "To partition a set A is to separate the elements of A into nonempty subsets ... which are called the classes of the partition." $\endgroup$ – Alan Ox Oct 14 '15 at 21:23
  • $\begingroup$ Yes classES plural. I figured out what you meant once I started writing my answer :) $\endgroup$ – BrianO Oct 14 '15 at 21:24

We have: $$ x \sim y \iff x - y=q \in \mathbb{Q} $$ Consider the equivalence class of $0$, this is the set of real numbers $x$ such that $ x-0=q \in\mathbb{Q}$ so the class $[0]$ is the set of rational numbers.

Now consider an irrational number $z$, since $z-0$ is irrational we have $z\notin [0]$ (it is not rational), So it represents another equivalence class $[z]$ that contains all the irrational numbers $y$ such that $y=z+q$ with $q \in \mathbb{Q}$.

For another irrational number $y \notin [z]$ ( i.e. $y-z \notin \mathbb{Q}$), we have another equivalence class $[y]$ .. and so on.

So, The equivalence classes are represented by $[0]$ and $[z_1],[z_2],\ldots$ where $z_i$ are all irrational numbers that differ by an irrational number, and there are uncountable infinitely many such classes.


1st example

The equivalence relation is: $$ x \sim y \iff x - y \in \mathbb{Q} $$ For every $x \in \mathbb{R}$, let $[x]_\sim$ be its equivalence class: $$ [x]_{\sim} = \{y \in \mathbb{R} \mid x \sim y\} $$

Then the partition, or quotient, is just: $$ \mathbb{R}/\!\!\sim \; = \{[x]_{\sim} \mid x \in \mathbb{R} \} $$ Two reals $x, y$ are $\sim$-equivalent when their difference is a rational. Notice, then, that the equivalence class of $0$ is $\mathbb{Q}$: all rationals are $\sim$-equivalent and none are equivalent to any irrational. The equivalence class of $\pi$ is just $\{\pi + q \mid q \in \mathbb{Q}\}$, and does not contain, for example, $e$, or $e + 17$. In general, the equivalence classes are of the form $r + \mathbb{Q} = \{r + q \mid q \in \mathbb{Q}\}$.

Each equivalence class is countable, so there are uncountably many equivalence classes.

2nd example

Here, differences are in $\mathbb{Z}$, and every equivalence class has a unique representative in $[0,1)$. The equivalence class of $\frac 1 2$ is $\{\dots - \frac 3 2, -\frac 1 2, \frac 1 2, \frac 3 2, \dots \}$; the equivalence class of $\pi$ is $\{\dots \pi - 4, \pi - 3, \pi - 2, \dots \}$.

Again, each equivalence class is countable, so there are uncountably many equivalence classes.

  • $\begingroup$ I think we're on the same page on the integer differences, you just put it much better than I did. My point with it though was that it's possible to associate a unique representative for any given real number. If r = 9.9987, then it goes into the equivalence class of 0.9987. But I don't see any way to do that if the difference is a rational number. All rational numbers go into the same equivalence class, but what designator would be used for, say, pi? pi - 1, pi - 4.552, etc are all in the same equivalence class. $\endgroup$ – Alan Ox Oct 14 '15 at 22:56
  • $\begingroup$ That's quite right, you can't, in the case of example 1. This "indiscernibility" of the equivalence classes is exploited in the construction of a nonmeasurable set. By the Axiom of Choice, there is a set that contains a unique representative for each equivalence class. This "choice set" is plucked out of thin air, you can't say anything about the representatives. A (any) set of representatives of this relation is called a Vitali set. It's the standard example of a set that is not measurable. $\endgroup$ – BrianO Oct 14 '15 at 23:20

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