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I have the relation $x\mathsf{R}y$ if $x - y \in \mathbf{Z}$ where $x, y$ are real numbers. I want to describe the equivalence classes of the relation.

I have proven to myself that $x-y$ is an equivalence relation if $x-y \in \mathbf{Z}$. But, how does this set partition out? Since an equivalence class of a set S containing a is defined as: $[a] = \{x \in S : x\mathsf{R}a\}$ isn't this just the set of all real numbers? i.e. doesn't this simply partition out into $\mathbf{R}$?

I guess I'm confused on how the definition for an equivalence class works. I read the questions and answers related to this question before I submitted it. I think, unfortunately, I need a bit more of an explanation for a complete understanding as I am a complete beginner in abstract algebra. Aren't there an infinite number of equivalence classes here?

Thank you in advance, very much appreciated.

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    $\begingroup$ What are $S$ and $Z$? $\endgroup$ – Bernard Sep 2 '17 at 15:20
  • $\begingroup$ One way of describing equivalence classes is by finding a simple way of describing an element of each class, so that every element of $\mathbb R$ is equivalent to one of the elements. For example, what can you say about the least non-negative member of each class. $\endgroup$ – Mark Bennet Sep 2 '17 at 15:30
  • $\begingroup$ Swapped out S in the question. R is Reals...Z is Integers. $\endgroup$ – Idle Math Guy Sep 2 '17 at 15:35
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$y$ is equivalent to $x$ if $x-y =k$ for some integer $k$, right? That means $x = y + k$ for some integer $k$.

So, the equivalence class of $y$ is the set of elements of the form $y+k$, where $k$ ranges over the integers, since it is exactly this set that satisfies the above property of $x-y=k$.

So, start with an element $y$. To get its equivalence class, simply shift $y$ by $k$ for each integer $k$, and you have the equivalence class of $y$.

In math, you would write $[y] = \{y +k | k \in \Bbb Z \}$.

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    $\begingroup$ The first line of this answer is worth it's weight in gold. Now I understand the answers to the other questions related to this one. Beginners to abstract algebra, take note to how layman derived the partition classes in the first line. $\endgroup$ – Idle Math Guy Sep 2 '17 at 15:29
  • $\begingroup$ @IdleMathGuy I'm glad it helped! $\endgroup$ – layman Sep 2 '17 at 17:52
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Pick any real number $x$. Its equivalence class will consist of elements $y$ such that $y-x=k$ for some $k\in\mathbb{Z}$, so $y=x+k$. This tells you that $[x]=x+\mathbb{Z}:=\{x+k:k\in\mathbb{Z}\}$. The quotient space is the set of all equivalence classes $\mathbb{R}\big/\mathcal{R}=\{[x]:x\in\mathbb{R}\}$, but for every equivalence class there must be one representative belonging to $[0,1)$ --why? think of the fractional part function-- so you can write $\mathbb{R}\big/\mathcal{R}$ without repetitions as $$\mathbb{R}\big/\mathcal{R}=\{[x]:x\in[0,1)\}=\{x+\mathbb{Z}:x\in[0,1)\},$$ which is indeed an infinite set.

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  • $\begingroup$ Would you still consider the answer $[x] = $ { $x + k | k \in \mathbb{Z} $} a correct answer or acceptable answer? $\endgroup$ – Idle Math Guy Sep 2 '17 at 15:59
  • $\begingroup$ @IdleMathGuy Yes, if the question is what is the equivalence class $[x]$ of a number $x$. $\endgroup$ – Jonatan B. Bastos Sep 2 '17 at 16:54
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Here $x$R$y$ if $x-y \in Z$. So if we consider equivalence classes for $a$, then it contain $a-1,a,a+1$, but it does not contain any number in between $(a,a+1)$. Also any two numbers in $[a,a+1)$ are not in equivalence relation to each other as their difference is not an integer. The equivalence classes are $\{[x]| x \in [a,a+1)\}$ where $a \in R$ is any real number. For example if we take $a=0$ we get the equivalence classes as $\{[x]| x \in [0,1)\}$ and this number of equivalence classes is uncountable.

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Note that $x R y$ iff $ \{x\} R \{y\}$ where $\{x\}=x-\lfloor x\rfloor$

The equivalence classes that partitions $\mathbb R$ are $\{class(x): x\in [0,1)\}$

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