I searched and searched about quotient set and cannot figure out what it is. At the beginning, I thought it was the same as partitions, but now I'm confused. Can someone show some examples and explain?

  • $\begingroup$ This is, imo, way too messy to explain it all in this site, but in very short: every equivalence relation on a set determines a unique partition of that set, from which we can form a quotient set with the equivalence classes, and the other way around is true, too. Many elementary algebra, or college algebra, books deal with this. Also set theory books. $\endgroup$
    – DonAntonio
    Dec 5, 2013 at 19:00
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    $\begingroup$ You have a set $S$. You also have an equivalence relation $\sim$ on $S$. You define the class of an element $x\in S$ by $\overline{x}=\{y\in S \mid y \sim x\}$. And then you define the quotient set $S/\sim \,= \{\overline{x}\mid x \in S\}$. $\endgroup$
    – xavierm02
    Dec 5, 2013 at 19:02
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    $\begingroup$ I don't think there's any difference between "quotient set" and "partition." $\endgroup$ Jul 17, 2014 at 17:25
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    $\begingroup$ You might have a look at Wikipedia article or definition at ProofWiki (and other related stuff). $\endgroup$ Jul 7, 2016 at 5:55

2 Answers 2


A quotient set is what you get when you "divide" a set $A$ by $B\subseteq A$, wherein you set all elements of $B$ to the identity in $A$. For example, if $A=\Bbb Z$ and $B=\{5n\mid n\in\Bbb Z\}$, then you're making all multiples of $5$ zero for all intents and purposes, so the quotient is $\{0,1,2,3,4\}$.

Another (and more correct) way of saying this is that a quotient set is all equivalence classes on the set $A$ under a given equivalence relation. In the example above, $aRb\iff 5|(a-b)$, so clearly the equivalence classes are $n\equiv 0,1,2,3,4\pmod 5$. In reality, you can select any number from each equivalence class, so $\{20,-34,77,63,-1\}$ would be a "correct" quotient set, just not canonical.

  • $\begingroup$ @TimRatigan and quocient set based on mod i understand, but based on another relation like, (x,y) where x + |x| = y + |y|, based com E = {-3,-2,-1,0,1,2,3}, i found the relation elements, how can i build the quotien set? i figured out that it works for mod, but cant figure out to another single relation $\endgroup$
    – PlayMa256
    Dec 6, 2013 at 0:59
  • $\begingroup$ i'm only able to find the quotient set if have a relation ? $\endgroup$
    – PlayMa256
    Dec 6, 2013 at 22:16
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    $\begingroup$ The quotient set is really just picking one element from each equivalence class and putting your chosen elements in a set. You can only get equivalence classes from an equivalence relation, so yes, you need a relation. $\endgroup$ Dec 6, 2013 at 22:47
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    $\begingroup$ The first part of your answer is very misleading. If I took a set $A$ and just took an arbitrary subset $B$, what would the quotient $A/B$ mean? If you are simply identifying all the points of $B$ as a single point, then that is not the same kind of quotient you are taking in your example of $A = Z$ and $B$ multiples of $5$. (In that example, you are implicitly using the addition structure, and you are partitioning $A$ with $B$, $B+1$, $B+2$, $B+3$, and $B+4$.) $\endgroup$
    – Braindead
    Jul 17, 2014 at 16:56
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    $\begingroup$ @Braindead Agreed: In fact, the equivalence relation $R$ to quotient $A$ by, need not be a subset of $A$, but a subset of $A\times A$. Another thing I think must be adressed is the fact that, according to the answer, the notation $A/R$ doesn't refer to a set of elements of $A$, but to a set of equivalence classes (that is, $\{0,1,2,3,4\}\neq \{0,1,2,3,44\}, $ but $\Bbb Z /5\Bbb Z:= \{[0],[1],[2],[3],[4]\}=\{[0],[1],[2],[3],[44]\}$). $\endgroup$ Jan 4, 2016 at 4:27

I'm way late to the party, but for anyone stumbling upon this like I did, a simple definition for a quotient set is "the set of all equivalence classes of a set under a given equivalence relation." An equivalence class IS the same as a partition, defined by using some equivalence relation. But the quotient is ALL of those equivalence classes (partitions) under that particular equivalence relation. You DO need an equivalence relation to build a quotient set, which is why the notation is S/~, which is read as "the quotient set of the set S under the equivalence relation ~."

At the risk of over-simplifying it, you could say that the quotient set under a particular equivalence relation is the same as the original set, but in partitions rather than all together. (This isn't strictly true, but it's a useful way to understand the basic idea)

For example, say the set $C$ is the set of all cars, and $\sim_c$ is an equivalence relation that means "is the same color as." So for some white car $w$ and some other white car $h$, $w\sim_ch$. With that kind of equivalence relation, $[w]$ is the equivalence class that means "all white cars," and is a partition of the set of all cars $C$. $[h]$ works just as well since both $w$ and $h$ are in the same equivalence class. (You could write $[w]_{\sim_c}$ to be specific about which relation you're using for the equivalence class) The quotient set $C/\sim_c$ would be the set of all equivalence classes in $C$ under $\sim_c$. I.e., it's the set of all partitions of $C$, partitioned by color. If you happen to be fully color blind so that all cars look either white, grey, or black, then the quotient set $C/\sim_c$ would be $\{[w],[g],[b]\}$ (given that $g$ and $b$ are grey and black cars just like $w$ is a white car). So to build a quotient set, you either list out all of the possible equivalence classes OR generalize it with set notation: $$C/\sim_c\ \ = \{[x]_{\sim_c}\mid x\in C\}$$

Since each equivalence class in this context represents all cars with a specific color and the quotient set contains all color groups (by definition), then the quotient set ultimately still represents the same group of objects. If you were to visualize this, then you'd see all the cars in the world gathered into different groups by color rather than all in one heap. More strictly speaking, the quotient set is more like a listing of the groups rather than a listing of all the cars in the groups (After all, the elements in $C$ are cars, but the elements in $C/\sim_C$ are sets of cars). Writing $[w]$ is like saying "all cars with the same color as this $w$ one," or "this car's color is what I mean by 'white.'" (the first one is more mathematically correct, but the second makes more contextual sense with the example)

I hope that helps to thoroughly visualize the concept. In general, a quotient set differs from a partition in that a quotient set contains partitions as elements, and those partitions are defined by an equivalence relation. The two are definitely distinct.

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    $\begingroup$ You are confusing the partition with the classes (parts) in the partition. The partition is the collection of all classes, not the classes themselves. $\endgroup$
    – user50229
    Mar 6, 2017 at 18:30
  • $\begingroup$ Very late to the party, but isn't it the case that the classes partition the set into two disjoint sets (the guys that are similar in some way and the guys that are not) and the quotient is the collection of all of those? So, in Halcyon's answer, we'd have a garage (a set of cars, I mean) divided by colors. Then, we'd say that this division (the quotient set) is a partition. What I mean here is that he seemed to mistake the fact that the classes partition the set with them being the partition. $\endgroup$ Nov 25, 2023 at 14:59

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