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I search and search 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?

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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. – DonAntonio Dec 5 '13 at 19:00
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 $S/\sim \,= \{\overline{x}\mid x \in S\}$. – xavierm02 Dec 5 '13 at 19:02
I don't think there's any difference between "quotient set" and "partition." – goblin Jul 17 '14 at 17:25
up vote 2 down vote accepted

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.

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@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 – Matheus Silva Dec 6 '13 at 0:59
i'm only able to find the quotient set if have a relation ? – Matheus Silva Dec 6 '13 at 22:16
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. – Tim Ratigan Dec 6 '13 at 22:47
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 $B+4$.) – Braindead Jul 17 '14 at 16:56
@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 $\Bbb Z /5\Bbb Z:= \{[0],[1],[2],[3],[4]\}=\{[0],[1],[2],[3],[44]\}$). – YoTengoUnLCD Jan 4 at 4:27

I thought $Z/nZ=\{a+nZ:a\in Z, n\in N\}$ symbols or $X/A=\{x+A:x\in X,a\in A\}$ and $x+A=\{x+a:x\in X,a\in A\}$.

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