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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

2 Answers 2

up vote 1 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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The expression is: "for all intents and purposes", not "for all intensive purposes". ;-) – amWhy Dec 5 '13 at 19:05
Good catch. I'm a bit tired. – Tim Ratigan Dec 5 '13 at 19:08
@amWhy: It’s a holy cross-eyed bear! Well, no — but it is a mondegreen. – Brian M. Scott Dec 5 '13 at 19:10
@BrianM.Scott I haven't heard that one! Thanks for the link! ;-) – amWhy Dec 5 '13 at 19:12
Also an eggcorn – Tim Ratigan Dec 5 '13 at 19:17

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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