2
$\begingroup$

Suppose we define an equivalence relation $\sim$ on a set $S$. The exact details of the relation and checking that it verifies reflexivity, symmetry, and transitivity is less important, so let's assume it satisfies those properties. Then $S/\sim$, read "S mod $\sim$ denotes the set of equivalence classes.

I am trying to make sense of this notation, specifically the use of "mod." In a quotient $R/S$, we say that $r_1, r_2 \in R$ are equivalent if $r_1 - r_2 \in S$. Similarly, in $\mathbb{Z}/2\mathbb{Z}$, $a$ is equivalent to $b$ if $a - b \in 2\mathbb{Z}$. I do not see any parallel in general to $S/\sim$. The only similarity is that they are equivalent "with respect to $\sim$."

Am I missing something? Is there a better way to think of this notation or even what it means to "quotient by $\sim$"?

$\endgroup$
3
  • 2
    $\begingroup$ The parallel is that those are both special cases of quotienting by an equivalence relation; it's unclear from your question whether you understand this. $\endgroup$ Dec 27 '20 at 0:55
  • $\begingroup$ @azif00 It's read that way, i.e. pronounced that way, not notated that way. $\endgroup$ Dec 27 '20 at 1:14
  • $\begingroup$ @QiaochuYuan This is a very helpful comment, and I think you are correct that I do not fully understand this. Would you mind explaining more about what it means to quotient by an equivalence relation? $\endgroup$
    – JeremyS
    Dec 27 '20 at 1:20
2
$\begingroup$

When you defined $a \equiv b \pmod n$ we use the notation $\pmod n$ BECAUSE $a- b\in n\mathbb Z$ is an equivalence relation that means that the relation $a-b\in n\mathbb Z$ "partitions" $\mathbb Z$ into equivalence classes and $\mathbb Z/n\mathbb Z$ is the set of equivalence classes it is partitioned into.

$\mathbb Z/n\mathbb Z = \{[0],[1],......., [n-1]\}$ where $[k]=\{m\in \mathbb Z| m\equiv k \pmod n\} = \{m \in \mathbb Z| m-k \in n\mathbb Z\} = \{....,k-n, k, k+n, k+2n,.....\}$.

But there is nothing magical about the relation $a\equiv b$. EVERY equivalence relation will partition a set, $S$ into equivalence classes $U_\alpha$ where each and every $s\in S$ is in exactly one, and only one, $U_\alpha$ and all the elements in $U_\alpha$ are related to each other. And $S/\sim$ is the set of all these equivalence classes.

So for instance; take any equivalence relation $\sim$ an a set $S$. For example let say $S= \mathbb N$ and $a \sim b\iff $ the highest power of $3$ that divides $a$ is the same highest power of $3$ that divides $b$.

for example $3\sim 6 \sim 12 \sim 51 etc. $ because $3|6, 12, 51$ but $3^2 \not\mid 6,12,51$.

If we let $U_1 = \{3, 6, 12, 15, 21, 24, .....\}=\{k\in \mathbb Z| 3\mid k, 9\not \mid k\}$ then that is one equivalence class. If we let $U_2=\{9,18,36,....\} = \{k\in \mathbb Z| 9|k, 27\not \mid k\}$ that's another equivalence class. And we can let $U_m =\{k\in \mathbb Z| 3^m|m, 3^{m+1}\not \mid k\}$ can be any other.

Also note $\{0\} = U_\infty = \{k\in \mathbb N| 3^j\mid k$ for all $k\in \mathbb N\}$ is the final equivalence class..

Then $\mathbb Z/\sim = \{U_\infty, U_1, U_2, U_3......\}$ is the set of all equivalence classes.

That is entirely equivalent to $\mathbb Z/(\pmod n) = \{[0],[1],.....[n-1]\}$ where $[k] = \{.... k-2n, k-n, k, k+n,...\} = \{k+mn|m\in \mathbb Z\} = \{z\in \mathbb Z| a\equiv z\pmod n\} = \{z\in \mathbb Z| a-z\in n\mathbb Z\}$ are the set of all equivalence classes.

=====

I do not see any parallel in general to S/∼. The only similarity is that they are equivalent "with respect to ∼."

Thats all the similarity that there is. but that's that's enough similarity for anyone. That is a HUGE similarity.

$\endgroup$
1
  • $\begingroup$ How many is a "bunch"? I only saw one. $\endgroup$
    – fleablood
    Dec 27 '20 at 3:32
1
$\begingroup$

Our experience with modular arithmetic includes a few features:

  • The formalization of $\mathbb Z / 2 \mathbb Z$, and more generally $\mathbb Z / n \mathbb Z$, as a set of equivalence classes.

  • The usage of the abbreviation "mod", and the notation of the division symbol "$/$", when referring to an individual equivalence class or to a whole set of equivalence classes

Because of the utility of this formalization/usage/notation, it has been generalized to all equivalence relations: an individual equivalence class, or a set of equivalence classes, is often referred to using the abbreviation "mod" and/or the division symbol "$/$".

What you are seeing in action here is mathematicians borrowing, extending, stretttttccccchhhhhhhing old notation and re-using it in new situations. It happens allllllllll the time in mathematics: usage and notation for some special case being stretched to apply to a much broader case. That's all that's happening here.

$\endgroup$
0
$\begingroup$

Good. In the case of a generic equivalence relationship, this is the same. In the case of $\mathbb{Z} /2\mathbb{Z}$ this mod notation is used to symbolize the congruence modulo p.

$\endgroup$
-1
$\begingroup$

There is in fact a much more general answer to your question that covers quotient structures in general. First, read this post, which explains how we can form quotient structures using a suitable partition of the larger structure into parts that respects the structure (the operations and predicates). Notice that it uniformly explains the isomorphism theorems for all kinds of structures. Next, observe that we write $S/\sim$ when $\sim$ is an equivalence relation that partitions $S$, but we also write $S/T$ when $T$ is an ideal of $S$, meaning that special part referred to in the linked post that we can use to construct a structure-respecting partition $S$. That is precisely why we write $ℤ/nℤ$, and $G/H$ for groups $G,H$ with $H ◁ G$, and $V/W$ for vector spaces $V,W$ with $W ⊆ V$, and $R/I$ for ring $R$ and ideal $I$ of $R$, and so on...

$\endgroup$
2
  • $\begingroup$ The notation is sensibly suggestive, with $S/T$ meaning $S$ divided up by $T$. $\endgroup$
    – user21820
    Dec 27 '20 at 14:36
  • $\begingroup$ Readers, ignore the downvote because it's likely from a troll, which was unable to point out any problem with this explanation despite it being general unlike the other answers. $\endgroup$
    – user21820
    Jan 8 at 9:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.