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The precise definition given in Dummit and Foote (sec 0.3, pg. 9) is the following:

The definition of elements in residue class $x$ are those that are between $\bar0, \bar1, ..., \overline{x-1}$.

Any integer congruent to any of these elements can be a representative of the element.

My answer to this question was severely downvoted and got deleted:

Hence, $\overline{768}$ CANNOT be an element of the class as many others have incorrectly pointed out - but $-768$ can be.

Yes, $-768 \mod 157 \equiv \overline{17}$.

And why is this if I just quoted a definition from a respectable math book? It appears that Dummit and Foote (3e) is incorrect, then.

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Jossie Calderon
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    Sorry, but your question is really not clear. As the definition you mentioned : The definition of elements in residue class are those that are between $\bar 0$, $\bar 1$,..., $\overline{x-1}$, that's definitely not correct. Element of residue class of $x$ (in $\mathbb Z/N\mathbb Z$) are element of ${x+Nk\mid k\in\mathbb Z}$ and not what you mentioned – user649261 Mar 04 '19 at 17:11
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    Your answer was wrong (and not consistent with Dummit & Foote): $\overline{768}$ is a perfectly good element of the quotient, and coincides with $\overline{768 \pmod{157}}.$ Moreover, $-768$ is not an element of the quotient, because it's not a coset. –  Mar 04 '19 at 17:15
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    I'm not sure the negative reaction to this question is justified. There's a rantiness here, but also a perfectly good question, namely whether Dummit/Foote is incorrect or the OP is misunderstanding the situation. I agree the polemic should be removed, but I think we're being a bit too negative here. – Noah Schweber Mar 04 '19 at 17:22
  • @MorganRodgers That's a fair point. – Noah Schweber Mar 04 '19 at 18:51

1 Answers1

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Dummit and Foote is correct, and your answer was incorrect; the issue is that you seem to be misunderstanding what the quotient group is, and consequently what exactly Dummit and Foote are saying in the quoted passage.

Below I write "$\overline{x}$" for the equivalence class of $x$ modulo $157$.

The key point is that $\overline{768}$ is literally the same thing as $\overline{140}$, and is definitely an element of $\mathbb{Z}/157\mathbb{Z}$. The equality $\mathbb{Z}/157\mathbb{Z}=\{\overline{0},\overline{1},...,\overline{156}\}$ is correct; but it's also true that $\overline{178}\in\mathbb{Z}/157\mathbb{Z}$.

Meanwhile, a straight-up integer like $-768$ isn't an element of $\mathbb{Z}/157\mathbb{Z}$; rather, it's an element of an element of $\mathbb{Z}/157\mathbb{Z}$ (specifically, $-768\in\overline{17}$). That is:

  • $\mathbb{Z}/157\mathbb{Z}$ is a set of sets of integers; as such, no integer is an element of $\mathbb{Z}/157\mathbb{Z}$.

  • On the other hand, every integer is an element of a (unique) element of $\mathbb{Z}/157\mathbb{Z}$; namely, $x\in\overline{x}$.

  • Moreover, every $x$ is in $\overline{y}$ for some (unique) $y\in\{0,1,2,..., 156\}$; in particular, combining this point with the previous one we have that if $x\not\in \{0,1,2,..., 156\}$ then "$\overline{x}$" is just a different name for some $\overline{y}$ with $y\in \{0,1,2,..., 156\}$.

Noah Schweber
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  • After, when there are no confusion, we often use the short writing $x$ for $\bar x$. – user649261 Mar 04 '19 at 17:26
  • @user649261 Quite right (and worth pointing out to the OP to forestall later confusion) - but that's notation you don't want to abuse too early! – Noah Schweber Mar 04 '19 at 17:26
  • Yes, I completely agree :) – user649261 Mar 04 '19 at 17:26
  • There are $n$ distinct equivalence classes $\mod n$ - Dummit and Foote. $\overline{178}$ cannot be by definition an equivalence class – Jossie Calderon Mar 04 '19 at 19:23
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    @JossieCalderon Not quite. "There are $n$ distinct equivalence classes mod $n$" That's correct. "$\overline{768}$ cannot be by definition an equivalence class" That is incorrect. "$\overline{768}$" is just a different name for $\overline{140}$, in the same way that $0.9999...$ is just a different name for $1$. There are indeed only $157$ distinct classes, and the thing we call $\overline{768}$ is one of them - namely, it's the same thing as $\overline{140}$. $a\not=b$ does not mean $\overline{a}\not=\overline{b}$, and this is in fact the key point about quotient groups. (continued) – Noah Schweber Mar 04 '19 at 19:40
  • It will help to shift from $157$ to something smaller, so you can think about this more concretely. Work mod $10$; looking at the definition of $\overline{a}$, do you see why $\overline{3}=\overline{13}$? (If not, what is something you think is in one set but not the other set?) – Noah Schweber Mar 04 '19 at 19:42
  • How can $\overline{768}$ exist by definition? – Jossie Calderon Mar 04 '19 at 19:56
  • @JossieCalderon By definition, $\overline{a}$ - for any (integer) $a$ - is the set ${b: b=a$ mod $n}$ (where for us $n=157$ of course). For example (in our setting) we have $\overline{0}=\overline{157}=\overline{-157}=\overline{314}=...$ Now for $a\not\in{0,1,2,...,n-1}$, the notation "$\overline{a}$" is in some sense redundant: we have $\overline{a}=\overline{s}$ (literally $=$: they are the same sets of integers) for some $s\in{0,1,2,..., n-1}$. But that doesn't make that notation nonsensical - it just means it's not going to be the usual notation we use to refer to that set. – Noah Schweber Mar 04 '19 at 20:13
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    @Jossie An analogy may help, the only positive rationals are $,\overline{1/1},\ \overline{1/2},\ \overline{1/3},\overline{2/3},, \overline{1/4},\overline{3/4}, \ldots$ but this does not imply that $,\overline{2/4}\not\in\Bbb Q.,$ It is in $\Bbb Q,$ since $, \overline{2/4} = \overline{1/2} = {n/2n : n\neq 0}$ all denote exactly the same equivalence class in $\Bbb Q.\ \ $ – Bill Dubuque Mar 04 '19 at 20:44
  • @JossieCalderon Do the explanations we've given help? – Noah Schweber Mar 07 '19 at 13:49
  • @Noah Yes. You're saying that there's $n$ sets, but infinite ways of saying it. – Jossie Calderon Mar 12 '19 at 00:27
  • @JossieCalderon Glad we could help! – Noah Schweber Mar 12 '19 at 01:42