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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    $\begingroup$ Sorry, but your question is really not clear. As the definition you mentioned : The definition of elements in residue class 𝑥 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 $\endgroup$ – user649261 Mar 4 at 17:11
  • $\begingroup$ If your question about why you got downvoted, you should post on Meta (since this is not itself a mathematical question). If your question is if whether Dummit and Foote is incorrect, you should include more context. I can't tell what Dummit and Foote says that you think is incorrect. In particular, what you have in your quotes block is not identical to the wording in Dummit & Foote. $\endgroup$ – Morgan Rodgers Mar 4 at 17:11
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    $\begingroup$ 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. $\endgroup$ – user296602 Mar 4 at 17:15
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    $\begingroup$ 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. $\endgroup$ – Noah Schweber Mar 4 at 17:22
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    $\begingroup$ @NoahSchweber I think if the question is whether Dummit & Foote is incorrect, the statement in question should appear exactly as written in the book. My copy is in my office, but from a search online, the block quote above is definitely not the "precise definition" from the book. $\endgroup$ – Morgan Rodgers Mar 4 at 18:47

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 $y\in \{0,1,2,..., 156\}$.

  • $\begingroup$ After, when there are no confusion, we often use the short writing $x$ for $\bar x$. $\endgroup$ – user649261 Mar 4 at 17:26
  • $\begingroup$ @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! $\endgroup$ – Noah Schweber Mar 4 at 17:26
  • $\begingroup$ Yes, I completely agree :) $\endgroup$ – user649261 Mar 4 at 17:26
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    $\begingroup$ @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) $\endgroup$ – Noah Schweber Mar 4 at 19:40
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    $\begingroup$ @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.\ \ $ $\endgroup$ – Bill Dubuque Mar 4 at 20:44

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