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