finding the elements of equivalence class that contains $(-5,8)$ on the relation $\sim$ in $D\times S$ as follows

Question

Let $D=\Bbb{Z}$ be an integral domain and $S = \Bbb{Z}\backslash p\Bbb{Z}$ for some prime $p \in \Bbb{Z}$. Define the relation $\sim$ on $D \times S$ as: $(a,b) \sim (c,d) \Leftrightarrow ad=bc, \forall (a,b),(c,d) \in D \times S$.

We know that $\sim$ is an equivalence relation. And then, we denote the equivalence classes that contains $(a,b)$ by $\frac{a}{b} = \{(c,d) \in D \times S \mid (c,d) \sim (a,b) \}$, for all $(a,b) \in D\times S$.

Now, let $p=7$. How to find all elements in the class that contains $(-5,8)$? If $p=7$, then $S = \Bbb{Z}_7$, right? It means, $S = \{0,1,2,\dots, 6\} \pmod 7$ which implies $(-5,8) = (-5,1)$, right? Thus, those all elements are in $\{(-5k,k) \mid 0\ne k \in \Bbb{Z} \}$ ?

Any ideas? Thanks in advanced.

Answer 1

The simplest way I can see to write the answer is $\{ (-5n, 8n) \mid n \in \Bbb Z \setminus 7 \Bbb Z \}$. These ordered pairs represent all fractions $\frac {-5n}{8n}$ that (a) reduce to $\frac{-5}{8}$ in lowest terms and (b) have a denominator that's relatively prime to $7$.

Answer 2

You are trying to compute the equivalence class of $(-5,8)$ by definition this is the set of elements $\{(a,b)\in \mathbb{Z}\times (\mathbb{Z}-7\mathbb{Z}): -5b=8a\}$. You can let $a$ range among the integers and $b$ among those integers that are not a multiple of 7. The class will be

$\{...,\frac{-40}{64},\frac{-30}{48},\frac{-25}{40},\frac{-20}{32},\frac{-15}{24},\frac{-10}{16},\frac{-5}{8},\frac{5}{-8},\dots\}$