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Define a relation $R$ on $\mathbb{Z}\times(\mathbb{Z}\setminus\{0\})$ by $(a,b)R(x,y)\iff ay=bx.$

$a)$ Prove that $R$ is an equivalence relation.

$b)$ Describe the equivalence classes corresponding to $R$.

I know that an equivalence relation satisfies transitivity, reflexivity, and symmetry. Also, I know that the Cartesian Product is $A\times B=\{(a,b):a\in A$ and $b\in B\}. $If I could get a hint for $a)$, that would be appreciated. For part $b)$, could someone explain what an equivalence class is? Please refrain from complete solutions. Thank you.

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    $\begingroup$ Please, try to make the title of your questions more informative (a glance on your profile shows this isn't the first one). E.g., Why does $a\le b$ imply $a+c\le b+c$? is much more useful for other users than A question about inequality. For more information on choosing a good title, see this post. $\endgroup$
    – Lord_Farin
    Oct 3, 2013 at 13:57
  • $\begingroup$ @Lord_Farin: I'll keep that in mind. Thanks for editing the title. $\endgroup$ Oct 3, 2013 at 16:25

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For a, you need to show all three elements of the definition are satisfied. So show that $(a,b)R(a,b)$, $(a,b)R(c,d) \implies (c,d)R(a,b)$ and transitivity. Go back to the definition of $R$ and they should fall out.

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Two elements $(a,b)$ and $(x,y)$ are in the same equivalence class iff they are related to each other. In this case, it happens iff $$ \frac{a}{b} = \frac{x}{y} $$ ie. They define the same rational number (For instance $(1,2) \sim (2,4)$.

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  • $\begingroup$ I think this exercise is a preliminary to defining the rational numbers, so using (the existence of) the rational numbers to show this is an equivalence relation would be a somewhat circular reasoning. $\endgroup$ Oct 3, 2013 at 14:03
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For (a), you really just need to go through each property.

e.g. to check if it's reflexive, is $(a,b)\text{~}(a,b)$ for all $(a,b)\in \mathbb{Z} \times\mathbb{Z} \backslash \{0\}$?

For (b), the equivalence class of an element is the set of all elements that are related to that element.

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