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The question states: Let $A = \mathbb{R} \setminus \{0\}$, the set of nonzero real numbers. Let $R$ be the relation on $A$ defined by:

$xRy$ if $x/y$ is rational

(a) Prove that $R$ is an equivalence relation on $A$.

(b) Determine equivalence class $[1]$.

So for part (a) I got:

Reflexive:

$\forall x \in \mathbb{R}, x/x \in \mathbb{Q} \implies xRx$

Symmetric:

Suppose $xRy$, then $x/y \in \mathbb{Q}$ is the same as $y/x \in \mathbb{Q}$, thus $yRx$

Transitive:

$x, y, z \in \mathbb{R}$

Let $xRy$ be $x/y \in \mathbb{Q}$ and $yRz$ be $y/z \in \mathbb{Q}$, thus $x/z \in \mathbb{Q}$ and $xRz$

(b) I'm not sure I understand correctly. I know that $[a] = \{x \in A| xR1\}$, but wouldn't that mean all nonzero real numbers? Since to get $1$, $x$ and $y$ should equal each other.

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  • $\begingroup$ Please improve your mathematical type setting. $\endgroup$
    – jMdA
    Commented Apr 8, 2021 at 22:05

4 Answers 4

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$$[1]=\{x\in \Bbb R-\{0\}\;:\: \frac x1\in \Bbb Q\}$$

$$=\Bbb Q-\{0\}$$

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  • $\begingroup$ Thank you. This answer made the most sense to me as it's basically exactly what I thought. I just didn't know how to write it. $\endgroup$ Commented Apr 8, 2021 at 22:08
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The equivalence class of $1\in \mathbb{R}\backslash\{0\}$ is given by all non zero real numbers $x$ such that $xR1$, i.e. $x/1=x\in \mathbb{Q}$ by definition of $R$. Therefore we have $[1] =\mathbb{Q}\backslash\{0\}$.

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An equivalence class for $1$ will be defined as follows:

$$[1]=\{x \in\mathbb{R}-\{0\}: x\mathrm{R}1\}=\{x \in\mathbb{R}-\{0\}: x/1=x \in\mathbb{Q}\}.$$

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I just want to add that this is a special case of a quite general phenomenon related to quotient groups.

Let $(G, \cdot)$ be a group, and let $H \subseteq G$ be a normal subgroup. Then consider the relation $R \subseteq G^2$ defined by $x R y$ if and only if $x y^{-1} \in H$. Then $R$ is an equivalence relation.

The proof is quite simple. We see that $x x^{-1} = e \in H$, so $xRx$ for all $x$; thus, we have reflexivity. If $x R y$, then we have $x y^{-1} \in H$; since $H$ is normal, we thus have $y^{-1} x \in H$. Then we have $y x^{-1} = (y^{-1} x)^{-1} \in H$; then $y R x$. Thus, we have symmetry. Finally, if we have $x R y$ and $y R z$, then we have $x y^{-1} \in H$ and $y z^{-1} \in H$; this means that $(x y^{-1})(y z^{-1}) = x z^{-1} \in H$; thus, $x R z$. This gives us transitivity.

This allows us to define the quotient group, a critical concept in group theory.

In this case, the group is $\mathbb{R} - \{0\}$ under multiplication, and the subgroup is $\mathbb{Q} - \{0\}$ under multiplication.

In particular, the equivalence class of $x$ is always $\{xz | z \in H\}$, since $y R x$ iff $y x^{-1} \in H$ iff there exists $z \in H$ such that $y = xz$. The special case of $x = e$ gives us the set $\{z | z \in H\} = H$. So the equivalence class of 1, the identity element, is going to be the full subgroup $\mathbb{Q} - \{1\}$.

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