Help me find the equivalence class for a relation of a set of nonzero real numbers

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

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

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 $[1] =\mathbb{Q}\backslash\{0\}$.

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

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

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 $\mathbb{Q} - \{1\}$.