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

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.

– jMdA
Commented Apr 8, 2021 at 22:05

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

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

• 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. Commented Apr 8, 2021 at 22:08

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