Let $\sim$ be a relation on $\mathbb R$ by $x \sim y \iff x-y \in\mathbb Q$. Show that $\sim$ is an equivalence relation. Show that the equivalence class $[0] = \mathbb Q$. What is $\left[\frac 13\right]$?
For equivalence relation we have
- $x\sim x \implies x - x = 0 \in \mathbb Q \implies \sim$ is reflexive
- $x\sim y \implies x - y \in \mathbb Q \implies y - x \in\mathbb Q \implies \sim$ is symmetric
- $(x\sim y) \wedge (y \sim z) \implies (x - y \in \mathbb Q) \wedge (y-z \in\mathbb Q) \implies \mathbb Q \ni (x-y)+(y-z) = x - z \implies x\sim z \implies \sim$ is transitive.
Then $\sim$ is an equivalence relation.
Second part: Note that $x \sim 0 \implies \mathbb Q\ni x - 0 = x$ for all $x$. Thus $[0] = \mathbb Q$.
Third part: Note that we have $0\sim \frac 13 \implies \mathbb Q \ni 0 - \frac 13 = -\frac 13 \implies 0 \in \left[\frac 13\right]$. Similarly $\frac 13 \sim 0 \implies \mathbb Q \ni \frac 13 - 0 = \frac 13 \implies \frac 13 \in [0]$. The two results combine to mean we have $[0] = \left[\frac 13\right]$.
Are these three valid? Feedback is welcomed!