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Define a relation on the set of all real numbers $x, y \in \mathbb{R}$:

$x ≃ y$ if and only if $x − y \in \mathbb{Z}$

Prove that this is an equivalence relation, and find the equivalence class of the number $1/3$.

I proved that the relation is:

reflexive

$$x-x = 0 \in \mathbb{Z}$$

symmetric

$$x-y = y-x \in \mathbb{Z}$$

transitive

$$x-y \in \mathbb{Z}$$

$$y-z \in \mathbb{Z}$$

$$x-z \in \mathbb{Z}$$

Are these proofs enough?

I'm stuck on the step where I need to find the equivalence class.

RobPratt
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ph-quiett
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2 Answers2

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Opps, carefull: $$x-y\ne y-x$$

You should write: if $x\sim y$ then $x-y\in Z$ so $-(x-y) = y-x\in Z$ so $y\sim x$.

Also, if $x\sim y$ and $y\sim z$ then $x-y,y-z\in Z$, so $(x-y)+(y-z) \in Z$ so $x-z\in Z$ so $x\sim z$.

And equivance class is $$Z+{1\over 3} = \{...-{5\over 3},-{2\over 3},{1\over 3}, {4\over 3},{7\over 3},...\}$$

nonuser
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    @ph-quiett Pay particular attention to the way this answer uses important words telling what the easy algebra means. You put the algebra in your question but not the reasoning. – Ethan Bolker Apr 07 '19 at 19:34
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An equivalence class of an element $x$ is the set of all elements which are in relation $\sim$ with $x$. In your case $x=1/3$, so the equivalence class is the set of all elements $y$ such that $\frac{1}{3}-y = k \in \mathbb Z$. It follows that $y=\frac{1}{3}-k = l+\frac{1}{3}$, $l \in \mathbb Z$. Therefore, the equivalence class of $\frac{1}{3}$ is

$$\left[\frac{1}{3}\right]_\sim = \left\{k+\frac{1}{3} : k\in \mathbb Z\right\}$$

Described in common language, this is the set of all fractions with $3$ in the denominator and their numerator divided by $3$ gives the remainder $1$. Some of these numbers are $\frac{1}{3}, \frac{4}{3}, \frac{7}{3}, \cdots$

Haris Gušić
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