Prove that this is an equivalence relation and give all the different equivalence classes

Question

Let $R$ be a relation defined on real numbers by letting $a\mathrel R b$ iff $\cos (a) = \cos (-b)$. Prove that this is an equivalence relation and give all the different equivalence classes. Also show that this is all the classes and that they are different.

Answer 1

$\cos(-b) = \cos b$ for all real $b$, hence, your equivalence relation is equivalent to $a\mathcal Rb \iff \cos a = \cos b$.

You need to show that $\mathcal R$ is reflexive. Since $\cos (a) = \cos(-a) $, then $a\mathcal{R} a$. Hence, $\mathcal R$ is reflexive.

You need to show that $\mathcal R$ is symmetric. Since $\cos(a) = \cos(-b) \iff \cos(a) = \cos (b) \iff \cos b = \cos a \iff \cos b = \cos(-a),$ we have that $a \mathcal{R} b \implies b \mathcal{R} a $, so $\mathcal R$ is symmetric.

You need to show that $\mathcal R$ is transitive. This means that if $a \mathcal{R} b$ and $b\mathcal{R} c$, then it must follow that $a \mathcal R c$. So supposing $\cos (a) = \cos (-b) = \cos(b) $ and also $\cos(b) = \cos( -c ) $, it certainly follows by transitivity of equality that $\cos (a) = \cos (-c).$ Hence $a \mathcal{R} c$.

Therefore, $\mathcal R$ is reflexive, transitive, and symmetric, and is therefore, an equivalence relation on the set of real numbers.

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An equivalence class $[a]$ of $\mathcal R$ consists of all real numbers $x$ such that $x \mathcal R a$. Recall that an equivalence class partitions the set on which it is defined, meaning the every real number is in one and only one equivalence class, so that the union of the equivalence classes IS the set of reals, and so that the intersection of any two equivalence classes is empty.

Let's just explore $a \in \mathbb R$. $\cos a = \cos(-a) = \cos (2\pi + a) = \cos (2\pi -a) = \cos (2k \pi + a) = \cos (2k\pi - a)$, where $k$ is any integer. So for any given $a\in \mathbb R$, the equivalence class of $a$ is given by $$[a] = \{x\in \mathbb R\mid x=2k\pi \pm a,\;k \in \mathbb Z\}.$$

Answer 2

Hint: Note that $\cos(-x)=\cos x$ for all $x$ and hence this relation is of type $aRb\iff f(a)=f(b)$.

Answer 3

Notice since $\cos (-x) = \cos x$, then $x \mathcal{R} x$

Secondly, if $x \mathcal{R} y$ and $y \mathcal{R} z$, then we have that $\cos (x) = \cos (-y) = \cos(y) $ and we also have $\cos(y) = \cos( -z ) $. Therefore, we must have $\cos (x) = \cos (-z) \implies x \mathcal{R} z$

Since $\cos(x) = \cos(-y) \iff \cos(y) = \cos(-x)$. Then we have that $x \mathcal{R} y \implies y \mathcal{R} x $

Therefore, we have checked that the relation on the reals is reflexive, transitive, and symmetric, therefore, $\mathcal{R}$ must be an equivalence relation.