How to show that the relation $xRy$ if $\sin(x-y)=0$ is transitive?

Question

"Logically" it seems transitive as if $x-y=k()$ and $y-z=k(\pi)$ then $x-z=k'(\pi)$ but how to put it into a good proof? also what would we be its equivalence classes since if it is transitive then it will be an equivalence relation (already proved it is reflexive and symmetric)

Answer 1

If $\sin(x-y)=0$ and $\sin(y-z)=0$, then\begin{align}\sin(x-z)&=\sin\bigl((x-y)+(y-z)\bigr)\\&=\sin(x-y)\cos(y-z)+\cos(x-y)\sin(y-z)\\&=0.\end{align}So, yes, your relation is transitive.

Answer 2

You want to prove $\sin(x-y) = 0 = \sin(y-z) = 0 \Rightarrow \sin(x-z) = 0$.

Now $\sin(x-y)=0 \Rightarrow \exists k_1 \in \mathbb{Z}$ such that $ x - y = k_1 \pi$. In the same way, $\exists k_2 \in \mathbb{Z}$ such that $ y - z = k_2 \pi$. So $x - z = x -y + y - z = (k_1 + k_2)\pi \Rightarrow \sin(x-z) = \sin((k_1 + k_2)\pi) = 0$ because $k_1 + k_2 \in \mathbb{Z}$.

So the equivalence classes are of the kind $x + \pi \mathbb{Z}$, with $x \in [0,\pi)$.

Answer 3

Here $xRy$ if $\sin(x-y)=0$, i.e., if $x-y$ is an integer multiple of $\pi$. So, if $x-y$ and $y-z$ are integer multiples of $\pi$, then $x-z$ will be again an integer multiple of $pi$, implying that the relation is transitive. The equivalence class corresponding to any $x$ will be $ {x+k \pi ; k is an integer}$.