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I have a problem with this task. If anyone had a similar problem it would help me.

The task is:

In the set $S\in[-\pi,\pi]$ a binary relation is defined ρ with $xρy⟺sgn(\sin(x-\pi))=sgn(\sin(y))$. Examine whether the relation is an equivalence relation.

I tried this:

Reflexivity: $(\forall x \in S) x\rho x$

$sgn(\sin(x-x))=sgn(\sin(x))\\sgn(0)=sgn(sin(x))\\sgn(\sin(x))=0\\\sin(x)=0\\x=arcsin(0)=0\in [-\pi,\pi]$

The relation is reflexive

Symmetry:

$(\forall x,y \in S) x\rho x \iff y\rho x\\sgn(\sin(x-\pi))=sgn(\sin(y))\iff sgn(\sin(y)))=sgn(\sin(x-\pi))$

The relation is symmetric.

And now I don't know how to prove transitivity ?

Thanks in advance !

LogicNotFound
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    I don't understand your proof of Reflexivity. Say $x=\frac {\pi}4$. Then $\sin(x)>0$ but $x-\pi=-\frac {3\pi}4$ has negative sin, no? – lulu Nov 07 '20 at 18:52
  • Now I see that my reflexivity is not good – LogicNotFound Nov 07 '20 at 18:55
  • Thw relation has $x-\pi$, but in your reflexive proof, you changed that to $x - x$. And in the end you ended up "proving" $\forall x \in [-\pi, \pi], x = 0$, and declared that must be true since $0$ is also in $[-\pi,\pi]$!? You need to keep better of what you are trying to prove! Where does $x - \pi$ lie on the circle relative to $x$ itself? What can you say about the sines of $x$ and $x-\pi$? – Paul Sinclair Nov 08 '20 at 04:51

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