Originally Posted by

**emakarov** Well, this depends on the definition of the sgn function. I would say $\displaystyle \text{sgn}(0)$ should be $\displaystyle 0$, so $\displaystyle \text{sgn}(\sin(0))=\text{sgn}(-\sin(0))=0$.

However, I don't see how this changes anything. If, for some function $\displaystyle g$ and some transitive relation $\displaystyle \equiv$, the relation $\displaystyle R(x,y)$ is defined as $\displaystyle g(x)\equiv g(y)$, then $\displaystyle R$ is transitive. If we have some unusual case, e.g., when $\displaystyle g(x)$ is not defined, then we cannot say $\displaystyle R(x,y)$ for any $\displaystyle y$. If $\displaystyle g(x)$ and $\displaystyle g(y)$ are defined but $\displaystyle g(x)\equiv g(y)$ is not, then again $\displaystyle x$ and $\displaystyle y$ are not related, so the premise of the transitivity property is false. What is important is for $\displaystyle g$ to be a function, i.e., return exactly one value for each $\displaystyle x$.

Do you have any counterexample in mind?