Is the absolute value function an equivalence relation?

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Let $f \colon \mathbb{R} \to \mathbb{R}$ be the well-known absolute value function: $f(x) = \lvert x \rvert$. Show that f, considered as a relation, is neither reflexive nor symmetric. Show also that it is transitive.

I'm not sure how to proceed on this one. Is the absolute value function not an equivalence relation?

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This relation is defined by

$$(\forall x,y\in\Bbb R)\; (x \mathcal R y\iff y=|x|)$$

So, it is not reflexive since $-1$ is not in relation with $ -1 $ because $|-1|\ne -1$.

It is not symetric since $-1 $ is in relation with $ 1$ but $ 1 $ is not.

It is transitive :

$$x\mathcal Ry \quad \text{ and } \quad y\mathcal R z\implies$$ $$|x|=y \quad \text{ and } \quad|y|=z \implies |x|=|y|=z$$

$$\implies x \mathcal R z$$