Absolute Value induces Equivalence Compatible with Integer Multiplication

Theorem
Let $\Z$ be the set of integers.

Let $\mathcal R$ be the relation on $\Z$ defined as:
 * $\forall x, y \in \Z: \left({x, y}\right) \in \mathcal R \iff \left\vert{x}\right\vert = \left\vert{y}\right\vert$

where $\left\vert{x}\right\vert$ denotes the absolute value of $x$.

Then $\mathcal R$ is a congruence relation for integer multiplication, but not for integer addition.

Proof
$\mathcal R$ is trivially shown to be an equivalence relation:


 * Reflexive:
 * $\forall x \in \Z: \left\vert{x}\right\vert = \left\vert{x}\right\vert$


 * Symmetric:
 * $\forall x, y \in \Z: \left\vert{x}\right\vert = \left\vert{y}\right\vert \implies \left\vert{y}\right\vert = \left\vert{x}\right\vert$


 * Transitive:
 * $\forall x, y, z \in \Z: \left\vert{x}\right\vert = \left\vert{y}\right\vert \land \left\vert{y}\right\vert = \left\vert{z}\right\vert \implies \left\vert{x}\right\vert = \left\vert{z}\right\vert$

Now suppose:
 * $\left\vert{x_1}\right\vert = \left\vert{x_2}\right\vert$
 * $\left\vert{y_1}\right\vert = \left\vert{y_2}\right\vert$

Then by definition of absolute value:

That is:
 * $\left({x_1 y_1, x_2 y_2}\right) \in \mathcal R$

That is, $\mathcal R$ is a congruence relation for integer multiplication.

Now consider that:
 * $\left\vert{-1}\right\vert = \left\vert{1}\right\vert$, so $-1 \mathcal R 1$
 * $\left\vert{2}\right\vert = \left\vert{2}\right\vert$, so $2 \mathcal R 2$

But $-1 + 2 = 1$ while $1 + 2 = 3$.

But it does not hold that $\left\vert{1}\right\vert = \left\vert{3}\right\vert$.

So, $\mathcal R$ is not a congruence relation for integer addition.