Absolute Value induces Equivalence not Compatible with Integer Addition

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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 not a congruence relation for integer addition.


Proof

From Absolute Value Function on Integers induces Equivalence Relation, $\mathcal R$ is an equivalence relation.


However, consider that:

\(\displaystyle \left\vert{-1}\right\vert = \left\vert{1}\right\vert\) \(\implies\) \(\displaystyle -1 \mathop {\mathcal R} 1\)
\(\displaystyle \left\vert{2}\right\vert = \left\vert{2}\right\vert\) \(\implies\) \(\displaystyle 2 \mathop {\mathcal R} 2\)

By conventional integer addition:

$-1 + 2 = 1$

while:

$1 + 2 = 3$

But it does not hold that:

$\left\vert{1}\right\vert = \left\vert{3}\right\vert$

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

Hence the result, by Proof by Counterexample.

$\blacksquare$


Sources