# Equivalence Relation on Integers Modulo 5 induced by Squaring

## Theorem

Let $\beta$ denote the relation defined on the integers $\Z$ by:

$\forall x, y \in \Z: x \mathrel \beta y \iff x^2 \equiv y^2 \pmod 5$

Then $\beta$ is an equivalence relation.

### Number of $\beta$-Equivalence Classes

The number of distinct $\beta$-equivalence classes is $3$:

 $\ds \eqclass 0 \beta$  $\ds$ $\ds \eqclass 1 \beta$ $=$ $\ds \eqclass 4 \beta$ $\ds \eqclass 2 \beta$ $=$ $\ds \eqclass 3 \beta$

### Addition Modulo $\beta$ is not Well-Defined

Let the $+_\beta$ operator ("addition") on the $\beta$-equivalence classes be defined as:

$\eqclass a \beta +_\beta \eqclass b \beta := \eqclass {a + b} \beta$

Then such an operation is not well-defined.

### Multiplication Modulo $\beta$ is Well-Defined

Let the $\times_\beta$ operator ("multiplication") on the $\beta$-equivalence classes be defined as:

$\eqclass a \beta \times_\beta \eqclass b \beta := \eqclass {a \times b} \beta$

Then such an operation is well-defined.

## Proof

Checking in turn each of the criteria for equivalence:

### Reflexivity

We have that for all $x \in \Z$:

$x^2 \equiv x^2 \pmod 5$

It follows by definition of $\beta$ that:

$x \mathrel \beta x$

Thus $\beta$ is seen to be reflexive.

$\Box$

### Symmetry

 $\ds x$ $\beta$ $\ds y$ $\ds \leadsto \ \$ $\ds x^2$ $\equiv$ $\ds y^2$ $\ds \pmod 5$ $\ds \leadsto \ \$ $\ds y^2$ $\equiv$ $\ds x^2$ $\ds \pmod 5$ $\ds \leadsto \ \$ $\ds y$ $\beta$ $\ds x$

Thus $\beta$ is seen to be symmetric.

$\Box$

### Transitivity

Let:

$x \mathrel \beta y$ and $y \mathrel \beta z$

for $x, y, z \in \Z$.

Then by definition:

 $\ds x^2$ $\equiv$ $\ds y^2$ $\ds \pmod 5$ $\ds y^2$ $\equiv$ $\ds z^2$ $\ds \pmod 5$ $\ds \leadsto \ \$ $\ds x^2$ $\equiv$ $\ds z^2$ $\ds \pmod 5$ $\ds \leadsto \ \$ $\ds x$ $\beta$ $\ds z$

Thus $\beta$ is seen to be transitive.

$\Box$

$\beta$ has been shown to be reflexive, symmetric and transitive.

Hence by definition it is an equivalence relation.

$\blacksquare$