Equivalence Relation on Integers Modulo 5 induced by Squaring

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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 $3$:

\(\displaystyle \eqclass 0 \beta\) \(\) \(\displaystyle \)
\(\displaystyle \eqclass 1 \beta\) \(=\) \(\displaystyle \eqclass 4 \beta\)
\(\displaystyle \eqclass 2 \beta\) \(=\) \(\displaystyle \eqclass 3 \beta\)


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 operator 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 operator 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

\(\displaystyle x\) \(\beta\) \(\displaystyle y\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle x^2\) \(\equiv\) \(\displaystyle y^2\) \(\displaystyle \pmod 5\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle y^2\) \(\equiv\) \(\displaystyle x^2\) \(\displaystyle \pmod 5\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle y\) \(\beta\) \(\displaystyle x\)

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:

\(\displaystyle x^2\) \(\equiv\) \(\displaystyle y^2\) \(\displaystyle \pmod 5\)
\(\displaystyle y^2\) \(\equiv\) \(\displaystyle z^2\) \(\displaystyle \pmod 5\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle x^2\) \(\equiv\) \(\displaystyle z^2\) \(\displaystyle \pmod 5\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle x\) \(\beta\) \(\displaystyle z\)

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.

$\blacksquare$


Sources