Equivalence Relation on Integers Modulo 5 induced by Squaring/Multiplication Modulo Beta is Well-Defined

Theorem

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

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

We have that $\beta$ is an equivalence relation.

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

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

Let:

$\ds x, x'$

$\in$

$\ds \eqclass x \beta$

$\ds y, y'$

$\in$

$\ds \eqclass y \beta$

We have:

$\ds x^2$

$\equiv$

$\ds \paren {x'}^2$

$\ds \pmod 5$

$\ds y^2$

$\equiv$

$\ds \paren {y'}^2$

$\ds \pmod 5$

$\ds \leadsto \ \ $

$\ds \paren {x y}^2$

$\equiv$

$\ds \paren {x' y'}^2$

$\ds \pmod 5$

Hence the result.

$\blacksquare$

1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $3$: Equivalence Relations and Equivalence Classes: Exercise $9 \ \text{(ii)}$