Congruence by Product of Moduli/Real Modulus

Theorem

Let $a, b, z \in \R$.

Let $a \equiv b \pmod z$ denote that $a$ is congruent to $b$ modulo $z$.

Then $\forall y \in \R, y \ne 0$:

$a \equiv b \pmod z \iff y a \equiv y b \pmod {y z}$

Let $y \in \R: y \ne 0$.

Then:

$\ds a$

$\equiv$

$\ds b$

$\ds \pmod z$

$\ds \leadstoandfrom \ \ $

$\ds a \bmod z$

$=$

$\ds b \bmod z$

Definition of Congruence

$\ds \leadstoandfrom \ \ $

$\ds y \paren {a \bmod z}$

$=$

$\ds y \paren {b \bmod z}$

Left hand implication valid only when $y \ne 0$

$\ds \leadstoandfrom \ \ $

$\ds \paren {y a} \bmod \paren {y z}$

$=$

$\ds \paren {y b} \bmod \paren {y z}$

$\ds y a$

$\equiv$

$\ds y b$

$\ds \pmod {y z}$

Definition of Congruence

Hence the result.

Note the invalidity of the third step when $y = 0$.

$\blacksquare$