Congruence by Product of Modulo
From ProofWiki
Theorem
Let $a, b, z \in \R$.
Let $a \equiv b \left({\bmod\, z}\right)$ denote that $a$ is congruent to $b$ modulo $z$.
Then $\forall y \in \R, y \ne 0$:
- $a \equiv b \left({\bmod\, z}\right) \iff y a \equiv y b \left({\bmod\, y z}\right)$
Proof
Let $y \in \R: y \ne 0$.
Then:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle a\) | \(\equiv\) | \(\displaystyle b\) | \(\displaystyle \) | \(\displaystyle \pmod z\) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \iff\) | \(\displaystyle \) | \(\displaystyle a \, \bmod \, z\) | \(=\) | \(\displaystyle b \, \bmod \, z\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Definition of congruence by Modulo operation | ||
| \(\displaystyle \) | \(\displaystyle \iff\) | \(\displaystyle \) | \(\displaystyle y \left({a \,\bmod\, z}\right)\) | \(=\) | \(\displaystyle y \left({b \,\bmod\, z}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Left hand implication valid only when $y \ne 0$ | ||
| \(\displaystyle \) | \(\displaystyle \iff\) | \(\displaystyle \) | \(\displaystyle \left({y a}\right) \bmod\, \left({y z}\right)\) | \(=\) | \(\displaystyle \left({y b}\right) \bmod\, \left({y z}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Product Distributes over Modulo Operation | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle y a\) | \(\equiv\) | \(\displaystyle y b\) | \(\displaystyle \) | \(\displaystyle \pmod {y z}\) | \(\displaystyle \) | Definition of congruence by Modulo operation |
$\blacksquare$
Note the invalidity of the third step when $y = 0$.
Sources
- Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (1968): $\S 1.2.4$: Law $\text{C}$, Exercise $24$