Congruence by Product of Moduli/Real Modulus
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Theorem
Let $a, b, z \in \R$.
Let $a \equiv b \pmod z$ denote that $a$ is congruent to $b$ modulo $z$.
Then $\forall y \in \R, y \ne 0$:
- $a \equiv b \pmod z \iff y a \equiv y b \pmod {y z}$
Proof
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 $y \ne 0$ | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \paren {y a} \bmod \paren {y z}\) | \(=\) | \(\ds \paren {y b} \bmod \paren {y z}\) | Product Distributes over Modulo Operation | ||||||||||
\(\ds y a\) | \(\equiv\) | \(\ds y b\) | \(\ds \pmod {y z}\) | Definition of Congruence |
Hence the result.
Note the invalidity of the third step when $y = 0$.
$\blacksquare$
Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $24$