Congruence by Product of Modulo

From ProofWiki
Jump to: navigation, search

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

Personal tools
Namespaces
Variants
Actions
Navigation
ProofWiki.org
ToDo
Toolbox
Google AdSense