Congruence by Factors of Modulo
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Theorem
Let $a, b \in \Z$.
Let $r$ and $s$ be coprime integers.
Then:
- $a \equiv b \pmod {r s}$ if and only if $a \equiv b \pmod r$ and $a \equiv b \pmod s$
where $a \equiv b \pmod r$ denotes that $a$ is congruent modulo $r$ to $b$.
Proof
Necessary Condition
This is proved in Congruence by Divisor of Modulus.
Note that for this result it is not required that $r \perp s$.
$\Box$
Sufficient Condition
Now suppose that $a \equiv b \pmod r$ and $a \equiv b \pmod s$.
We have by definition of congruence:
- $a \equiv b \pmod r \implies \exists k \in \Z: a - b = k r$
From $a \equiv b \pmod s$ we also have that:
- $k r \equiv 0 \pmod s$
As $r \perp s$, we have from Common Factor Cancelling in Congruence:
- $k \equiv 0 \pmod s$
So:
- $\exists q \in \Z: a - b = q s r$
Hence by definition of congruence:
- $a \equiv b \pmod {r s}$
$\blacksquare$
Examples
$n \equiv 7 \pmod {12}$ implies $n \equiv 3 \pmod 4$
Let $n \equiv 7 \pmod {12}$.
Then:
- $n \equiv 3 \pmod 4$
Warning
Let $r$ not be coprime to $s$.
Then it is not necessarily the case that:
- $a \equiv b \pmod {r s}$ if and only if $a \equiv b \pmod r$ and $a \equiv b \pmod s$
where $a \equiv b \pmod r$ denotes that $a$ is congruent modulo $r$ to $b$.
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: Law $\text{D}$
- 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 $18$