Formula: the Chinese Remaindering Theorem

Let $m$ and $n$ be two relatively prime elements of an Euclidean domain $R$ . (You may think of $R = {\mbox{${\mathbb{Z}}$}}$ or $R = {\mbox{${\mathbb{Q}}$}}[x]$ .) Let $s, t \in R$ be such that $s \, m + t \, n = 1$ . For every $a, b \in R$ there exists $c \in R$ such that

$$(\forall x \in R) \ \ \ \ \ \left\{ \begin{array}{l} x \equiv a \... ...uiv b \mod{\, n} \\ \end{array} \right. \ \ \iff \ \ x \equiv c \mod{\, m \, n}$$ (1)

where a convenient $c$ is given by

$$c \ = \ a + (b - a) \, s \, m = b + (a - b) t\, n$$ (2)

Therefore, for every $a, b \in R$ the system of equations

$$\left\{ \begin{array}{l} x \equiv a \mod{\, m} \\ x \equiv b \mod{\, n} \\ \end{array} \right.$$ (3)

has a solution.