partial fractions

Every fractional number, i. e. such a rational number $\frac{m}{n}$ that the integer $m$ is not divisible by the integer $n$, can be decomposed to a sum of partial fractions as follows:

$$\frac{m}{n}=\frac{m_1}{p_1^{{\nu }_1}}+\frac{m_2}{p_2^{{\nu }_2}}+\mathrm{\cdots }+\frac{m_t}{p_t^{{\nu }_t}}$$

Here, the $p_i$’s are distinct positive prime numbers, the ${\nu }_i$’s positive integers and the $m_i$’s some integers.  Cf. the partial fractions of expressions.

Examples:

$$\frac{6}{289}=\frac{6}{{17}^2}$$

$$-\frac{1}{24}=-\frac{3}{2^3}+\frac{1}{3^1}$$

$$\frac{1}{504}=-\frac{1}{2^3}+\frac{32}{3^2}-\frac{24}{7^1}$$

How to get the numerators $m_i$ for decomposing a fractional number $\frac{1}{n}$ to partial fractions?  First one can take the highest power $p^{\nu }$ of a prime $p$ which divides the denominator $n$.  Then  $n=p^{\nu }u$,  where  $\gcd (u,p^{\nu })=1$.  Euclid’s algorithm gives some integers $x$ and $y$ such that

$$1=xu+y{p}^{\nu }.$$

Dividing this equation by $p^{\nu }u$ gives the

$$\frac{1}{n}=\frac{1}{p^{\nu }u}=\frac{x}{p^{\nu }}+\frac{y}{u}.$$

If $u$ has more than one distinct prime factors, a similar procedure can be made for the fraction $\frac{y}{u}$, and so on.

Note.  The numerators  $m_1$, $m_2$, …, $m_t$  in the decomposition are not unique.  E. g., we have also

$$-\frac{1}{24}=-\frac{11}{2^3}+\frac{4}{3^1}.$$

Cf. the programme “Murto” (in Finnish) or “Murd” (in Estonian) or “Bruch” (in German) or “Bråk” (in Swedish) or “Fraction”(in French) http://www.wakkanet.fi/ pahio/ohjelmi.htmlhere.