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: $$\frac{m}{n} = \frac{m_1}{p_1^{\nu_1}}+\frac{m_2}{p_2^{\nu_2}}+\cdots+\frac{m_t}{p_t^{\nu_t}}$$ 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 $$1 = xu+yp^{\nu}.$$ Dividing this equation by $p^{\nu}u$ gives the decomposition $$\frac{1}{n} = \frac{1}{p^{\nu}u} = \frac{x}{p^{\nu}}+\frac{y}{u}.$$ 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) here.