Partial Fraction decomposition of rational functions plays a role in calculus. Do the ideas extend to rational numbers?

Let me try to ask it precisely, though the answer I'm looking for may ultimately be to a different phrasing.

Given a rational number (in lowest terms) $a/b$ where $b$ factors as a product of distinct primes $b=p_1p_2\cdots p_n$ and $a<b$, does there always exist a decomposition of the form $$\frac{a}{b}=\frac{q_1}{p_1}+\cdots+\frac{q_n}{p_n}$$ where the $q_i$s are integers (and probably satisfying some sort of condition, perhaps only $|q_i|<p_i$).

Easy example: 1/6 = 1/2 - 1/3

Better example: 29/70 = 1/2 - 4/5 + 5/7

Of course it would be extra nice if we could extend to $b$ being composite and handle prime powers in the factorization the same as in the rational function case.

  • $\begingroup$ Hint: consider the case where $a=1$ and see if the result mentioned here helps. math.stackexchange.com/questions/1264619/… $\endgroup$
    – Steve Kass
    Jul 26, 2020 at 21:42
  • 1
    $\begingroup$ @Aeryk Without any restrictions on the $q_i$ such as $|q_i| \lt p_i$, there's a solution such as shown in Generalization/Extension of Bezout's Lemma where if you multiply both sides by $b$, the result in that link shows there's always a solution for $a = 1$, so you can then multiply by whatever $a$ you actually have to get the specific solution. $\endgroup$ Jul 26, 2020 at 22:10

2 Answers 2


I did not find a formal definition for the PFD of a fraction anywhere, but it is quite logical that it is possible to find it for any fraction $x = A/B \in \mathbb Q$, and I think the following definition should work fine and make the choice of coefficients unique, assuming that A/B is in lowest terms:

$$x = \text{round}(x) + \sum_{\text{prime }p~\mid~B} ~ \sum_{k=1}^{v_p(B)} \frac{c(p,k)}{p^k}$$

where the integers $|c(p,k)|<p$ are chosen such that the partial sum with only $c(p,j)$, $j\ge k$ is as close as possible to $x = A/B$. ($v_p$ is of course the p-valuation.)

Indeed, you get these $c(p,k)$ by simply computing $x\cdot p^k \pmod p$ (proceeding by decreasing k, and incrementally subtracting the determined terms from the "remaining" x to decompose), but you can choose the representative in the open interval $(-p,p)$. The additional condition (...as close as possible...) makes it unique, except for $\pm1/2$ where we can choose as to get the "round" we want, floor or ceiling, when the fractional part is 0.5.

With this definition, I rather get 29/70 = 1/2 + 1/5 - 2/7 (which is certainly nicer than 1/2 - 4/5 + 5/7).


You're correct. You don't even need the condition $a<b$.

The question amounts to asking if every $a$ can be expressed as a linear combination (integer scalars) of the numerators $$\frac{p_1p_2\cdots p_n}{p_1},\: \frac{p_1p_2\cdots p_n}{p_2},\: ...,\: \frac{p_1p_2\cdots p_n}{p_n}$$ This is true because the numerators above have a collective greatest common denominator of $1$.


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