Let $\bf N$ be the set of positive integers and let $\bf Q$ be the set of all rational numbers. Consider all functions $f:{\bf Z}\to{\bf Q}$. We say $f$ is a sum of $q\_1,q\_2,\dots,q\_s$ if for all positive integer $n$ the equality $f(n)=q\_1(n)+q\_2(n)+\dots+q\_s(n)$ holds. How can one prove that for each $f:{\bf Z}\to{\bf Q}$ there exist bijections $q\_1,q\_2,q\_3:{\bf Z}\to{\bf Q}$ such that $f=q\_1+q\_2+q\_3$? Is there an easy example of $f$ which is not presentable as a sum of two bijections?



Sum of three bijections
One key observation is that with $3$ functions, we are free to have *one* of them assume any rational value at any Natural number. This is not possible when we only have $2$ functions where after we select the value for $q\_1(n)$, we have $q\_2(n)$ completely determined by $f(n) - q\_1(n)$ and visa versa. The other key observation is that we can split $\mathbb{N}$ into the $3$ disjoint infinite subsets $A\_1, A\_2, A\_3$ with each subset consisting of the set of indices where we make the $q\_i$ assume the "next" rational value it has not already assumed according to some bijective enumeration. Specifically:


Fix an arbitrary function $f: \mathbb{N} \rightarrow \mathbb{Q}$ and an arbitrary bijection
$e: \mathbb{N} \rightarrow \mathbb{Q}$. The function $e$ induces a well-order, $<\_e$ on the set of rational numbers defined by $r\text{ }<\_e\text{ }s$ exactly when $e^{-1}(r) < e^{-1}(s)$ (i.e., $r$ is listed before $s$). It also induces a well-order $<\_e^\*$ on pairs of rationals defined by $\langle r, s\rangle <\_e^\* \langle t, u\rangle$ exactly when $r\text{ }<\_e\text{ }t$ or both $r = t$ and $s\text{ }<\_e\text{ }u$ (so-called lexicographical ordering).


We can then define each of the $q\_i$ by induction as follows:


If $n \equiv i \pmod 3$, define $q\_i(n)$ to be the $<\_e$-least rational value not already assumed by $q\_i(m)$ for $m < n$.


Then for the $j$ and $k$ such that $n \not\equiv j \pmod 3$ and $n \not\equiv k \pmod 3$, let $\langle r\_j, r\_k\rangle$ be the $<\_e^\*$-least pair such that $r\_j$ was not assumed by any of the $q\_j(m)$ and $r\_k$ was not assumed by any of the $q\_k(m)$ for $m < n$ and $r\_j + r\_k = f(n) - q\_i(n)$. Note that there is such a pair since there are infinitely many pairs $\langle r\_j, r\_k\rangle$ satisfying the equality and only finitely many pairs excluded from consideration. Then define $q\_j(n) = r\_j$ and $q\_k(n) = r\_k$.


