Is the set of all sums-of-rationals-that-give-one countable? Some (but not all) sums of rational numbers gives us 1 as a result. For instance:
$$\frac12 + \frac12 = 1$$
$$\frac13 + \frac23 = 1$$
$$\frac37 + \frac{3}{14} + \frac{5}{14}  = 1$$
Is the set of all of these sums countable?
Sums that differ just by their addends ordering are the same. So $(\frac25+\frac35)$ is the same element as $(\frac35+\frac25$), and shouldn't be counted twice. 
Also, for avoiding duplicity, all fractions should be in their irreducible form. No no need to count $(\frac26 + \frac23)$ if we already have $(\frac13 + \frac23)$.
If the set is countable, what would be a mapping function from $\mathbb N$ to this set?
 A: Since the set of rationals is countable, the set of all finite sets of rationals is countable, and therefore the set of all finite sums of rationals is countable. In particular, the set of finite sets of rationals with sum $1$ is countable. An explicit bijection between this set and $\Bbb N$ would be messy.
Added: If you limit yourself to finite sums of positive rationals, you can order the sums as follows. For $n\in\Bbb Z^+$ let $\mathscr{F}_n$ be the set of all multisets $F$ of positive rationals such that $\sum F=1$, $|F|\le n$, and whenever $\frac{p}q\in F$ is in lowest terms, then $q\le n$. For example, the only member of $\mathscr{F}_1$ is $\{\!\{1\}\!\}$, and the only members of $\mathscr{F}_2$ are $\{\!\{1\}\!\}$ and $\left\{\!\!\left\{\frac12,\frac12\right\}\!\!\right\}$, and
$$\mathscr{F}_3=\left\{\{\!\{1\}\!\},\left\{\!\!\left\{\frac12,\frac12\right\}\!\!\right\},\left\{\!\!\left\{\frac13,\frac13,\frac13\right\}\!\!\right\},\left\{\!\!\left\{\frac23,\frac13\right\}\!\!\right\}\right\}\;.$$
If $F=\{\!\{r_1,\ldots,r_\ell\}\!\},G=\{\!\{s_1,\ldots,s_m\}\!\}\in\mathscr{F}_n$, where $r_1\ge\ldots\ge r_\ell$ and $s_1\ge\ldots\ge s_m$, define $F\prec_n G$ iff either $\ell<m$, or $\ell=m$ and $r_k>s_k$, where $k=\min\{i:r_i\ne s_i\}$; $\prec_n$ is a strict linear order on $\mathscr{F}_n$.
Let $\mathscr{F}=\bigcup_{n\in\Bbb Z^+}\mathscr{F}_n$, and for each $F\in\mathscr{F}$ let $n(F)=\min\{k\in\Bbb Z^+:F\in\mathscr{F}_k\}$. For $F,G\in\mathscr{F}$ define $F\prec G$ iff either $n(F)<n(G)$, or $n(F)=n(G)$ and $F\prec_{n(F)} G$; then $\langle\mathscr{F},\prec\rangle$ is a strict linear order that is order-isomorphic to $\langle\Bbb N,<\rangle$. I’d not care to try to write down the order isomorphism explicitly, but this does at least give the explicit ordering of $\mathscr{F}$ corresponding to such a bijection.
