Can I split $\frac{1}{a-b}$ into the form $f(a)+f(b)$? I was wondering if I can split apart the following fraction
\begin{align}
\frac{1}{a-b}
\end{align}
into the form:
\begin{align}
f(a)+f(b)
\end{align}
where $f(a)$ and $f(b)$ is some function in terms of $a$ and $b$

In fact, we can't even split it as
$$\frac{1}{a-b}=f(a)+g(b)$$
for any functions $f,g$.

Suppose instead that functions $f,g$ from $\mathbb{R}$ to $\mathbb{R}$ are such that
$$f(a)+g(b)=\frac{1}{a-b}$$
for all $a,b\in\mathbb{R}$ with $a\ne b$.

Then we would have
\begin{align*}
&
\begin{cases}
f(x+1)+g(x)={\Large{\frac{1}{(x+1)-x}}}={\Large{\frac{1}{1}}}=1\\[4pt]
f(x-1)+g(x)={\Large{\frac{1}{(x-1)-x}}}={\Large{\frac{1}{-1}}}=-1\\
\end{cases}
\\[6pt]
\implies\!\!\!\!&\;\;\;\;f(x+1)-f(x-1)=2
\qquad\qquad\qquad\qquad\qquad
\\[4pt]
\end{align*}
but we would also have
\begin{align*}
&
\begin{cases}
f(x+1)+g(x-2)={\Large{\frac{1}{(x+1)-(x-2)}}}={\Large{\frac{1}{3}}}\\[4pt]
f(x-1)+g(x-2)={\Large{\frac{1}{(x-1)-(x-2)}}}={\Large{\frac{1}{1}}}=1\\
\end{cases}
\\[6pt]
\implies\!\!\!\!&\;\;\;\;f(x+1)-f(x-1)=\frac{1}{3}-1=-\frac{2}{3}\\[4pt]
\end{align*}
contradiction.
