# 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 $$b$$

• You may also wish to have a think about why writing it in the form $f(a) - f(b)$ is not possible either. – Theo Bendit Jul 30 '19 at 6:28
• Did you really mean $f(a)+f(b)$ or did you mean $f(a)+g(b)$? – bof Jul 30 '19 at 6:57
• We are not here to do your homework problems!!! – JaberMac yesterday

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.
• An easier argument is to note that if $h(x,y)=f(x)+g(y)$ then $h(a,b)+h(c,d)=h(a,d)+h(b,c)$. – Lord Shark the Unknown Jul 30 '19 at 8:20
• @Lord Shark the Unknown: Did you mean:$\;h(a,b)+h(c,d)=h(a,d)+h(c,b)\,$? – quasi Jul 30 '19 at 8:29
$$f(a)+f(b)$$ is symmetric: it does not change if you switch $$a$$ and $$b$$. But $$\frac 1 {a-b}$$ is not symmetric. So you cannot do this.