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 $b$

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

  • $\begingroup$ An easier argument is to note that if $h(x,y)=f(x)+g(y)$ then $h(a,b)+h(c,d)=h(a,d)+h(b,c)$. $\endgroup$ – Lord Shark the Unknown Jul 30 '19 at 8:20
  • $\begingroup$ @Lord Shark the Unknown: Did you mean:$\;h(a,b)+h(c,d)=h(a,d)+h(c,b)\,$? $\endgroup$ – quasi Jul 30 '19 at 8:29
  • $\begingroup$ More than likely! $\endgroup$ – Lord Shark the Unknown Jul 30 '19 at 8:45
  • $\begingroup$ @Lord Shark the Unknown: As a followup to your comment, I have asked: math.stackexchange.com/questions/3308222 $\endgroup$ – quasi Jul 30 '19 at 10:40

$f(a)+f(b)$ 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.


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