7
$\begingroup$

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 $b$

$\endgroup$
  • 3
    $\begingroup$ 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
6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
26
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.