Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm trying to find functions $f(x)$ and $g(y)$ such that $$f(x)\cdot g(y) = x + y$$

I can't seem to find a single solution to this problem. Anything I try becomes of the form $f(x,y) \cdot g(y) or f(x) \cdot g(x,y)$

Here is my work so far:

$$f(x)g(y) = x + y$$

$$f'(x)g(y) + f(x)g'(y)\frac{dy}{dx} = 1 +\frac{dy}{dx} $$

$$f'(x)g(y) - 1 = \frac{dy}{dx} - f(x)g'(y)\frac{dy}{dx} $$

$$f'(x)g(y) - 1 = \frac{dy}{dx} (1 - f(x)g'(y))$$

$$\frac{f'(x)g(y) - 1}{1 - f(x)g'(y)} = \frac{dy}{dx} $$

But since $g(y)$ doesn't have a known or restricted order, I have no idea what $f(x)$ would be:

I am considering taking higher derivatives and then using substitutions (ex: $g(y) = (x+y)/f(x))$ but I'm not sure if that will work.

share|cite|improve this question
up vote 23 down vote accepted

You have that $$\tag 1 g(0)f(0)=0$$ while $$f(1)g(0)=1$$ $$f(0)g(1)=1$$ The first equations says that either $g(0)=0$ or $f(0)=0$, or both. But this contradicts the last two equations.

ADD Note that what I wrote holds also when $x=-y$, and when $x=0$, $y=\alpha$ a constant, and vice-versa. In fact, the last observation means your function cannot be defined for any value of $x$ or $y$. Indeed, we have that for any $\alpha,\beta\in\Bbb R$

$$f(0)g(\alpha)=\alpha$$ $$f(\beta)g(0)=\beta$$

But beacuse of $(1)$, the last relations are impossible, so we cannot define either $f$ or $g$ for any $x\in\Bbb R$.

share|cite|improve this answer
The last line can be simplified to "multiplying the last two relations you get $0=1$." :) – N. S. May 2 '13 at 19:40
But can the following possible? – frogeyedpeas May 2 '13 at 19:41
@N.S. "Simplified"? Is it really that complicated? =) – Pedro Tamaroff May 2 '13 at 19:41
@frogeyedpeas The point of my answer is that no such decomposition can exist. – Pedro Tamaroff May 2 '13 at 19:41
15 seconds of posting delay is worth commenting about, which is why I'm doing it too! And it's taking me 30+ seconds to do so, too. :) – Kaz May 2 '13 at 23:20

This would mean that $\frac{a+y}{b+y}$ would have to be a constant function of $y$ for all $a,b$. But $\frac{a+y}{b+y}=1-\frac{b-a}{b+y}$ is only constant when $a=b$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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