# How to show that there exist no function f,g:R->R satisfying f(x)g(y)=x+y, for all re

1. Jan 11, 2013

### thchian

1. The problem statement, all variables and given/known data
How to show that there exist no function f,g:R$\nearrow$R satisfying f(x)g(y)=x+y, for all real x,y?

2. Relevant equations

Any equation for showing function?

3. The attempt at a solution

how about starting with sumthing like let h(x,y)=x+y ??

Last edited: Jan 11, 2013
2. Jan 11, 2013

### HallsofIvy

Staff Emeritus
Re: How to show that there exist no function f,g:R->R satisfying f(x)g(y)=x+y, for al

I would recommend exploring what would have to be true if such a formula were true. For example, if f(x)g(y)= x+ y, then f(0) g(y)= y for all y so if we set a= f(0), and $a\ne 0$, g(y)= y/a. Similarly, f(x)g(0)= x so if b= g(0), and $b\ne 0$, then f(x)= x/b. That means that we must have f(x)g(y)= xy/ab for some numbers a and b, provided f(0) and g(0) are not equal to 0. You will need to determine what happens if f(0)= 0 and g(0)= 0.

3. Jan 11, 2013

### Staff: Mentor

Re: How to show that there exist no function f,g:R->R satisfying f(x)g(y)=x+y, for al

You could try to construct f(x) and g(y) via their derivatives and lead this to a contradiction.
Alternatively, consider f(0)g(a), f(a)g(0) and so on.

4. Jan 11, 2013

### Ray Vickson

Re: How to show that there exist no function f,g:R->R satisfying f(x)g(y)=x+y, for al

This assumes continuity, differentiability, etc., which was not part of the problem statement.

5. Jan 11, 2013

### Staff: Mentor

Re: How to show that there exist no function f,g:R->R satisfying f(x)g(y)=x+y, for al

We know that f(x)g(y) is differentiable. Anyway, we don't need it, as the result of the integral is already there:

Consider $x\neq-y$ => $g(y)\neq 0$:
f(x)=(x+y)/g(y) (you could derive both sides with respect to x here)
That leads to the same concept as HallsofIvy proposed.

6. Jan 12, 2013

### thchian

Re: How to show that there exist no function f,g:R->R satisfying f(x)g(y)=x+y, for al

so, i get this...
a=0 when $b\ne 0$
b=0 when $a\ne 0$
ab=0 when $a\ne 0$ and $b\ne 0$
then, i get nothing here... so, what does it mean? Am i approaching the wrong way?

7. Jan 13, 2013

### Staff: Mentor

Re: How to show that there exist no function f,g:R->R satisfying f(x)g(y)=x+y, for al