# The function $f : \mathbb{R} \rightarrow \mathbb{R}$ satisfies $f(x) f(y) - f(xy) = x + y$ for all $x$, $y \in \mathbb{R}$. Find $f(x)$.

The function $f : \mathbb{R} \rightarrow \mathbb{R}$ satisfies $f(x) f(y) - f(xy) = x + y$ for all $x$, $y \in \mathbb{R}$. Find $f(x)$.

With $y=1$, we can establish that $$f(x)f(1)-f(x)=x+1\implies f(x)=\frac{x+1}{f(1)-1}\cdot\tag{*}$$ Note that the division by $f(1)-1$ is valid because $f(1)$ cannot be $1$ (if it were, $f(x)f(1)-f(x)$ would be identically $0$ whereas $x+1$ would vary with $x$.) It remains to find $f(1)$. First, $$x=0,y=0\implies [f(0)]^2-f(0)=0\implies f(0)=0\text{ or }f(0)=1.$$ But $f(0)$ cannot be $0$, as we can see below: $$x=1,y=0\implies f(1)f(0)-f(0)=1\implies f(0)=1, f(1)=2.$$ We conclude that $f(x)=x+1$.

• It wasn't obvious to me at first why f(0) couldn't be 0. So I tried to apply it to the $f(x)=\frac{x+1}{f(1)-1} \text{ and got } f(0)=0=\frac{0+1}{f(1)-1}$ But 1 cannot be 0. Awesome answer @yurnero. P.S. Tried to up your answer again but found out it only takes away if you do it again :p – randomgirl Mar 7 '15 at 0:39
• You can also just let $x=1$ and $y=1$ and solve the quadratic. No need for the funny reasoning about $f(0)$. – DanielV Mar 7 '15 at 1:48
• @DanielV I thought about that but it wasn't clear to me how to eliminate $f(1)=-1$ without further substitution. – yurnero Mar 7 '15 at 2:42

Setting $y = 0$ gives $f[x] f[y] - f[x y] = f[0] (f[x] - 1)$ and also $f[x] f[y] - f[x y] = x$

$f[0] (f[x] - 1) == x$

Setting $x=0$ now gives

$f[0] (f[0] - 1) = 0$

If $f[0]$ is not zero, then $f[0] - 1$ must be zero.

So we have $f[0]=1$.

But then:

$f[x] - 1 = x$

and

$f[x] = x + 1$

• Please see this tutorial on how to format mathematics on this site. – N. F. Taussig Mar 7 '15 at 1:24
• Thank's. Try to do better next time. – Frieder Mar 7 '15 at 1:37

Setting $y = 0$, we get [f(0) f(x) - f(0) = x,] so $f(0) [f(x) - 1] = x$ for all $x$. In addition, setting $x = 0$, we get $f(0) [f(0) - 1] = 0$, so $f(0) = 0$ or $f(0) = 1$. But $f(0) [f(x) - 1] = x$ for all $x$, so $f(0)$ cannot be 0, which means $f(0) = 1$. Then $f(x) - 1 = x$, or $f(x) = \boxed{x + 1}$ for all $x$. It is easy to verify that this solution works.