# How can I complete my solution in function problem?

Let $$f:\mathbb R\to \mathbb R$$ with $$f(x)f(y) - f(xy) = x + y$$ for every $$x,y \in R$$.

Prove that:

a)$$f(0) = 1$$

b)$$f(x) = x + 1$$

My solution:

a) $$f(x)f(y) - f(xy) = x + y$$

$$f(0)f(0) - f(0\cdot0) = 0 + 0$$

$$f(0)^2 - f(0) = 0$$

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

Therefore $$f(0) = 0$$ or $$f(0) = 1$$

Since I must prove that $$f(0) = 1$$, I should reject the first solution. I can't figure out how though...

b) $$f(x)f(y) - f(xy) = x + y$$

By replacing $$y$$ with $$0$$

$$f(x)f(0) - f(0\cdot0) = x + 0$$

$$f(x)\cdot1 - 1 = x$$

$$f(x) = x + 1$$

• Look what happens in your calculations in bart b) if you assume $f(0)=0$. Can you derive a contradiction? Aug 27 '15 at 22:00
• I thought about that, $x = 0$ comes up. Aug 27 '15 at 22:02

If you take f(0) to be 0, replacing x by 0 and y by 1 we get:

f(1)·f(0) - f(0·1) = 0 + 1

f(1)·0 - 0 = 1

0 = 1

Which is a contradiction. Therefore, f(0) must be 1.

• 5 seconds faster :( Aug 27 '15 at 22:02
• I'm really sorry :O Aug 27 '15 at 22:07

Your solution looks fine. For the rest of (a), take $x=0$ and $y=1$. Then $f(0)f(1)-f(0)=1$, which does not work for $f(0)=0$.