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 \(f(x)\).

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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 \(f(x)\).

Mathematics
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let \(y = 0\), \[f(x)f(0)-f(0) = x\tag{1}\] let \(x=0\) in above equation : \[f(0)f(0)-f(0) = 0\implies f(0) = 0\lor 1\] however \(f(0)=0\) is not possible (why?) therefore \(f(0)=1\) plug that back in equation \((1)\) to get \(f(x) = x+1\)
I was just fixing to ask why :p
I also noticed I get two possibilities for f(1)

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I don't know how to know which one is the right one for f(1)
\[ \text{Assume }f(0)=0\\ \begin{align*} f(0)f(x)-f(0)&=x\\ 0&=x \end{align*} \]
\[f^2(1)-f(1)-2=0 \\ (f(1)-2)(f(1)+1)=0 \\ f(1)=2 \text{ or } f(1)=-1\]
I also did a similar thing then ganeshie8 posted his answer.
just an example to convince myself on why there are false values : \(f = -1 \implies f^2 = 1 \implies f = \pm 1 \) ofcourse \(f=1\) is extraneous. since we're squaring the function, we are increasing the degree and thus increasing the number of solutions, so we must check for extraneous stuff in the end..
\[ \text{Assume }f(1)=-1\\ \begin{align*} f(1)f(x)-f(x)&=x+1\\ -2f(x)&=x+1\\ f(x)&=-\frac{x+1}{2}\\ f(x)f(y)-f(xy)&=\frac{x+1}{2}\frac{y+1}{2}-\frac{xy+1}{2}\\ &=\frac{xy+x+y+1}{4}-\frac{xy+1}{2}\\ &\neq x+y \end{align*} \]
ok cool and if f(1)=2 \[f(1)f(x)-f(x)=x+1 \\ 2f(x)-f(x)=x+1 \\ f(x)=x+1 \\ \text{ checking } \\ f(x)f(y)-f(xy)=(x+1)(y+1)-(xy+1) \\ = xy+x+y+1-xy-1 \\ =x+y\] great stuff @thomas5267
One of the question on the only mathematics competition I have participated is to find the value of \(f(0)\) with \(f(x)f(y)=f(x+y)\). It can be solved with a similar method.
@zmudz can you give us some background on these problems? Are you looking for hints, solution methods or the solution?
i assumed its one to one and onto function :3 f(x) f(y) - f(xy) = x + y yx-xy=f(y)+f(x) xy-xy=f(x)+f(y) 0=f(x)+f(y) f(x)+f(y)=0 f(x)=-f(y) so f(x)=-x satisfy :)
I can't figure out how you got any of that. But checking your solution... we see that \[-x \cdot (-y)-(-xy) \neq x+y\]
hmmm you are right

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