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Define a function $f\colon\mathbb{R}\to\mathbb{R}$ which is continuous and satisfies
$f(xy)=f(x)f(y)-f(x+y)+1$ for all $x,y\in\mathbb Q$. With a supp condition $f(1)=2$. (I didn't notice that.)

How to show that $f(x)=x+1$ for all $x$ that belong to $\mathbb{R}$?i got the ans from Paul that it is ture for all rationals x but i still cannot show that for $x,y\in\mathbb R$ is correct.

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    You know that the function is continuous. Using the fact that for $x\in\mathbb R$ there is a sequence of rational numbers such that $x_n\to x$ you get $f(x_n)=x_n+1 \to f(x)$. See also [this question](http://math.stackexchange.com/questions/505/can-there-be-two-distinct-continuous-functions-that-are-equal-at-all-rationals).2012-01-04

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I assume that $f(x)$ is continuous.

Any function of the form

$f(x)=\begin{cases} x+1 &\mbox{if } x \epsilon \mathbb{Q} \\ \not=x+1 & \mbox{if } x\not\epsilon \mathbb{Q} \end{cases} $ is discontinous.

It is already proved that $f(x)=x+1$ for $x \epsilon \mathbb{Q}$.

About finding the function such that $f(xy)=f(x)f(y)-f(x+y)+1$

Therefore, $f(x)=x+1$ for $\forall x \epsilon \mathbb{R}$.

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    o i get it now,thx and it is straight forward.2012-01-04