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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 true for all rationals x but I still cannot show that for $x, y \in\mathbb R$ is correct.

ViktorStein
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johnny
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  • You probably want $f$ to be continuous. Am I right? (Or at least a way to force the real values of $f$ given the rational ones...) – Giovanni De Gaetano Jan 04 '12 at 12:28
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    A duplicate of this question for $\mathbb{Q}$ by the same user

    http://math.stackexchange.com/questions/96316/about-finding-the-function

     – nb1 Jan 04 '12 at 12:29
  • In this new question he asks to prove for real values what he "proved" for rational values in the other question. – Giovanni De Gaetano Jan 04 '12 at 12:31
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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. – Martin Sleziak Jan 04 '12 at 12:38

2 Answers2

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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 $\forall x \epsilon \mathbb{R}$.

Community
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nb1
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You have $f(1)=2$ so $f(x)=2f(x)-f(x+1)+1\iff f(x)=f(x+1)-1$ which implies $f(0)=1$ and $f(-1)=0$ so the function has three collinear points $(-1,0),(0,1)$ and $(1,2)$.

The equation of the line passing through these three points is $y=x+1$. It follows

$$\begin{cases}f(x)=x+1\\f(y)=y+1\\f(xy)=xy+1\\f(x+y)=x+y+1\end{cases}\Rightarrow f(xy)=f(x)f(y)-f(x+y)+1$$ All the conditions are verified.

Piquito
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