Functional Equation Analysis 4
Let $ f : \mathbb{Q} \to \mathbb{Q} $ be a function such that $ f(1) = 2 $ and
\[f(xy) = f(x) f(y) - f(x + y) + 1\]for all $ x, $ $ y \in \mathbb{Q} $.
Let $ n $ be the number of possible values of $ f \left( \frac{1}{2} \right), $ and let $ s $ be the sum of all possible values of $ f \left( \frac{1}{2} \right) $. Find $ n \times s $.
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
Setting $ y = 1, $ we get
\[f(x) = 2f(x) - f(x + 1) + 1,\]so $ f(x + 1) = f(x) + 1 $ for all $ x \in \mathbb{Q} $. Then
\begin{align*}
f(x + 2) &= f(x + 1) + 1 = f(x) + 2, \\
f(x + 3) &= f(x + 2) + 1 = f(x) + 3,
\end{align*}and so on. In general,
\[f(x + n) = f(x) + n\]for all $ x \in \mathbb{Q} $ and all integers $ n $.
Since $ f(1) = 2, $ it follows that
\[f(n) = n + 1\]for all integers $ n $.
Let $ x = \frac{a}{b}, $ where $ a $ and $ b $ are integers and $ b \neq 0 $. Setting $ x = \frac{a}{b} $ and $ y = b, $ we get
\[f(a) = f \left( \frac{a}{b} \right) f(b) - f \left( \frac{a}{b} + b \right) + 1.\]Since $ f(a) = a + 1, $ $ f(b) = b + 1, $ and $ f \left( \frac{a}{b} + b \right) = f \left( \frac{a}{b} \right) + b, $
\[a + 1 = (b + 1) f \left( \frac{a}{b} \right) - f \left( \frac{a}{b} \right) - b + 1.\]Solving, we find
\[f \left( \frac{a}{b} \right) = \frac{a + b}{b} = \frac{a}{b} + 1.\]Therefore, $ f(x) = x + 1 $ for all $ x \in \mathbb{Q} $.
We can check that this function works. Therefore, $ n = 1 $ and $ s = \frac{3}{2}, $ so $ n \times s = \boxed{\frac{3}{2}} $.
Freeflows