We show first that $ f(1) = 1$. Taking $ x = y = 1$, we have f(f(1)) = f(1). Hence $ f(1) = f(f(1)) = f(1 f(f(1)) ) = \frac{f(1)}{f(1)} = 1.$ Next we show that $ f(xy) = f(x)f(y)$. For any y we have $ 1 = f(1) = f(\frac{1}{f(y)} f(y)) = f(\frac{1}{f(y)})/y$, so if $ z = \frac{1}{f(y)}$ then $ f(z) = y$. Hence $ f(xy) = f(xf(z)) = \frac{f(x)}{z} = f(x) f(y)$. Finally, $ f(f(x)) = f(1 f(x)) = \frac{f(1)}{x} = \frac{1}{x}$. We are not required to find all functions, just one. So divide the primes into two infinite sets S = {p_1, p_2, ... } and T= {q_1, q_2, ... }. Define f(p_n) = q_n, and f(q_n) = 1/p_n. We extend this definition to all rationals using $ f(xy) = f(x) f(y); f(p_{i_1}p_{i_2}...q_{j_1}q_{j_2}.../(p_{k_1}...q_{m_1}...)) = p_{m_1}...q_{i_1}.../(p_{j_1}...q_{k_1}...)$. It is now trivial to verify that $ f(x f(y)) = \frac{f(x)}{y}$.

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