Exercise 16 solution: Case 2. Suppose i − k < 0. Then α ≤ (j − l)/(k − i). By Exercise 6, F^{i}(x_{0}) + j < F^{k}(x_{0}) + l implies F^{i}(x_{0} + j) < F^{k}(x_{0} + l). By Exercise 3 we have x_{0} + j < F^{k−i}(x_{0} + l), so that F^{k−i}(x_{0}) > x_{0} + j − l. As in the solution of Exercise 12, this implies F^{m(k−i)}(x_{0}) > x_{0} + m(j − l) for all integers m. Hence

α = lim_{m→∞} \frac{F^{m(k−i)}(x_{0}) − x_{0}}{m(k − i)} ≥ \frac{j − l}{k − i}.

This implies α = (j − l)/(k − i), contradicting the fact α is irrational.