Exercise 14 solution: Suppose ρ(f) = p/q. By Theorem 1, F has a p/q-periodic point x_{0} which satisfies F^{q}(x_{0}) − p = x_{0}. Note that, for any integer k, F^{q}(x_{0} + k) − p = F^{q}(x_{0}) + k − p = x_{0} + p + k − p = x_{0} + k, so x_{0} + k is a p/q-periodic point for all integers k. Set G(x) = F^{q}(x) − p, so that fixed points for G are p/q-periodic points for F.

Let x ∈ R. Pick k ∈ Z with x_{0} + k ≤ x < x_{0} + k + 1. If x is a fixed point for G then x is a p/q-periodic point for F. So suppose G(x) ≠ x.

(i) If G(x) < x then, since G is strictly increasing, {G^{n}(x)} is a decreasing sequence. Suppose G^{n}(x) < x_{0} + k for some n; let n_{0} be the smallest such n. Then G^{n_{0}}(x) < x_{0} + k and G^{n_{0}−1}(x) ≥ x_{0} + k. Since G preserves the ordering of points in R, x_{0} + k = G(x_{0} + k) ≤ G(G^{n_{0}−1}(x)) = G^{n_{0}}(x), a contradiction. Hence {G^{n}(x)} is bounded below by x_{0}+k and must converge to a point y by the Monotone Convergence Theorem. By Exercise 13, G(y) = y, so that y is a p/q-periodic point for F.

(i) If G(x) > x then {G^{n}(x)} is an increasing sequence. If, for some n, G^{n}(x) > x_{0} + k + 1, then let n_{0} be the smallest such n. We have G^{n_{0}}(x) > x_{0} + k + 1 and G^{n_{0}−1}(x) ≤ x_{0} + k + 1. Since G is increasing, x_{0} + k + 1 = G(x_{0} + k + 1) ≥ G(G^{n_{0}−1}(x)) = G^{n_{0}}(x), a contradiction. Hence {G^{n}(x)} is bounded above by x_{0}+k+1 and, as in (i), must converge to a point y satisfying G(y) = y, which is a p/q-periodic point for F.