Math 300: Final Exam Practice Solutions

Proof. First, a 1 = 1 ≤ 1 = φ 0 so the inequality holds for n = 1. For a second base case (we’ll
see why we need two base cases), we have a 2 = 1 ≤ φ 1 so the inequality holds for n = 2 as well.
Suppose now that for some k, a i ≤ φ i−1 for all 1 ≤ i ≤ k. (So, not only for n = k but for all
positive integers smaller than k.) We want to show that a k+1 ≤ φ k . Since we’ve already check the
base cases n = 1 and n = 2, we may assume here that k ≥ 3, which we need to do in order to write
a k+1 = a k + a k−1
according to the recursive definition. Both a k and a k−1 are elements to which the induction hypothesis
applies to, so
a k ≤ φ k−1 and a k−1 ≤ φ k−2 .
Then, noting that φ 2 = φ + 1 as a result of the hint, we have:
a k+1 = a k + a k−1 ≤ φ k−1 + φ k−2 = φ k−2 (φ + 1) = φ k−2 φ 2 = φ k .
Thus a k+1 ≤ φ k as required, so by induction we conclude that a n ≤ φ n−1 for all positive n.
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