Page 306 - HOW TO PROVE IT: A Structured Approach, Second Edition

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292 Mathematical Induction
surprising, therefore, that proofs involving the Fibonacci numbers often require
strong induction rather than ordinary induction.
To illustrate this we’ll prove the following remarkable formula for the
Fibonacci numbers:

√ n √ n
1+ 5 1− 5
2 − 2
F n = √ .
5
It is hard at first to believe that this formula is right. After all, the Fibonacci
numbersareintegers,anditisnotatallclearthatthisformulawillgiveaninteger
√
value. And what do the Fibonacci numbers have to do with 5? Nevertheless,
a proof by strong induction shows that the formula is correct. (To see how this
formula could be derived, see exercise 8.)

th
Theorem 6.4.3. If F n is the n Fibonacci number, then
√ n √ n
1+ 5 1− 5
2 − 2
F n = √ .
5

Scratch work
Because F 0 and F 1 are defined separately from F n for n ≥ 2, we check the
formula for these cases separately. For n ≥ 2, the definition of F n suggests that
we should use the assumption that the formula is correct for F n−2 and F n−1
to prove that it is correct for F n . Because we need to know that the formula
works for two previous cases, we must use strong induction rather than ordinary
induction. The rest of the proof is straightforward, although the algebra gets a
little messy.
Proof. We use strong induction. Let n be an arbitrary natural number, and
suppose that for all k < n,
√ k √ k
1+ 5 1− 5
2 − 2
F k = √ .
5
Case 1. n = 0. Then
n n 0 0
√ √ √ √
1+ 5 1− 5 1+ 5 1− 5
2 − 2 2 − 2
√ = √
5 5
1 − 1
= √ = 0 = F 0 .
5