Page 307 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Strong Induction 293
Case 2. n = 1. Then
n n 1 1
√ √ √ √
1+ 5 − 1− 5 1+ 5 − 1− 5
2 2 2 2
√ = √
5 5
√
5
= √ = 1 = F 1 .
5
Case 3. n ≥ 2. Then applying the inductive hypothesis to n − 2 and n − 1,
we get
F n = F n−2 + F n−1
√ n−2 √ n−2 √ n−1 √ n−1
1+ 5 1− 5 1+ 5 1− 5
2 − 2 2 − 2
= √ + √
5 5

√ n−2 √ n−1 √ n−2 √ n−1
1+ 5 1+ 5 1− 5 1− 5
2 + 2 − 2 + 2
= √
5
√ n−2 √ √ n−2 √
1+ 5 1+ 5 1− 5 1− 5
2 1 + 2 − 2 1 + 2
= √ .
5
Now note that
2 √ √ √ √
√
1 + 5 1 + 2 5 + 5 6 + 2 5 3 + 5 1 + 5
= = = = 1 + ,
2 4 4 2 2
and similarly
2 √

√
1 − 5 1 − 5
= 1 + .
2 2
Substituting into the formula for F n ,weget
n−2 2 n−2 2
√ √ √ √
1+ 5 1+ 5 1− 5 1− 5
2 2 − 2 2
F n = √
5
n n
√ √
1+ 5 1− 5
2 − 2
= √ .
5
Notice that in the proof of Theorem 6.4.3 we had to treat the cases n = 0
and n = 1 separately. The role that these cases play in the proof is similar to
the role played by the base case in a proof by ordinary mathematical induction.
Although we have said that proofs by strong induction don’t need base cases,
it is not uncommon to find some initial cases treated separately in such proofs.