1.8 Induction

A Bogus Induction

The Fibonacci numbers

$$0, 1, 1, 2, 3, 5, 8, 13, \ldots$$

are defined as follows: let $$F(n)$$ be the $$n^{th}$$ Fibonacci number,

• $$F(0) ::= 0$$
• $$F(1) ::= 1$$
• $$F(n) ::= F(n-1) + F(n-2), \;\; \text{ for } n \ge 2$$

Bogus Claim: Every Fibonacci number is even.

Which step(s) in the proof contain logical errors?

Exercise 1

Lines 1 through 6 contain no logical errors. The fatal flaw is in line 7. Using strong induction, we can conclude that a predicate $$P(n)$$ holds for all $$n \geq 0$$ provided that we show all of the following:

• $$P(0)$$
• $$P(0) \to P(1)$$
• $$[P(0) \wedge P(1)] \to P(2)$$
• $$[P(0) \wedge P(1) \wedge P(2)] \to P(3)$$
• etc.

The first assertion is proved in line 3. The third and subsequent assertions are proved on lines 4-6. However, the second assertion, $$P(0) \to P(1)$$, is proved nowhere (and is actually false). Therefore, we cannot apply the strong induction principle in line 7.

If you said that line 4 contains a logical error, you were on the right track. The natural place to prove the second assertion would have been in line 4. But by saying, "suppose $$n \geq 2$$" instead of "suppose $$n \geq 1$$", the proof explicitly avoided doing so.

Technically, there is no logical error in line 4: It is simply the beginning of a proof for the case when $$n \geq 2$$. On the other hand, it's reasonable to say that line 4 is the place where the proof makes a strategic error because it skips the $$n = 1$$ case.