# 1.3 Well Ordering Principle

## A Bogus Well Ordering Principle Proof

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. Then

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

Identify which step(s) contain the logical error!

Bogus Claim: Every Fibonacci number is even.

Exercise 1

Steps 1 through 10 contain no logical errors. The fatal flaw is in the final step 11. The proof only shows that there is a minimum $$m$$ in $$C$$ and it is not 0. The assumption that $$m \geq 2$$ leads to a contradiction; it leaves the case $$m = 1$$ unexamined, while in fact, $$1 \in C$$. (The supposition that "$$m \geq 2$$ so the definition (*) of $$F(m)$$ applies" is no excuse for ignoring the case $$m = 1$$.)

If you said that step 7 contains a logical error, you were on the right track. The natural place to handle the case $$F(1)$$ would have been right after step 6. But the the proof explicitly avoided the case $$m = 1$$, by saying, "suppose $$m \geq 2$$." However, there is no logical error in line 7: it is simply the beginning of a proof for the case when $$m \geq 2$$. On the other hand, it's reasonable to say that line 7 is the place where the proof makes an organizational, or perhaps strategic, error because it skips the $$m = 1$$ case.