Page 293 - HOW TO PROVE IT: A Structured Approach, Second Edition

P1: Oyk/
0521861241c06 CB996/Velleman October 20, 2005 1:8 0 521 86124 1 Char Count= 0

Recursion 279
such that ∀x ∈ A 1 ( f (x) = c 1 ). If we can show that f (a 1 ) = c 1 then we
will be done, since then it will follow that ∀x ∈ A( f (x) = c 1 ).
Let a 2 be an element of A that is different from a 1 , and let A 2 = A \
{a 2 }. Applying the inductive hypothesis again, we can choose a number
c 2 ∈ R such that ∀x ∈ A 2 ( f (x) = c 2 ). Notice that sincea 1 = a 2 ,a 1 ∈ A 2 ,
so f (a 1 ) = c 2 . Now let a 3 be an element of A that is different from both
a 1 and a 2 . Then a 3 ∈ A 1 and a 3 ∈ A 2 ,so f (a 3 ) = c 1 and f (a 3 ) = c 2 .
Therefore c 1 = c 2 ,so f (a 1 ) = c 1 , as required.

6.3. Recursion

In Chapter 3 we learned to prove statements of the form ∀nP(n) by letting n
be arbitrary and proving P(n). In this chapter we’ve learned another method
for proving such statements, when n ranges over the natural numbers: Prove
P(0), and then prove that for any natural number n,if P(n) is true then so is
P(n + 1). Once we have proven these statements, we can run through all the
natural numbers in order and see that P must be true of all of them.
We can use a similar idea to introduce a new way of defining functions. In
Chapter 5, we usually defined a function f by saying how to compute f (n) for
any n in the domain of f. If the domain of f is the set of all natural numbers, an
alternative method to define f would be to say what f (0) is and then, for any
natural number n, say how we could compute f (n + 1) if we already knew the
value of f (n). Such a definition would enable us to run through all the natural
numbers in order computing the image of each one under f.
For example, we might use the following equations to define a function f
with domain N:

f (0) = 1;
for every n ∈ N, f (n + 1) = (n + 1) · f (n).
The second equation tells us how to compute f (n + 1), but only if we already
know the value of f (n). Thus, although we cannot use this equation to tell us
directly what the image of any number is under f,we can use it to run through
all the natural numbers in order and compute their images.
We start with f (0), which we know from the first equation is equal to 1.
Plugginginn = 0inthesecondequation,weseethat f (1) = 1 · f (0) = 1 · 1 =
1, so we’ve determined the value of f (1). But now that we know that f (1) = 1,
we can use the second equation again to compute f (2). Plugging in n = 1inthe
second equation, we find that f (2) = 2 · f (1) = 2 · 1 = 2. Similarly, setting
n = 2 in the second equation we get f (3) = 3 · f (2) = 3 · 2 = 6. Continuing