• 沒有找到結果。

# THREE FUNDAMENTAL PROOF TECHNIQUES

Problems for Section 1.4

1.4.1. Prove that the following are countable.

(a) The union of any three countable sets, not necessarily infinite or dis-joint.

(b) The set of all finite subsets of N.

1.4.2. Explicitly give bijections between each of the following pairs.

(a) N and the odd natural numbers.

(b) N and the set of all integers.

(c) Nand N x N x N.

(We are looking for formulas that are as simple as possible and involve only such operations as addition and multiplication.)

1.4.3. Let C be a set of sets defined as follows, 1.0EC

2. If Sl E C and S2 E C, then {Sl,S2} E C.

3. If Sl E C and S2 E C, then Sl x S2 E C.

4. Nothing is in C except that which follows from (1), (2), and (3).

(a) Explain carefully why it is a consequence of (1-4) that {0, {0}} E C.

(b) Give an example of a set S of ordered pairs such that SEC, and

### lSI>

1.

(c) Does C contain any infinite sets? Explain.

(d) Is C countable or uncountable? Explain.

1.4.4. Show that the dovetailing method of Figure 1-8 visits the pair (i, j) mth, where

m= 2[(i+j)2+3i+j]. 1

1.5 THREE FUNDAMENTAL PROOF TECHNIQUES

Every proof is different, since every proof is designed to establish a different result. But like games of chess or baseball, observation of many leads one to realize that there are patterns, rules of thumb, and tricks of the trade that can be found and exploited over and over again. The main purpose of this section is to introduce three fundamental principles that recur, under various disguises, in many proofs: mathematical induction, the pigeonhole principle, and diagonalization.

The Principle of Mathematical Induction: Let A be a set of natural num-bers .mch that

24 Chapter 1: SETS, RELATIONS, AND LANGUAGES (1)

### °

E A, and

(2) for each natural number n, if {O, 1, ... , n} ~ A, then n

### +

1 E A.

Then A = N.

In less formal terms, the principle of mathematical induction states that any set of natural numbers containing zero, and with the property that it contains n

### +

1 whenever it contains all the numbers up to and including n, must in fact be the set of all natural numbers.

The justification for this principle should be clear intuitively; every natural number must wind up in A since it can be "reached" from zero in a finite succession of steps by adding one each time. Another way to argue the same idea is by contradiction; suppose (1) and (2) hold but A

### f::.

N. Then some number is omitted from A. In particular, let n be the first number among 0,1,2, ... that is omitted from

### N.t

Then n cannot be zero, since

### °

E A by (1);

and since 0,1, ... , n - 1 ~ A by the choice of n, then n E A by (2), which is a contradiction.

In practice, induction is used to prove assertions of the following form: "For all natural numbers n, property P is true." The above principle is applied to the set A = {n : P is true of n} in the following way.

(1) In the basis step we show that

### °

E A, that is, that P is true of 0.

(2) The ind'uction hypothesis is the assumption that for some fixed but arbitrary n ~ 0, P holds for each natural number 0,1, ... , n.

(3) In the induction step we show, using the induction hypothesis, that P is true of n

### +

1. By the induction principle, A is then equal to N, that is, P holds for every natural number.

Example 1.5.1: Let us show that for any n ~ 0, 1

2

n =

### n2:tn.

Basis Step. Let n = 0. Then the sum on the left is zero, since there is nothing to add. The expression on the right is also zero.

Induction Hypothesis. Assume that, for some n ~ 0, 1

2

m =

whenever m ~ n.

### t

This is a use of another principle, called the least number principle, that is ac-tually equivalent to the principle of mathematical induction, so we are not really

"proving" the principle of mathematical induction. The least number principle is:

If A <:;:; N and A

### i=

N, then there is a unique least number n E N - A; that is, a unique number n such that n 1:. A but 0,1, ... ,n - 1 E A. A somewhat frivolous example of the least number principle is the fact that there are no uninteresting numbers. For s:Ippose there were; then there would have to be a least such num-ber, say n. But then n would have the remarkable property of being the least uninteresting number, which would surely make n interesting. "

1.5: Three Fundamental Proof Techniques 25

Induction Step.

1

2

71

(71

1) = (1

2

71)

(71

1) 712 +71

- 2 -

(71

### +

1) (by the induction hypothesis) 712

71

271

### +

2

2

(71+1)2+(71+1) 2

as was to be shown. <>

Example 1.5.2: For any finite set A, 12AI = 21AI; that is, the cardinality of the power set of A is 2 raised to a power equal to the cardinality of A. We shall prove this statement by induction on the cardinality of A.

Basis Step. Let A be a set of cardinality 71 = 0. Then A = 0, and 21AI = 2° = 1;

on the other hand, 2A = {0}, and 12AI = 1{0}1 = 1.

Induction Hypothesis. Let 71

### >

0, and suppose that 12A I = 21AI provided that IAI ::;71.

Induction Step. Let A be such that IAI

71

1. Since 71

### >

0, A contains at least one element a. Let B

### =

A - {a}; then IBI

### =

n. By the induction hypothesis, 12BI

21BI

### =

2n. Now the power set of A can be divided into two parts, those sets containing the element a and those sets not containing a. The latter part is just 2B, and the former part is obtained by introducing a into each member of 2B. Thus

2A

### =

2B U {CU {a}: C E 2B}.

This division in fact partitions 2A into two disjoint equinumerous parts, so the cardinality of the whole is twice 21B1 , which, by the induction hypothesis, is 2· 2n = 2n+1, as was to be shown.<>

We next use induction to establish our second fundamental principle, the pigeonhole principle.

The Pigeonhole Principle: If A and B are finite sets and IAI

### >

IBI, then there is no one-to-one function from A to B.

In other words, if we attempt to pair off the elements of A (the "pigeons") with elements of B (the "pigeonholes"), sooner or later we will have to put more than one pigeon in a pigeonhole.

Proof: Basis Step. Suppose IBI

0, that is, B

### =

0. Then there is no function f : A r-+ B whatsoever, let alone a one-to-one function.

26 Chapter 1: SETS, RELATIONS, AND LANGUAGES Induction Hypothesis. Suppose that f is not one-to-one, provided that f : A r-+

Outline