Chapter 2: Mathematical Preliminaries );J, 57

2.4 PRINCIPLE OF INDUCTION

The process of reasoning from general observations to specific truths is calledinduction.

The following propelties apply to the set N of natural numbers~theprinciple of induction.,

Property 1 Zero is a natural number.

Property 2 The successor of any natural number is also a natural number.

Property 3 Zero is not the successor of any natural number.

Property 4 No two natural numbers have the same successor.

Property 5 Let a property pen) be defined for every natural number n. If(i) P(O) is true. and (ii) P(successor of II) is true whenever pen) is true, thenP(n) is true for all II.

A proof by complete enumeration of all possible combinations is calledperfect induction. e.g. proof by truth table.

The method of proof by induction can be used to prove a property pen) forall n.

2.4.1 METHOD OF PROOF BY INDUCTION

This method consists of three basic steps:

Step 1 Prove P(n) for II = 0/1. This is called the proof for the basis.

Step 2 Assume the result/properties for pen). This is called the inductionhypothesis.

Step 3 Prove P(II + l) using the induction hypothesis.

EXAMPLE 2.20

Prove that 1 + 3 + 5 + ... + r = n~. for all n > O. where r is an odd integerand n is the number of terms in the sum. (Note: r = 211 - 1.)

Solution

(a) Prooffor the basis. For n =1. L.H.S. = 1 and R.H.S. = 12 = 1. Hencethe result is true for n = 1.

(b) By induction hypothesis. we have 1 + 3 + 5 + ... + r = Il~. As r =2n - 1.

L.H.S. = 1 + 3 + 5 + ... + (2n - 1) = 11 2

(e) We have to prove that 1 + 3 + 5 + .. , + r + r + 2 = (n + 1)2:

L.R.S. = (1 + 3 + 5 + '" + r + (r + 2))

= 112 + r + 1 = n: + 2n - 1 + 2 = (ll + 1)= = R.H.S.

58 !i2 Theory ofComputer Science

EXAMPLE 2.21

Prove the following theorem by induction:

1 + 2 + 3 + ... + n = n(n + 1)/2

Solution

(a) Proof for the basis. For n = L L.H.S. = 1 andR.H.5. = 1(1 + 1)/2 = 1

(b) Assume 1 + 2 + 3 + ... + n = n(1l + 1)/2.(c) We have to prove:

1 + 2 + 3 + + (n + 1) = (n + 1)(n + 2)/2

1 + 2 + 3 + + n + (n + 1)

=n(n + 1)/2 + (n + 1) (by induction hypothesis)

= (n + 1)(n + 2)/2 (on simplification)

The proof by induction can be modified as explained in the followingsection.

2.4.2 MODIFIED METHOD OF INDUCTION

Three steps are involved in the modified proof by induction.

Step 1 Proof for the basis (n = 0/1).

Step 2 Assume the result/properties for all positive integers < n + 1.

Step 3 Prove the result/properties using the induction hypothesis (i.e. step 2),forn+l.

Example 2.22 below illustrates the modified method of induction. Themethod we shall apply will be clear once we mention the induction hypothesis.

EXAMPLE 2.22

Prove the following theorem by induction: A tree with n vertices has (n - 1)edges.

Solution

For n = 1, 2, the following trees can be drawn (see Fig. 2.15). 50 the theoremis true for n = 1, 2. Thus, there is basis for induction.

o In=1 n=2

Fig. 2,15 Trees with one or two vertices.

Chapter 2: Mathematical Preliminaries ~ 59

Consider a tree T with (n + 1) vertices as shown in Fig. 2.16. Let e bean edge connecting the vertices Vi and 1} There is a unique path between Vi

and vi through the edge e. (Property of a tree: There is a unique path betweenevery pair of vertices in a tree.) Thus, the deletion of e from the graph willdivide the graph into two subtrees. Let nl and n: be the number of verticesin the subtrees. As 111 ::; 11 and 11: ::; n. by induction hypothesis, the totalnumber of edges in the subtrees is 11] - 1 + n: - 1. i.e. n - 2. So, the numberof edges in T is n - 2 + 1 = 11 - 1 (by including the deleted edge e). Byinduction. the result is true for all trees.

e

Fig. 2.16 Tree T with (n + 1) vertices.

EXAMPLE 2.23

Two definitions of palindromes are given below. Prove by induction that thetwo definitions are equivalent.

Definition 1 A palindrome is a string that reads the same forward andbackward.

Definition 2 (i) A is a palindrome.(ii) If a is any symboL the string a is a palindrome.

(iii) If a is any symbol and x is a palindrome. then axa is a palindrome.(iv) Nothing is a palindrome unless it follows from (i)-(iii).

Solution

Let x be a string which satisfies the Definition L i.e. x reads the same forwardand backward. By induction on the length of x we prove that x satisfiesthe Definition 2.

If Ix I ::; 1. then x =a or A. Since x is a palindrome by Definition L i\and a are also palindromes (hence (i) and (ii», i.e. there is basis for induction.If !x I > 1. then x =mva, where w. by Definition 1. is a palindrome: hence therule (iii). Thus. if x satisfies the Definition L then it satisfies the Definition 2.

Let x be a string which is constructed using the Definition 2. Weshow by induction on i x I that it satisfies the Definition 1. There is basisfor induction by rule (ii). Assume the result for all strings with length < n.Let x be a string of length n. As x has to be constructed using therule (iii). x = aya. where y is a palirrlr'Jme. As y is a palindrome byDefinition 2 and Iy I < 71, it satisfies the Definition 1. So, x = aya also satisfiesthe Definition 1.

60 J;l Theory of Computer Science

EXAMPLE 2.24

Prove the pigeonhole principle.

Proof We prove the theorem by induction on m. If m = 1 and 11 > L thenall these 11 items must be placed in a single place. Hence the theorem is truefor III = 1.

Assume the theorem for m. Consider the case of III + 1 places. We provethe theorem for 11 =In + 2. (If 11 > In + 2. already one of the In + 1 places willreceive at least two objects from m + 2 objects, by what we are going to prove.)Consider a particular place. say, P.

Three cases arise:

(i) P contains at least two objects.(ii) P contains one object

(iii) P contains no object.

In case (i). the theorem is proved for n = 111 + 2. Consider case (ii). As P

contains one object, the remaining m places should receive 111 + 1 objects. Byinduction hypothesis, at least one place (not the same as P) contains at least twoobjects. In case (iii), III + 2 objects are distributed among In places. Once again,by induction hypothesis, one place (other than P) receives at least two objects.Hence. in aD the cases, the theorem is true for (m + 1) places. By the principleof induction. the theorem is true for all Ill.

2.4.3 SIMULTANEOUS INDUCTION

Sometimes we may have a pair of related identities. To prove these, we mayapply two induction proofs simultaneously. Example 2.25 illustrates thismethod.

EXAMPLE 2.25

A sequence Fo, F], F2, ... called the sequence of Fibonacci numbers (namedafter the Italian mathematician Leonardo Fibonacci) is defined recursively asfollows:

Prove that:Fo = O.

P" : (2.2)

(2.3)

Proof We prove the two identities (2.2) and (2.3) simultaneously bysimultaneous induction. PI and Q] are F12 + F0

2 = F1 and F2F 1 + F]Fo = Fc

Chapter 2: Mathematical Preliminaries );! 61

respectively. As Fo = 0, F I = 1, F~ =1, these are true. Hence there is basisfor induction. Assume Pn and Qw So

(2.2)

(2.3)Now.

= (F,~-l + F,~) + F"-lF,, + F,; + F"-lF,,

=F,;-l + F,~ + (F,,-l + F" )F" + F"F,,-l

=(F,~-l + F,;) + F,'~lF" + F"F,'-l

(by (2.3»)

This proves Pn+ l •

Also.

(By P'1+1 and (2.3))

This proves Qn+l'

So. by induction (2.2) and (2.3) are true for all 11.

We conclude this chapter with the method of proof by contradiction.

2.5 PROOF BY CONTRADICTION

Suppose we want to prove a property P under certain conditions. The methodof proof by contradiction is as follows:

Assume that property P is not true. By logical reasoning get a conclusionwhich is either absurd or contradicts the given conditions.

The following example illustrates the use of proof by contradiction andproof by induction.

EXAMPLE 2.26

Prove that there is no string x in {a. b}* such that (L-r: = xb. (For the definitionof strings. refer to Section 2.3.)