Chapter 3 Mathematical Induction

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Chapter 3 Mathematical Induction




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Presentations text content in Chapter 3 Mathematical Induction

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Chapter 3Mathematical Induction

Discrete Mathematics: A Concept-based Approach

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Introduction

The mathematical Induction is a technique for proving results over a set of positive integers. It is a process of inferring the truth from a general statement for particular cases. A statement may be true with reference to more than hundred cases, yet we cannot conclude it to be true in general. It is extremely important to note that mathematical induction is not a tool for discovering formulae or theorems.

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A proof by mathematical induction has three parts, namely, basis clause, induction clause and external clause.

Basis clause: We show that S(n) is true for initial values taken by ‘n’.Inductive clause :We assume that the statement is true for some arbitrary k, ( k ) i.e. S(k) is true, and we try to prove that also S(k+1) is true.

External clause: We make the statement that the statement is true if and only if it is proved by basis or induction clause.

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We have two types of mathematical induction, namely, weak induction and strong induction.In case of strong induction, we prove the statement S(n) for all the values of ‘n’ and then generalize. But in case of weak induction, we prove for basis and induction steps and then generalize. As an illustration can be drawn with respect to temperature measurement in a city. If we measure the temperature in each locality of city and then average of all the measurements is taken as the temperature of city. This is strong induction and often not a viable method.

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On the other hand, we make measurements at few specifiedlocalities, where weather stations are placed, and the average

of measurements at these is given as the temperature of city.This is what called weak induction.

In proofs we use weak induction than strong induction.

Example :

Show by mathematical induction to prove the formula for the sum of a finite number of terms of a geometric progression.

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Where , and n is a non-negative integer.Proof:

Basis step, for It is true for the basis.

ii) Induction step: For inductive hypothesis, we assume that

s(k) is true for some arbitrary positive integer k

------(1)

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For completeness, if s(k) is true, then we have to prove p(k+1) is also true. Using (1) for n=k+1

Solving RHS of the above equation

Which is true for n= k+1. Hence based on Mathematical

Induction hypothesis sum to n-terms of a GP is correct.

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Example: Use the principle of mathematical induction to

Prove thatProof:

i

) Basis step: For n = 1, we have,

ii)Induction step: We assume that the statement S(n) is true for some arbitrary positive integer k, where k is an integer

greater than or equal to 1.

Let

Now, we have to show that the result is true for n = k+1.

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We know that,

Using the above equation is written as

Hence, by mathematical induction the result is true for all

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Example : Prove that for all positive integers ‘ n’.

Proof: Basis step: for n =1. f(1) is given by

So the formula holds for (n =1).

Induction Step: Suppose the formula holds for (n = k),

i.e

,

Discrete Mathematics: A Concept-based Approach

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Consider the case for (n = k+1). By mathematical induction

Discrete Mathematics: A Concept-based Approach

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After simplifying we get

By the principle of Mathematical Induction, the formula holds

for all positive ‘n’.

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Example : Prove by mathematical Induction that the

arithmetic mean (A.M) is always greater than or equal to the geometric mean(G.M). Or Prove that for allpositive integers (a

1

,a

2,-------a

n).Proof:Basis step: Consider , we obtain

the inequality holds good for (n = 2). Since the arithmetic

mean is to be obtained for two numbers so we have started for

(n = 2).

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Induction Step: Suppose the inequality hold good for n = k,

for some positive integer k. Consider the case (n = 2k) , using the case (n = 2), and the induction hypothesis, we have,

The inequality also holds good for (n =2k). Hence, the inequality

holds for all positive powers of 2.

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Example. Let ( x > -1) be a real number, prove that

(1+ x)n ≥ 1+ nx for all natural number n.

Proof: Basis step : for n=1,

(1+ x)

1

= 1+ 1(x). The inequality hold good for n = 1.Induction Step: Suppose the inequality hold good for some

positive k, therefore (1+ x)

k

≥ 1+

kx

is true.

For the case n= k+1, we have,

(1+x)

k+1

= (1+x)

k (1+x) = (1+kx) (1+x) = 1+ (k+1) x + k x2

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= 1+ [k(1+x)+1] x, taking k(1+x) = k’

≥ 1 + (k’+1) x.

Hence, if the inequality holds for the case (n = k), it also holds

for (n = k+1).

It follows that (1+ x)

n ≥ 1+

nx

for all natural numbers ‘n’.

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Example: consider the following code which yields the square of the given positive integer.

Function square (x)Begin:

a = 0

b= 0

while (b< >x) do

a=a + x

b= b+1

return (a)

end

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Consider the predicate p(n) representing We shall prove it for p(k+1).

Basis step:

Induction step :

If p(k) is true, then p(k+1) is true. Hence by mathematical induction the predicate is true for any integer n.

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Jokes and Paradoxes.

Let the predicate P (n) represent the statement that one can drink any amount (n units) of water when feeling thirsty. We will prove the statement ‘when one feels thirsty one will be able to drink n drops of water’ using mathematical Induction.

Basis step : Clearly, the statement holds for n=1 because one certainly wants to drink some water when feeling thirsty.

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Induction step :

Suppose the statement holds for n=k, i.e, when one feels thirsty one is able to drink k-drops of water. Consider the case n= k+1. By assumption, when one feels thirsty one is able to drink k drops of water. Being thirsty, one certainly is able to drink one more drop. So the statement holds for n = k+1, as well.

By the principle of Mathematical Induction, the statement holds for all natural number n. In other words, when one feels thirsty, one can drink any amount of water.

Thus,

o

ne is able to swallow up all the water in oceans! Is it not a Joke?

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Consider, an example from the book on Calculus by. M. Spivak

, 17 Professors in the Department of Mathematics of a certain University. They meet every week for a discussion on research. They have a rule that if any professor finds mistakes in his own papers, then he must resign. For long time no Professor has ever resigned. This does not mean that no Professor has ever made any mistakes in his papers. Rather, everyone has made mistakes, and every other Professor has discovered it. In other words, everyone knows that every other Professor has made mistakes, but does not know his own mistakes.

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During the last meeting of a year, a research assistant said that he wants to tell everyone one thing that amongst the Professors, at least one mistake is made in his papers and is discovered by someone else. All the Professors are made to resign at the 17

th meeting of the subsequent year.

One interesting question is that what the research assistant has pointed out is something that every professor knows. Why did nobody resign without the remark from the research assistant? Why did every professor resign after this ‘well- known’ remark?

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The mathematics professors were busy in finding the mistakes of other fellow professors and did not resign until the research

assistant pointed out mistakes in the papers written by eachProfessor.

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Summary

Mathematical induction (MI) is a useful method of giving proofs to mathematical statements. The method can be extended to prove statements about more general well founded structures, such as trees. This generalization is also known as structural induction.

There are variations of mathematical induction, namely, complete induction and transfinite induction. MI is useful in computer science for verifying the algorithms, wherein we write the loop invariants. It is also used in proving theorems in Graph theory with much ease compared to analytical proofs.

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