Mathematics

Principle of Mathematical Induction Session Principle of Mathematical Induction

Session Objectives

Session Objective 1. Introduction 2. Steps involved in the use of mathematical induction 3. Principle of mathematical induction.

Statement Statement:- A sentence which can be judged as true or false. Example:1. ‘2 is only even prime number’ 2. ‘Bagdad is capital of Iraq’ 3. ‘2n+5 is always divisible by 5 for all nN Mathematical statement: Example 1 and 3.

Induction Induction :It’s a process Particular  General Example: Statement- ’2n+1’ is odd number. n=1 2.1+1=3 is odd. True n=2 2.2+1=5 is odd. True n=3 2.3+1=7 is odd. True Observation  tentative conclusion (‘2n+1 is odd’) let its true for n=m. i.e 2m+1 is odd.

Induction for n=m+1 2(m+1)+1 =2m+1+2 odd +2=odd Now it is Generalized ’2n+1 is odd for all n’

Induction Steps Involved: 1. Verification 2. Induction 3. Generalization. Important: Process of Mathematical Induction (PMI) is applicable for natural numbers. Usage: 1. to prove mathematical formula Ex:1.3+2.32+3.33+......+n.3n= 2. to check divisibility of a expression by a number Ex: Prove n3+5n is divisible by ‘3’.

Algorithm Let P(n) be the given statement. Step 1: Prove P(1) is true Verification Step 2: Assume P(n) is true for some n=mN i.e. P(m) is true. Step 3: Using above assumption prove P(m+1) is true. i.e P(m)  P (m+1) Step 4: Above steps lead us to generalize the fact P(n) is true for all n N.

Questions

Illustrative Example Principle of mathematical induction is applicable to set of integers (b) set of real numbers (c) set of positive integers (d) None of these Solution : (c) Principle of mathematical induction is applicable to natural numbers or set of positive integers only.

Illustrative Example Show by PMI that 1.3+2.32+3.33+......+n.3n= Solution: Step 1. for n=1, p(1)=1. 3=3 (LHS) L.H.S=R.H.S P(1) is true Step2. Assume that P(m) is true

Solution Continued Step3: To prove P(m + 1) holds true Adding. (m + 1).3m+1 to both sides P(m)  P (m+1)

Solution Continued Step4. As P(m)  P (m+1) P(n) is true for all n N (Proved)

Illustrative Example Prove n3+5n is divisible by ‘3’ for n N (By PMI or Otherwise) Solution: P(n) : ‘n3+5n is divisible by 3’ Step1: P(1) = ‘6 divisible by 3’ which is true Step2: For some n=m, P(m) holds true i.e. m3+5m=3k, k N step3: To prove P(m+1) holds true, we have to prove that (m+1)3+5(m+1) is divisible by 3. = 3k´ (m+1)3+5(m+1)=(m3+5m)+(3m2+3m+6) = 3k+3(m2+m+2) ( m2 + m + 2 I )

Solution Continued (m+1)3+5(m+1)=3k’ P(m+1) is divisible by 3  P(m)  P (m+1) Step4: P(n) is true for all n N  n3+5n is divisible by ‘3’ for n N

Class Exercise - 6 Prove that mathematical induction that 72n + 3n – 1(23n – 3) is divisible by 25, . Solution : Let P(n) : 72n + (23n – 3)3n – 1 is divisible by 25. Step I: n = 1 P(1) = 72 + (23 – 3)31 – 1 = 72 + 20 · 30 = 49 + 1 = 50 As P(1) is 50 which is divisible by 25, hence P(1) is true.

Solution Continued Step II: Assuming P(m) is divisible by 25, P(m) = 72m + (23m – 3)3m – 1 = 25(K) ... (i) (K is a positive integer.) Now P(m + 1) = 72(m + 1) + (23(m + 1) – 3)3m + 1 – 1 = 72m + 2 + (23m + 3 – 3)3m + 1 – 1 = 49 × 72m + 8(23m – 3)3m – 1 × 3 = 49 × 72m + 24(23m – 3)3m – 1 With the help of equation (i), we can write the above expression as

Solution Continued Now from the above equation, we can conclude that P(m + 1) is divisible by 25. Hence, P(n) is divisible by 25 for all natural numbers.

Alternative Solution Alternative Method: without PMI n3+5n = n(n2+5) Product of three consecutive numbers

Illustrative Example P(n) is the statement ‘n2 – n + 41 is prime’ Verify it. Solution: For n = 1 P(1) = ‘41 is a prime’ True. For n = 2 P(2) = ‘43 is a prime’  True. But for n = 41 P(41) = ‘412 is a prime’  False. False Statement

Class Exercise -3 Prove by PMI that 1.2.3. + 2.3.4 + 3.4.5 + ... + n(n + 1) (n + 2) = Solution: step1. P(1): LHS=1.2.3=6  L.H.S=R.H.S Step2. Assume P(m) is true

Solution Continued Adding (m+1)(m+2)(m+3) 1.2.3.+2.3.4+...+m(m+1)(m+2)+(m+1)(m+2)(m+3)=

Solution Continued 1.2.3.+2.3.4+...+(m+1)(m+2)(m+3) Step4. As P(m)  P (m+1) P(n) is true for all n N  1.2.3.+2.3.4+3.4.5+...+n(n+1)(n+2)=

Class Exercise -4 If a+b=c+d and a2+b2=c2+d2 , then show by mathematical induction, an+bn = cn+dn Solution: Let P(n) : an + bn = cn + dn n = 2, P(2) : a2+b2 = c2+d2 For n = 1, P(1) : a+b=c+d P(1) and P(2) hold true. Assume P(m) and P(m + 1) hold true am+bm = cm+dm am+1+bm+1= cm+1+dm+1

Solution Continued a+b = c+d; a2+b2=c2+d2 am+bm = cm+dm; am+1+bm+1= cm+1+dm+1 P(m+2) : am+2+bm+2 = (a+b)(am+1+bm+1)–ab(am+bm) = (c+d)(cm+1+dm+1)–cd(cm+dm) = cm + 2 + dm + 2  P(m+2) holds true. an+bn = cn+dn holds true for n N

Class Exercise - 7 Solution: For n = 1, LHS = Cos Assume P(m) holds true

Solution Continued multiplying both sides by Cos2m :P(m+1) holds true As P(m)  P(m+1) P(n) is true for all n N

Class Exercise - 8 Solution: For n = 1, LHS = 1;RHS =9/8 LHS < RHS Let assume P(m) in true

Solution Continued P (m+1) holds true  P(n) holds true  n N

Class Exercise -9 Solution: For n = 1, LHS =7 RHS = 7 LHS = RHS Let P(n) holds true for n = m P(m):7+77+777+...+77...7(m times)

Solution Continued 7+77+777+...+77...7(m times) Adding77...7(m+1)times to both sides 7+77+...+77...7(m times)+77...7(m + 1) times

Solution Continued 7+77+...+77...7(m + 1) times  P(m + 1) holds true  P(n) is true 7+77+777+...+77...7(n times)

Class Exercise -10 Prove that Solution :- For n = 2,

Solution Continued P(m + 1) holds true. P(n) holds true , n > 1.

Thank you