Principle of Mathematical Induction

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Session
Principle of Mathematical Induction
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Session Objective
1. Introduction 2. Steps involved in the use of
mathematical induction 3. Principle
of mathematical induction.
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Statement
Statement- A sentence which can be judged as
true or false.
Example1. 2 is only even prime number 2.
Bagdad is capital of Iraq 3. 2n5 is
always divisible by 5 for all n?N
Mathematical statement
Example 1 and 3.
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Induction
Induction Its a process Particular ? General
Example Statement- 2n1 is odd number.
n1 2.113 is odd.
True
n2 2.215 is odd.
True
n3 2.317 is odd.
True
Observation ? tentative conclusion (2n1 is
odd)
let its true for nm. i.e 2m1 is odd.
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Induction
for nm1
2(m1)1 2m12
odd 2odd
Now it is Generalized
?2n1 is odd for all n
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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
2. to check divisibility of a expression by a
number
Ex Prove n35n is divisible by 3.
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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 nm?N
i.e. P(m) is true.
Step 3 Using above assumption prove P(m1) is
true. i.e P(m) ? P (m1)
Step 4 Above steps lead us to generalize the
fact P(n) is true for all n ?N.
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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.
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Illustrative Example
Solution
Step 1.
for n1, p(1)1. 33 (LHS)
?L.H.SR.H.S
P(1) is true
Step2. Assume that P(m) is true
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Solution Continued
Step3 To prove P(m 1) holds true Adding.
(m 1).3m1 to both sides
?P(m) ? P (m1)
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Solution Continued
Step4. As P(m) ? P (m1) P(n) is true
for all n ?N
?1.32.323.33......n.3n
(Proved)
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Illustrative Example
Prove n35n is divisible by 3 for n ?N
(By PMI or Otherwise)
Solution
P(n) n35n is divisible by 3
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Solution Continued
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Class Exercise - 6
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.
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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
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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.
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Alternative Solution
Alternative Method without PMI
n35n n(n25)
n(n2 -16)
n(n2-1)6n
n(n-1)(n1)6n
Product of three consecutive numbers
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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
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Class Exercise -3
Solution
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Solution Continued
Adding (m1)(m2)(m3)
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Solution Continued
Step4. As P(m) ? P (m1)
P(n) is true for all n ?N
? 1.2.3.2.3.43.4.5...n(n1)(n2)
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Class Exercise -4
If abcd and a2b2c2d2 , then show by
mathematical induction, anbn cndn
Solution
?P(1) and P(2) hold true.
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Solution Continued
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Class Exercise - 7
Solution
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Solution Continued
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Class Exercise - 8
Solution
For n 1, LHS 1RHS 9/8 ?LHS lt RHS
Let assume P(m) in true
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Solution Continued
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Class Exercise -9
Solution
Let P(n) holds true for n m
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Solution Continued
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Solution Continued
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Class Exercise -10
Solution -
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Solution Continued
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