1. MATHEMATICAL INDUCTION

2. SYNOPSIS Introduction Types of mathematical statements
Induction Method
Deduction method
Principle of mathematical induction
Problems

3. INTRODUCTION
The key basis for mathematical thinking is deductive and inductive reasoning.
There are mainly three methods to prove a mathematical statement. They are:
Inductive Method
Deductive method
Contradiction method

4. DEDUCTION METHOD
It is an application of general case to a particular case.
Example:
A number which is divisible by 9 is divisible by 3.
918 is divisible by 9 implies 9+1+8=18 and hence a multiple of 3.

5. Induction Method
Method of proving a statement from particular to general is called Induction.
Example:
2 is divisible by 2
4 is divisible by 2
6 is divisible by 2.
Every even number is divisible by 2.

6. ILLUSTRATION
To understand the basic principle of mathematical induction, the best example is that of thin rectangular tiles placed as shown in the next slide:

8. When the first tile is pushed, all the tiles will fall
When the first tile is pushed, all the tiles will fall. To be absolutely sure that the tiles will fall, it is optimum to know that the first tile falls and the event that any tile falls, its successor necessarily falls.
This is the major principle of mathematical induction.

9. Principle of mathematical induction
Let P(n) be a mathematical statement involving the natural number n such that
(i) P(1) is true and
(ii) P(k) is true P(k+1) is true.
P(n) is true for all natural numbers.

10. THE PRINCIPLE
THERE ARE 3 MAJOR STEPS IN SOLVING PROBLEMS ON INDUCTION:
1. Proving that the given statement P(n) is true for n=1.
2. Supposing that P(n) is true for n=k.
3. then proving that the statement is true for n=k+1.

11. Examples (i) Prove that 1+3+5+…+(2n-1)=n2 Solution:
Let P(n) be the given statement.
Put n=1. LHS=1=RHS. Hence P(1) is true.
Let P(k) be true.That is, 1+3+…+(2k-1)=k2
To Prove:P(k+1) is true.
Now,1+3+…+(2k-1)+(2k+1)=k2+(2k+1)
=(k+1)2
Hence it is true for all n.

12. Example
(ii) Prove by principle of mathematical induction that 25n>33n for all n N
Solution:
Let P(n):25n>33n
Here P(1):25>33 or 32>27, which is true.
Let P(m) be true. That is,25m>33m
Now, in P(m+1),LHS=2 5(m+1 )
=25m . 25
= m (1)

13. RHS=33(m+1)
=33m . 33
=27 .33m (2)
As 32>27,LHS>RHS and hence P(m+1) is true.
Hence P(n) is true for all values of n N

14. Example
(iii) Prove by principle of mathematical induction that n(n+1)(2n+1) is divisible by 6 for all natural numbers n.
Solution:
Let a=n(n+1)(2n+1)
now for n=1, a= 1(1+1)(2+1)
=1X2X3=6, which is divisible by 6.
Hence P(1) is true.

15. Let us suppose that P(m) is true.
Hence m(m+1)(2m+1) is divisible by 6.
Let m(m+1)(2m+1) =6k
Consider a for n=m+1
a= (m+1)(m+2)(2m+3)
= (m+1)(m+2)((2m+1)+2)
=(m+1)m(2m+1)+(m+1).2(2m+1)+2(m+1)(m+2)
= m(m+1)(2m+1)+2(m+1)(3m+3)
= m(m+1)(2m+1)+6(m+1)2

16. 6k+6(m+1)2
6(k+(m+1)2), which is divisible by 6.
Hence P(m+1) is true when P(m) is true.
Hence n(n+1)(2n+1) is divisible by 6

17. QUESTIONS FOR PRACTICE
Using principle of mathematical induction, prove that
(1) …+n=
(2) …+(2n-1)2 =
(3) …+n2>
(4) Prove that 2n>n for all n N
(5) For each natural number n, 32n-1 is divisible by 8.
(6) For each natural number n, 152n-1+1 is divisible by 16.
(7) Prove that n(n+1) is an even integer for all natural number n.
(8) Prove that (a b)n=anbn, for all natural number n.