1. CSNB 143 Discrete Mathematical Structures
Chapter 6 – Induction

2. OBJECTIVES
Student should be able to understand the meaning of Principle of Mathematical Induction.
Students should be able to understand clearly each steps involved in different type of induction.
Students should know how to use induction in daily lives.

3. What, Which, Where, When
Concept of induction (domino) (Clear / Not Clear)
Difference between types of induction
Summation type (Clear / Not Clear)
Division type (Clear / Not Clear)
Identify basic step (Clear / Not Clear)

4. Identify induction steps for
Summation type (Clear / Not Clear)
Division type (Clear / Not Clear)

5. MATHEMATICAL INDUCTION
One of the proof techniques.
It acts like a domino.
Let say the statement to be proved can be written as n  no P(n), where no is some fixed integer.
Suppose we wish to show that P(n) is true for all integers n  no.

6. Suppose also,
P(no) is true, and
If P(k) is true for some k  no, then P(k + 1) must also be true.
Then P(n) is true for all n  no.
This result is called the Principle of Mathematical Induction.

7. Steps involved are:
Prove that P(no) is true.
This is called as basic step.
If this step is not true, then the next step is not relevant anymore.
Prove that P(k)  P(k + 1) is a tautology for all k  no.
This is called as induction steps. .
This step will prove that implication will always be true.

8. Types of induction.
There are several types of induction:
Summation type
Division type
Comparison type

9. Summation type Basic Step Prove that P(no) is true. Induction Steps
Find P(k).
Find P(k + 1). Concentrate on the left side.
Identify P(k) left side in P(k + 1) left side.
Replace P(k) left side with P(k) right side from b).
Get the right side of P(k+1).
Conclusion
Simplify.

10. Ex 1: Summation type.
Show by mathematical induction, for all n  1; … + n = n (n+1) 2
Basic step
Prove that P(no) is true.
no = 1, so
P(1) = 1 (1 + 1) = 1 (first element)
Therefore, it is true.

11. Consider for any number k  1, P(k) = 1 + 2 + 3 + … + k = k (k+1) 2
Induction steps
Find P(k).
Consider for any number k  1,
P(k) = … + k = k (k+1) 2

12. P(k+1) = 1 + 2 + 3 + …. + (k+1) = (k+1)[(k+1)+1] 2 = (k+1) (k+2) 2
Find P(k + 1). Concentrate on the left side.
P(k+1) = …. + (k+1) = (k+1)[(k+1)+1]
= (k+1) (k+2) 2
left side right side

13. Identify P(k) left side in P(k + 1) left side.
P(k+1) = …..+ k + (k+1)
P(k + 1) Left side
P(k) left side

14. Replace P(k) left side with P(k) right side from b).
P(k) = … + k = k (k+1) 2
P(k+1) = … + k + (k+1) =
= k (k+1) + (k+1)
P(k) left side P(k) right side

15. Get the right side of P(k+1).
P(k+1) = …. + k + (k+1) =
k (k+1) + 2 (k+1) 2
= k2 + k + 2k + 2
= k2 + 3k + 2
= (k+1) (k+2)

16. That is, the right side of P(k + 1).
Conclusion
So, with Principle of Mathematical Induction, P(n) is true for all n  1.

17. Exercise 1: Show that 2 + 4 + 6 + ….. + 2n = n(n+1) for all n  1.
Show that …. + (2n – 1) = n2; for all n  1.
Show that …. + (4n – 3) = n(2n – 1); n  1.
Show that … + 2n = 2n+1 – 1; n  0.

18. Division Type Basic Step Prove that P(no) is true. Induction Steps
Get the P(k).
Get the P(k + 1).
Separate P(k+1) to any form, close to P(k).
Identify P(k) from b).
Conclusion
Different parts.

19. Show by mathematical induction, for n  1, 4n – 1 is divisible by 3.
Basic step
Prove that P(no) is true.
no = 1, so
P(1) = 41 – 1
= 3
Therefore, it is divisible by 3, so it is true.

20. Induction steps
Get the P(k).
P(k) = 4k – 1
Get the P(k + 1)
P(k+1) = 4k+1 – 1
= 4.4k - 1

21. Separate P(k+1) to any form, close to P(k). = 4.4k - 1
Understood that 4 is 3 + 1, so
= 4.4k - 1
= (3 + 1).4k – = 3.4k + 1.4k – 1
Identify P(k) from b).
= 3.4k + 4k –1

22. From different parts
4k – 1 is divisible by 3 because P(k) is true.
3.4k is also divisible by 3 because of whatever the value of k,
3.4k is always divisible by 3.
Therefore, 4n – 1 is divisible by 3 for all n  1.

23. Exercise 2:
Show that 22n – 1 is divisible by 3, for all integers n  1.
Show that 7n – 2n is divisible by 5, for all integers n  1.
Show that 6(7n) – 2(3n) is divisible by 4, for all n  1.