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CSNB 143 Discrete Mathematical Structures

CSNB 143 Discrete Mathematical Structures. Chapter 6 – Induction. 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.

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CSNB 143 Discrete Mathematical Structures

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  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; 1 + 2 + 3 + … + n = n (n+1) 2 Basic step • Prove that P(no) is true. no = 1, so • P(1) = 1 (1 + 1) = 1 (first element) 2 • Therefore, it is true.

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

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

  13. Identify P(k) left side in P(k + 1) left side. • P(k+1) = 1 + 2 + 3 + …..+ 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) = 1 + 2 + 3 + … + k = k (k+1) 2 • P(k+1) = 1 + 2 + 3 + … + k + (k+1) = = k (k+1) + (k+1) 2 P(k) left side P(k) right side

  15. Get the right side of P(k+1). • P(k+1) = 1 + 2 + 3 + …. + k + (k+1) = k (k+1) + 2 (k+1) 2 = k2 + k + 2k + 2 2 = k2 + 3k + 2 2 = (k+1) (k+2) 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 1 + 3 + 5 + …. + (2n – 1) = n2; for all n 1. • Show that 1 + 5 + 9 + …. + (4n – 3) = n(2n – 1); n 1. • Show that 1 + 2 + 22 + 23 + … + 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). Understood that 4 is 3 + 1, so = 4.4k - 1 = (3 + 1).4k – 1 = 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.

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