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CSNB143 – Discrete Structure

CSNB143 – Discrete Structure. MATHEMATICAL INDUCTION. MATHEMATICAL INDUCTION. Learning Outcomes 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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CSNB143 – Discrete Structure

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  1. CSNB143 – Discrete Structure MATHEMATICAL INDUCTION

  2. MATHEMATICAL INDUCTION Learning Outcomes • 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. 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. • 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.

  4. MATHEMATICAL INDUCTION • 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 Types of induction. • There are several types of induction: • Summation type • Division type • Comparison type

  5. MATHEMATICAL INDUCTION 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.

  6. MATHEMATICAL INDUCTION 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.

  7. MATHEMATICAL INDUCTION Induction steps • Find P(k). Consider for any number k  1, P(k) = 1 + 2 + 3 + … + k = k (k+1) 2 • Find P(k + 1). Concentrate on the left side. P(k+1) = 1 + 2 + 3 + …. + (k+1) = (k+1)[(k+1) + 1] = (k+1) (k+2) 22 left side right side • 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

  8. MATHEMATICAL INDUCTION • Replace P(k) left side with P(k) right side from b). P(k+1) = 1 + 2 + 3 + … + k + (k+1) = k (k+1) + (k+1) 2 P(k) left side P(k) right side • 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 That is, the right side of P(k + 1). Conclusion: • So, with Principle of Mathematical Induction, P(n) is true for all n 1.

  9. MATHEMATICAL INDUCTION • 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.

  10. MATHEMATICAL INDUCTION 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.

  11. MATHEMATICAL INDUCTION • Ex 2: Division type. 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.

  12. MATHEMATICAL INDUCTION • Induction steps • Get the P(k). P(k) = 4k – 1 • Get the P(k + 1) P(k+1) = 4k+1 – 1 = 4.4k– 1 • 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 (same as if (a + b)c = ac + ab) = 3.4k + 1.4k – 1

  13. MATHEMATICAL INDUCTION • Identify P(k) from b). = 3.4k + 4k –1 Conclusion • From different parts 4k – 1 is divisible by 3 because P(k) is true. 3.4kis 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. • 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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