ppt02.ppt
- 1. 1 PMAS ARID AGRICULTURE UNIVERSITY RAWALPINDI BS SE (1B)EVENING. COURSE TITLE: DISCRETE STRUCTURE Group member: 1.ASIM SHAH(22-Arid-1019) 2.MASOOD-UR-REHMAN(22-Arid-1018) 3.SALIHA ALTAF(22-Arid-1054) UNIVERSITY INSTITUTE OF INFORMATION TECHNOLOGY
- 2. AGENDA OF TODAY Principle of mathematical induction 2.
- 3. INDUCTION •The principle of mathematical induction is a useful tool for proving that a certain predicate is true for all natural numbers. •It cannot be used to discover theorems, but only to prove them. •If we have a propositional function P(n), and we want to prove that P(n) is true for any natural number n, we do the following: •Show that P(1) is true. (basis step) • Show that if P(n) then P(n + 1) for any n N. (inductive step) • Then P(n) must be true for any n N. (conclusion)
- 4. • Example: • Show that n < 2n for all positive integers n. • Let P(n) be the proposition “n < 2n.” • 1. Show that P(1) is true. (basis step) • P(1) is true, because 1 < 21 = 2.
- 5. INDUCTION • 2. Show that if P(n) is true, then P(n + 1) is true. (inductive step) • Assume that n < 2n is true. • We need to show that P(n + 1) is true, i.e. • n + 1 < 2n+1 • We start from n < 2n: • n + 1 < 2n + 1 2n + 2n = 2n+1 • Therefore, if n < 2n then n + 1 < 2n+1
- 6. INDUCTION Then P(n) must be true for any positive integer.(conclusion) • n < 2n is true for any positive integer. • End of proof. • Another Example • 1 + 2 + … + n = n (n + 1)/2 1. Show that P(0) is true. (basis step) • For n = 0 we get 0 = 0. True.
- 7. INDUCTION 2. Show that if P(n) then P(n + 1) for any n N. (inductive step) • 1 + 2 + … + n = n (n + 1)/2 • 1 + 2 + … + n + (n + 1) = n (n + 1)/2 + (n + 1) • = (2n + 2 + n (n + 1))/2 • = (2n + 2 + n2 + n)/2 • = (2 + 3n + n2 )/2 • = (n + 1) (n + 2)/2 • = (n + 1) ((n + 1) + 1)/2
- 8. INDUCTION 3. Then P(n) must be true for any n N. (conclusion) • 1 + 2 + … + n = n (n + 1)/2 is true for all n N. • End of proof. 8
- 9. INDUCTION •There is another proof technique that is very similar to the principle of mathematical induction. •It is called the second principle of mathematical induction. •It can be used to prove that a propositional function P(n) is true for any natural number n.
- 10. INDUCTION • The second principle of mathematical induction: • Show that P(0) is true. (basis step) • Show that if P(0) and P(1) and … and P(n), then P(n + 1) for any n N. (inductive step) • Then P(n) must be true for any n N. (conclusion)
- 11. INDUCTION •Example: Show that every integer greater than 1 can be written as the product of primes. • Show that P(2) is true. (basis step) •2 is the product of one prime: itself.
- 12. INDUCTION • Show that if P(2) and P(3) and … and P(n),then P(n + 1) for any n N. (inductive step) • Two possible cases: • If (n + 1) is prime, then obviously P(n + 1) is true. • If (n + 1) is composite, it can be written as the product of two integers a and b such that 2 a b < n + 1. • By the induction hypothesis, both a and b can be written as the product of primes. • Therefore, n + 1 = a b can be written as the product of primes.
- 13. INDUCTION • Then P(n) must be true for any n N. (conclusion) •End of proof. •We have shown that every integer greater than 1 can be written as the product of primes. •Another example: 1+3+5+…+(2n-1)=n2
- 14. INTRODUCTION 14 Step 1: It is true for initial value. So for n=1 P(1):1(1)2 1=1 So p(n) for n=1 is true. Step 2: We assume that P(n) is true for n=1,2,3……k Step: Prove it is true Then we show that p(n) is true for n=k+1 P(k):1+3+5+…..+(2k-1)=k2 For n=k+1 P(k+1):1+3+5+……+(2K-1)+2(k+1)-1=(k+1)2 K +2k+2-1=(k+1) K +2k+1=(k+1) (k+1) =(k+1) 2 2 2 2 2 2