Induction: Part 2

1
Induction Part 2

• Sections 4.2-4.3

2
Proof Techniques

• Direct
• Assume Hypothesis is True, prove truth of
  conclusion
• Method of Exhaustion
• Method of Generalizing
• Counterexample
• Indirect
• Assume conclusion is False, prove premise is
  False
• Contradiction
• Contrapositive
• Induction
• Used to conjecture about outcomes that occur
  repeatedly according to a pattern
• E.g., P(n) is true for all positive integers
• The sum of the first n positive integers 123 .
  . . n n(n1)/2

3
Steps in an Induction Proof

• Basis Step The proposition P(n) is shown to be
  true for n1 (or, more generally, the first
  element in the set)
• Inductive Step The implication P(n)?P(n1) is
  shown to be true for every positive integer n
• P(0)
• P(n)?P(n1)
• ? ?n P(n)

For n?Z P(1)??n(P(n)?P(n1)) ??nP(n)
4
Proof Sum of first n integers

• If p(n) is the proposition that the sum of the
  first n positive integers is n(n1)/2, prove p(n)
  for n?Z
• Basis Step We will show p(1) is true
• p(1) 1(11)/2 2/2 1
• Inductive Step We want to show that p(n) ?
  p(n1)
• Inductive Hypothesis - Assume p(k) is True
• P(k) 1234. . . k k(k1)/2
• Show p(k1) is True
• 1234. . . k (k1) (k1)(k11)/2
• k(k1)/2 by IH so..
• k(k1)/2 (k1) (k1)(k2)/2 

5
Sum of first n integers (contd)

• k(k1)/2 (k1) (k1)(k2)/2 
• k(k1)/2 2(k1)/2 (k1)(k2)/2
• (k2k2k2)/2 (k23k2)/2
• (k23k2)/2 (k23k2)/2
• Since p(1) is true and p(n) ? p(n1), then p(n)
  is true for all positive integers n

6
Proof Sum of first n odd integers

• If p(n) is the proposition that the sum of the
  first n odd integers is n2, prove p(n) for n?Z
• P(n) 1 3 5 (2n-1) n2 for n 1
• Basis Step We will show p(1) is true
• p(1) 1 12
• Inductive Step We want to show that p(n) ?
  p(n1)
• Inductive Hypothesis - Assume p(k) is True
• P(k) 1234. . . (2k-1) k2
• Show p(k1) is True
• 1234. . . (2k-1) (2k1) (k1)2
• k2 by IH so..

7
Sum of first n odd integers (contd)

• k2 (2k1) (k1)2  
• k2 2k 1 k2 2k 1
• Since p(1) is true and p(n) ? p(n1), then p(n)
  is true for all positive integers n

8
Proof 8n-2n is divisible by 6

• Let p(n) be the statement that all numbers of the
  form 8n-2n for n?Z are divisible by 6
• Basis Step We will show that p(1) is true
• 81-21 6 which is clearly divisible by 6
• Inductive Step We want to show that p(n) ?
  p(n1)
• Inductive Hypothesis - Assume p(k) is True
• p(k) 8k-2k is divisible by 6
• Show p(k1) is True
• p(k1) 8k1 - 2k1 is divisible by 6

9
Divisible by 6 Example (cont.)

• 8k1 - 2k1 8(8k) - 2k1
• 8(8k) - 8(2k) 8(2k) - 2k1
• 8(8k-2k) 8(2k) - 2k1
• 8(8k-2k) 8(2k) 2(2k)
• 8(8k-2k) 6(2k)
• By the inductive hypothesis 8(8k-2k) is
  divisible by 6 and clearly 6(2k) is divisible by
  6. Thus 8k1 - 2k1 is divisible by 6. Since p(1)
  is true and p(n) ? p(n1), then p(n) is true for
  all nonnegative integers n.

10
Proof 21 divides 4n1 52n-1

• Prove that 21 divides 4n1 52n-1 whenever n is
  a positive integer
• Basis Step We will show that p(1) is true
• P(1) 4n1 52n-1 411 52(1)-1 425 21
  which is clearly divisible by 21
• Inductive Step We want to show that p(k) ?
  p(k1)
• Inductive Hypothesis - Assume p(k) is True
• p(k) 4k1 52k-1 is divisible by 21
• Show p(k1) is True
• p(k1) 4k11 52(k1)-1 is divisible by 21

11
21 divides 4n1 52n-1 (contd)

• 4k11 52(k1)-1 44k1 52k2-1
• 44k1 2552k-1
• 44k1 (421)2k-1
• 4(4k1 52k-1) 2152k-1
• The first term 4(4k1 52k-1) is divisible by 21
  by the induction hypothesis. The second term
  2152k-1 is clearly divisible by 21. Therefore
  their sum is divisible by 21.

12
Inequality Proof

• Prove that 2n1 lt 2n for all integers n 3
• P(n) 2n1 lt 2n
• Basis Step We will show that p(3) is true
• P(3) 23 1 lt 23 or 7 lt 8 which is clearly true
• Inductive Step We want to show that p(k) ?
  p(k1)
• Inductive Hypothesis - Assume p(k) is True
• P(k) 2k1 lt 2k
• Show P(k1) is True
• P(k1) 2(k1)1 lt 2(k1)

13
Inequality (cont)

• 2(k1)1 lt 2(k1)
• 2k21 lt 2(k1)
• (2k1)2 lt 22k
• (2k1)2 lt 2k 2k
• By the Inductive Hypothesis, we know that 2k1 lt
  2k
• And 2 lt 2k for integers k 2
• So, 2k3 lt 2(k1) since n 3
• Since P(3) is True and P(k) ? P(k1), then P(n)
  is true for all integers n 3

14
Another Inequality Proof

• Prove that 3n5 n2 for all integers n 5
• P(n) 3n5 n2
• Basis Step We will show that P(5) is true
• P(5) 15 5 52 or 20 25 which is clearly
  true
• Inductive Step We want to show that P(k) ?
  P(k1)
• Inductive Hypothesis - Assume p(k) is True
• P(k) 3k5 k2
• Show P(k1) is True
• P(k1) 3(k1)5 (k1)2

15
Proof continued

• 3(k1)5 (k1)2
• 3k35 k2 2k 1
• (3k5)3 k2 2k 1
• By the Inductive Hypothesis, we know that 3k5
  k2
• And 3 2k 1 for integers k 1
• So, 3k8 k2 2k 1 since n 5
• Since P(5) is True and P(k) ? P(k1), then P(n)
  is true for all integers n 5