Proof Technique - Discrete Mathematics - Lecture Slides, Slides of Discrete Mathematics

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CSE115/ENGR160 Discrete Mathematics 03/29/

4.1 Mathematical induction

• Want to know whether we can reach every step of this ladder - We can reach first rung of the ladder - If we can reach a particular run of the ladder, then we can reach the next run
• Mathematical induction : show that p(n) is true for every positive integer n

Analogy

Example

• Show that 1+2+… +n=n(n+1)/2, if n is a positive integer - Let p(n) be the proposition that 1+2+… +n=n(n+1)/ - Basis step: p(1) is true, because 1=1*(1+1)/ - Inductive step: Assume p(k) is true for an arbitrary k. That is, 1+2+…+k=k(k+1)/ We must show that 1+2+…+(k+1)=(k+1)(k+2)/ From p(k), 1+2+…+k+(k+1)=k(k+1)/2+(k+1)=(k+1)(k+2)/ which means p(k+1) is true - We have completed the basic and inductive steps, so by mathematical induction we know that p(n) is true for all positive integers n. That is 1+2+…+n=n(n+1)/

Example

• Let p(n) denote the proposition
• Basic step: p(1)=1^2 =
• Inductive steps: Assume that p(k) is true, i.e., 1+3+5+…+(2k- 1)=k 2 We must show 1+3+5+…+(2k+1)=(k+1) 2 is true for p(k+1) Thus,1+3+5+…+(2k-1)+(2k+1)=k^2 +2k+1=(k+1) 2 which means p(k+1) is true (Note p(k+1) means 1+3+5+…+(2k+1)=(k+1) 2 )
• We have completed both the basis and inductive steps. That is, we have shown p(1) is true and p(k)p(k+1)
• Consequently, p(n) is true for all positive integers n

Example

• Use mathematical induction to show that 1+2+2^2 +…+2n^ =2n+1-
• Let p(n) be the proposition: 1+2+2^2 +…+2n^ =2n+1-
• Basis step: p(0)=20+1-1=
• Inductive step: Assume p(k) is true, i.e., 1+2+2^2 +…+2k=2k+1- It follows (1+2+2^2 +…+2k)+2k+1=(2k+1-1)+2k+1=2*2k+1^ -1=2k+2-1 which means p(k+1): 1+2+2^2 +…+2k+1=2k+2-1 is true
• We have completed both the basis and inductive steps. By induction, we show that 1+2+2^2 +…+2n^ =2n+1-

Example

• Use induction to show
• Basis step: p(0) is true as
• Inductive step: assume
• So p(k+1) is true. By induction, p(n) is true for all nonnegative integers (^10)

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Example

• Use induction to show that n<2 n^ for n>
• Basis step: p(1) is true as 1<2 1 =
• Inductive step: Assume p(k) is true, i.e., k<2 k

We need to show k+1<2 k+ k+1<2k+1≤2k^ +2k^ =2k+ Thus p(k+1) is true

• We complete both basis and inductive steps, and show that p(n) is true for all positive integers n (^) Docsity.com 11

Example

• Show that n^3 -n is divisible by 3 when n is positive
• Basis step: p(1) is true as 1-1=0 is divisible by 3
• Inductive step: Suppose p(k)= k^3 -k is true, we must show that (k+1)^3 -(k+1) is divisible by 3 (k+1) 3 -(k+1)=k^3 +3k 2 +3k+1-(k+1)=(k^3 -k)+3(k^2 +k) As both terms are divisible by 3, (k+1) 3 -(k+1) is divisible by 3
• We have completed both the basis and inductive steps. By induction, we show that n^3 -n is divisible by 3 when n is positive

Example

• Show that if S is a finite set with n elements, then S has 2 n^ subsets
• Let p(n) be the proposition that a set with n elements has 2 n^ subsets
• Basis step: p(0) is true as a set with zero elements, the empty set, has exactly 1 subset
• Inductive step: Assume p(k) is true, i.e., S has 2 k^ subsets if |S|=k.

Example

• Use mathematical induction to show one of the De Morgan’s law: where A 1 , A 2 , …, An are subsets of a universal set U, and n≥
• Basis step: (proved Section 2.2, page 125)
• Inductive step: Assume is true for k≥

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Axioms for the set of positive

integers

• Axiom 1: The number 1 is a positive integer
• Axiom 2: If n is a positive integer, then n+1, the successor of n, is also a positive integer
• Axiom 3: Every positive integer other than 1 is the successor of a positive integer
• Axiom 4: Well-ordering property Every non- empty subset of the set of positive integers has a least element

Why mathematical induction is

valid?

• Because m-1 is less than m, it is not in S
• So p(m-1) must be true
• As the conditional statement p(m-1)p(m) is also true, it must be the case that p(m) is true
• This contradicts the choice of m
• Thus, p(n) must be true for every positive integer n

4.2 Strong induction and well-

ordering

• Strong induction : To prove p(n) is true for all positive integers n, where p(n) is a propositional function, we complete two steps
• Basis step: we verify that the proposition p(1) is true
• Inductive step: we show that the conditional statement (p(1)∧p(2) ∧… ∧p(k))→p(k+1) is true for all positive integers k