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Mathematical induction Isaac Fung. Announcement ► Homework 1 released ► Due on 6 Oct 2008 (in class)

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Presentation on theme: "Mathematical induction Isaac Fung. Announcement ► Homework 1 released ► Due on 6 Oct 2008 (in class)"— Presentation transcript:

1 Mathematical induction Isaac Fung

2 Announcement ► Homework 1 released ► Due on 6 Oct 2008 (in class)

3 Overview ► Induction  equation  divisibility  inequality ► Strong induction ► Well ordering principle

4 Principle of induction ► Last time, we saw how to prove some predicate P(n) is true for all integers n by showing that P(n) is true for arbitrary n e.g. If n is any integer that is not a perfect square, then is irrational ► But this does not always work e.g. ► We may not know how to build a tower with n floors directly, yet knowing how to build the ground floor and how to build one new floor on top of it, we can build a tower with any number of floors as we like ► Advantage of induction:  usually easy to build the ground floor  easy to build a new floor on top of another

5 Principle of induction ► Suppose the followings are true, for what values of n is P(n) true?  P(1) and P(n)=>P(n+1)  P(3) and P(n)=>P(n+1)  P(1) and P(n)=>P(n+2)  P(1), P(2) and P(n)=>P(n+2)  P(1) and P(n)=>P(2n)  P(1) and P(n)^P(n+1)=>P(n+2)  P(1), P(2) and P(n)^P(n+1)=>P(n+2)  P(1) and P(1)^ … ^P(n)=>P(n+1)  1, 2, 3, 4, 5, …  3, 4, 5, 6, 7, …  1, 3, 5, 7, 9 …  1, 2, 3, 4, 5, …  1, 2, 4, 8, 16, … 1111  1, 2, 3, 4, 5, …

6 Induction (Proving equation) ► For any integer n>=0, ► Proof: ► We prove by induction on n. ► Let P(n) be the proposition that

7 ► Base case, n=0: So P(n) is true for n=0. ► Inductive step: Suppose that P(n) is true for some n>=0, that is. Can we change some n to all n?

8 ► Then, for n>=0, ► By induction, P(n) is true for all integers n>=0. By the inductive hypothesis

9 Induction (Proving equation) ► For any integer n>=2, ► Proof: ► We prove by induction on n. ► Let P(n) be the proposition that

10 ► Base case, n=2: So P(2) is true. So P(2) is true. ► Inductive step: Suppose that P(n) is true for some n>=2. So,

11 ► Then, for n>=2, ► By induction, P(n) is true for all integers n>=2. By the inductive hypothesis

12 Induction (Divisibility) ► For any integer n>=1, is divisible by 6 ► Proof: ► We prove by induction on n. ► Base case, n=1: is divisible by 6. is divisible by 6. So it is true for n=1.

13 ► Inductive step: Suppose that for some n>=1, is divisible by 6 is divisible by 6 ► Then, ► Either n+1 or n+2 is even, so the last term is divisible by 6. Therefore is divisible by 6. ► By induction, is divisible by 6 for all integers n>=1 By the inductive assumption

14 Induction (Proving inequality) ► For any integer n>=4, ► Proof: ► We prove by induction on n. ► Base case, n=4: So the claim is true for n=4.

15 ► Inductive step: Assume that for some n>=4 ► Then, ► By induction, for all integer n>=4. By the inductive hypothesis By assumption, n>=2 By assumption, n>=1

16 Induction (Alternative proof of infinitude of primes) ► For any integer n>=1, is divisible by at least n distinct primes ► Proof: ► We prove by induction on n. ► Base case, n=1: is divisible by 3. is divisible by 3. So the claim is true for n=1.

17 ► Inductive step: Assume that for some n>=1, is divisible by n distinct prime numbers. is divisible by n distinct prime numbers. ► Then, ► Let and ► By the inductive assumption, q has n distinct prime factors. ► Also, note that the p, q differ by 2, so they can have no common factors except 2. But both of them are odd, so they are relatively prime. ► Since p has at least 1 prime factor and it is not a factor of q, p*q has at least n distinct prime factors. ► By induction, is divisible by n distinct primes for all n>=1 x 2 -1=(x+1)(x-1)

18 Strong induction ► If sequence {a n } is defined as follows then a n =1 ► Proof: ► We proceed by strong induction. ► Let P(n) be the proposition that a n <= (7/4) n Can we use weak induction?

19 ► Base case, n=1, 2: By definition, a 1 = 1 <= (7/4) 1, a 2 = 3 <=(7/4) 2 ► Inductive step: Assume that P(k) is true for 1 =2 ► a n+1 = a n + a n-1 <= (7/4) n + (7/4) n-1 <= (7/4) n + (7/4) n-1 = (7/4) n-1 (7/4+1) = (7/4) n-1 (7/4+1) <= (7/4) n-1 (7/4) 2 <= (7/4) n-1 (7/4) 2 = (7/4) n+1 = (7/4) n+1 ► By the principle of strong induction, P(n) is true for all n>=1 by the inductive assumption Can we assume P(n)^P(n+1) and then derive P(n+2)? Why we can’t just check the case n=1?

20 Strong induction ► If sequence {a n } is defined as follows then for any integer n>=0 ► Proof: ► We proceed by strong induction. ► Let P(n) be the proposition that for n = 0 for n > 0

21 ► Base case, n=0: By definition, a 0 = 1 ► Inductive step: Suppose that P(k) is true for 0 =0 ► Consider a n+1 ► By induction, P(n) is true for all integers n>=0 by the inductive hypothesis GP sum

22 Strong induction ► Every integer n>=1 can be expressed as the sum of distinct Fibonacci numbers ► Proof: ► We proceed by strong induction ► Let P(n) be the statement that “ n can be written as the sum of distinct Fibonacci numbers ” ► Base case, n=1: 1 is a Fibonacci number, so P(1) is true

23 ► Inductive step: Assume that P(k) holds for 1 =1 ► Consider n+1. If n+1 is a Fibonacci number, then we are done If not, we have, for some m>=1 f m < n+1 < f m+1 ► This gives n+1 = f m + (n+1-f m ) ► Now n+1-f m < n+1, so by the inductive hypothesis, n f m = f j1 + f j2 + f j3 + … for some distinct f j1, f j2, …

24 ► (cont) ► Now n+1-f m < n+1, so by the inductive hypothesis, n f m = f j1 + f j2 + f j3 + … for some distinct f j1, f j2, … ► Moreover, none of these f ji is f m as f m+1 < 2f m. If some f ji =f m, then 2f m <=n+1 and f m cannot be the largest Fibonacci number <=n+1 ► Therefore, n + 1 = f m + f j1 + f j2 + f j3 + … So P(n+1) is true. ► By the principle of strong induction, P(n) is true for all integers n>=1

25 Strong induction ► ► To divide a chocolate bar with m × n squares into unit squares, we need mn − 1 cuts ► ► Proof: ► ► Let A = m × n ► ► We prove by strong induction on A. ► ► Base case, A=1: 0 cut is needed Should we induct on m or n?

26 ► Inductive step: ► Assume that a chocolate bar of area B needs B-1 cuts for 1 =1 ► Consider a chocolate bar of area A+1 ► We can cut it once and divide it into two pieces of area A 1 and A 2, where A 1 +A 2 = A+1 ► By the inductive hypothesis, we need A 1 -1 and A 2 -1 cuts to divide them. So in total, we need A A = A cuts ► By the principle of strong induction, we ► By the principle of strong induction, we need mn − 1 cuts to divide a chocolate bar with m × n squares

27 Well ordering principle ► The equation has no non-zero integer solution ► Proof: ► Assume to the contrary that there are non-zero integers u, v, w and z satisfying this equation ► By the well ordering principle, there are non-zero integers u, v, w and z satisfying this equation, such that they do not share any common prime factors

28 ► Since the left hand side is even, x is also even, i.e. x = 2k for some integer k ► This implies w is even, i.e. w = 2l for some integer l ► Repeat the same argument, we can deduce that u and v are also even ► This contradicts our assumption that u, v, x and w do not have common prime factors

29 Additional references ► ► ► /training-induc.pdf /training-induc.pdf /training-induc.pdf ► Science/6-042JSpring-2005/8C9DC16C DA2-B58E- 850E90431AE1/0/rec5.pdf Science/6-042JSpring-2005/8C9DC16C DA2-B58E- 850E90431AE1/0/rec5.pdf Science/6-042JSpring-2005/8C9DC16C DA2-B58E- 850E90431AE1/0/rec5.pdf ► Science/6-042JSpring-2005/F592ADD6-24E C- A12A545EB2FB/0/rec4.pdf Science/6-042JSpring-2005/F592ADD6-24E C- A12A545EB2FB/0/rec4.pdf Science/6-042JSpring-2005/F592ADD6-24E C- A12A545EB2FB/0/rec4.pdf ►


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