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Mathematical induction

Mathematical induction. Isaac Fung. Announcement. Homework 1 released Due on 6 Oct 2008 (in class). Overview. Induction equation divisibility inequality Strong induction Well ordering principle. Principle of induction.

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Mathematical induction

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  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, 2, 3, 4, 5, … • 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. • 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. So it is true for n=1.

  13. Inductive step: Suppose that for some n>=1, 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. So the claim is true for n=1.

  17. Inductive step: Assume that for some n>=1, 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 x2-1=(x+1)(x-1)

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

  19. Why we can’t just check the case n=1? • Base case, n=1, 2: By definition, a1 = 1 <= (7/4)1, a2 = 3 <=(7/4)2 • Inductive step: Assume that P(k) is true for 1 <= k <=n for some n>=2 • an+1 = an + an-1 <= (7/4)n + (7/4)n-1 = (7/4)n-1 (7/4+1) <= (7/4)n-1 (7/4)2 = (7/4)n+1 • By the principle of strong induction, P(n) is true for all n>=1 Can we assume P(n)^P(n+1) and then derive P(n+2)? by the inductive assumption

  20. Strong induction • If sequence {an} 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, a0 = 1 • Inductive step: Suppose that P(k) is true for 0<=k<=n, for some n>=0 • Consider an+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 <= k <= n for some n>=1 • Consider n+1. If n+1 is a Fibonacci number, then we are done If not, we have, for some m>=1 fm < n+1 < fm+1 • This gives n+1 = fm + (n+1-fm) • Now n+1-fm < n+1, so by the inductive hypothesis, n + 1 - fm = fj1 + fj2 + fj3 + … for some distinct fj1, fj2, …

  24. (cont) • Now n+1-fm < n+1, so by the inductive hypothesis, n + 1 - fm = fj1 + fj2 + fj3 + … for some distinct fj1, fj2, … • Moreover, none of these fji is fm as fm+1 < 2fm . If some fji=fm, then 2fm<=n+1 and fm cannot be the largest Fibonacci number <=n+1 • Therefore, n + 1 = fm + fj1 + fj2 + fj3 + … 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<=B<=A for some A>=1 • Consider a chocolate bar of area A+1 • We can cut it once and divide it into two pieces of area A1 and A2, where A1+A2 = A+1 • By the inductive hypothesis, we need A1-1 and A2-1 cuts to divide them. So in total, we need A1-1 + A2-1 + 1 = A cuts • 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 • http://hkn.eecs.berkeley.edu/~min/cs70/sect2/solution2a.pdf • http://hkn.eecs.berkeley.edu/~min/cs70/sect3/solution3.pdf • http://www.math.northwestern.edu/~mlerma/problem_solving/putnam/training-induc.pdf • http://ocw.mit.edu/NR/rdonlyres/Electrical-Engineering-and-Computer-Science/6-042JSpring-2005/8C9DC16C-1328-4DA2-B58E-850E90431AE1/0/rec5.pdf • http://ocw.mit.edu/NR/rdonlyres/Electrical-Engineering-and-Computer-Science/6-042JSpring-2005/F592ADD6-24E6-4797-800C-A12A545EB2FB/0/rec4.pdf • http://www.math.udel.edu/~lazebnik/papers/invariants.pdf

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