3.3 Mathematical Induction 1 Follow me for a walk through...

Mathematical Induction Mathematical Induction The principle of mathematical induction 2 … ….. If Person #1 knows the secret For all n, if person #n knows the secret, then so does person #(n+1) Then For all n, person #n knows the secret

Mathematical 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. 3

Mathematical Induction 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(0) is true.(basis step) Show that if P(k) then P(k + 1) for any k (k  N).(inductive step). In the inductive step, we usually assume that P(k) is true (this assumption is called the inductive hypothesis) and try to show that then P(k+1) is also true. Then P(n) must be true for any n (n  N). (conclusion) 4

Mathematical Induction Examples of proofs by mathematical induction 5

Mathematical Induction Example1 Use mathematical induction to prove that the sum of the first m odd positive integers is n². Solution: Let P(n) be the proposition “ ….. + (2n - 1) = n².” 1. Show that P(1) is true. (basis step) P(1) is true, because 1 = 1². 6

Mathematical Induction 2. Show that if P(k) is true, then P(k + 1) is true. (inductive step) Assume that ….. + (2k - 1) = k² is true. We need to show that P(k + 1) is true, i.e ….. + (2k - 1) + (2k + 1) = (k + 1)² Indeed, By the inductive hypothesis ….. + (2k - 1) + (2k + 1) = k² + (2k + 1) = k² + 2k + 1 = (k + 1)² 7

Mathematical Induction 3. Then P(n) must be true for any positive integer. (conclusion) ….. + (2n - 1) = n² is true for any positive integer. End of proof. 8

Mathematical Induction Example2: Show that n < 2 n for all positive integers n. Let P(n) be the proposition “n < 2 n.” 1. Show that P(1) is true. (basis step) P(1) is true, because 1 < 2 1 = 2. 9

Mathematical Induction 2. Show that if P(k) is true, then P(k + 1) is true. (inductive step) Assume that k < 2 k is true. We need to show that P(k + 1) is true, i.e. k + 1 < 2 k+1 We start from k < 2 k : k + 1 < 2 k + 1  2 k + 2 k = 2 k+1 Therefore, if k < 2 k then k + 1 < 2 k+1 10

Mathematical Induction 3. Then P(n) must be true for any positive integer. (conclusion) n < 2 n is true for any positive integer. End of proof. 11

Mathematical Induction Example3 Use mathematical induction to prove that: … + n = n (n + 1)/2 Let P(n) be the proposition “ 1+2 +…+n = n (n+1)/2.” 1. Show that P(1) is true. (basis step) For n = 1 we get 1 ?1(1+1)/2. P(1) is true. 12

Mathematical Induction 2. Show that if P(k) then P(k + 1) for any k  N. (inductive step) P(k): … + k = k (k + 1)/2 is true P(k + 1): … + k + (k + 1) = k (k + 1)/2 + (k + 1) = (k + 1)(k/2 + 1) = (k + 1)(k + 2)/2 = (k + 1) ((k + 1) + 1)/2 13

Mathematical Induction 3. Then P(n) must be true for any positive integer. (conclusion) … + n = n (n + 1)/2 is true for any positive integer. End of proof. 14

Mathematical Induction Example4 Use mathematical induction to prove that: n³ - n is divisible by 3, for all positive integers n. Let P(n) be the proposition “ n³ - n is divisible by 3.” 1. Show that P(1) is true. (basis step) For n = 1 we get 1³ - 1 = 1 – 1 = 0 is divisible by 3 P(1) is true. 15

Mathematical Induction 2. Show that if P(k) then P(k + 1) for any k  N. (inductive step) P(k): k³ - k is divisible by 3 is true P(k + 1): ( k + 1)³ - (k + 1) = (k + 1)((k + 1)² - 1) = (k + 1)(k² +2k + 1 – 1) = (k + 1)(k² +2k) = k³ + 3 k² + 2k = k³ + 3 k² + 3k – k = k³ – k + 3 k² + 3k By the inductive hypothesis, k³ - k is divisible by 3. Thus, every term of the sum k³ – k + 3 k² + 3k is divisible by 3, and so is the sum itself. 16

Mathematical Induction 3. Then P(n) must be true for any positive integer. (conclusion) n³ - n is divisible by 3 is true for any positive integer. End of proof. 17

Mathematical Induction Example5 Use mathematical induction to prove that: n² > 2n + 1 for all n ≥ 5 18

Mathematical Induction Can principle of mathematical induction Be used to show that for any integer n, n = n + 1? 19

Mathematical 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 (Strong Induction). It can be used to prove that a propositional function P(n) is true for any natural number n. 20

Mathematical Induction Characterized by another inference rule: 1. ( basis step) P(0) is true. 2. (inductive step)  n  0: (  k’:0  k’  k P(k’))  P(k+1) 3. (conclusion)  n  0: P(n) The only difference between this and the 1st principle is that: ◦ the inductive step here makes use of the stronger hypothesis that P(k) is true for all smaller numbers k<n+1, not just for k=n. 21

Mathematical Induction Example1: Show that every integer greater than 1 can be written as the product of primes. Let P(n) be the proposition “ 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. 22

Mathematical Induction Show that if P(2) and P(3) and … and P(k), then P(k + 1) for any k  N. (inductive step) Two possible cases: If (k + 1) is prime, then obviously P(n + 1) is true. If (k + 1) is composite, it can be written as the product of two integers a and b such that 2  a  b < k + 1. By the induction hypothesis, both a and b can be written as the product of primes. Therefore, k + 1 = a  b can be written as the product of primes. 23

Mathematical 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. 24

Mathematical Induction Example2: Prove that every amount of postage of 12 cents or more can be formed using just 4-cent and 5-cent stamps. 25

Well-Ordering Principle. Every nonempty set S of nonnegative integers contains a least element; that is, there is some integer a in S such that a < b for all b's belonging to S. 26

Archimedean property Theorem: If a and b are any positive integers, then there exists a positive integer n such that na > b. Proof. Assume that the statement of the theorem is not true, so that for some a and b, na < b for every positive integer n. Then the set S = {b — na \ n a positive integer} consists entirely of positive integers. 27

Archimedean property By the Well-Ordering Principle, S will possess a least element, say, b - ma. Notice that b -(m +1)a also lies in S, because S contains all integers of this form. Furthermore, we have b - (m + 1)a = (b - ma) - a < b - ma contrary to the choice of b - ma as the smallest integer in S. This contradiction arose out of our original assumption that the Archimedean property did not hold; hence, this property is proven true. 28

The First Principle of Finite Induction Theorem: Let S be a set of positive integers with the following properties: (a) The integer 1 belongs to S. (b) Whenever the integer k is in S, the next integer k + 1 must also be in S. Then S is the set of all positive integers. 29

The First Principle of Finite Induction Proof. Let T be the set of all positive integers not in S, and assume that T is nonempty. The Well-Ordering Principle tells us that T possesses a least element, which we denote by a. Because 1 is in S, certainly a > 1, and so 0 < a - 1 < a. The choice of a as the smallest positive integer in T implies that a - 1 is not a member of T, or equivalently that a - 1 belongs to S. By hypothesis, S must also contain (a - 1) + 1 = a, which contradicts the fact that a lies in T. We conclude that the set T is empty and in consequence that S contains all the positive integers. 30

THE BINOMIAL THEOREM 31

THE BINOMIAL THEOREM Theorem: Proof. 32

THE BINOMIAL THEOREM 33

THE BINOMIAL THEOREM 34

THE BINOMIAL THEOREM Theorem: For any positive integer n Where 35

THE BINOMIAL THEOREM Proof. We use mathematical induction 36