6.1 The Basics of Counting Apr 02, 2018 Shawn Math 240 No comments yet 5.4 Recursive Algorithms Mar 23, 2018 Shawn Math 240 No comments yet 5.3 Recursive Definitions and Structural Induction Mar 19, 2018 Shawn Math 240 No comments yet 5.2 Strong Induction and Well-Ordering Mar 14, 2018 Shawn Math 240 No comments yet 5.1 Mathematical Induction Mar 14, 2018 Shawn Math 240 No comments yet 1 2 3 4 5 6 … 9
![Strong Induction • To prove that P(n) is true for all positive integers n • where P(n) is a propositional function, complete two steps: • Basis Step ○ Verify that the proposition P(1) is true. • Inductive Step ○ Show the conditional statement § [P(1)∧P(2)∧…∧P(k)]→ P(k+1) ○ holds for all positive integers k. • Strong Induction is sometimes called ○ the second principle of mathematical induction ○ complete induction Proof using Strong Induction • Prove that every natural number n 7 can be written as 3p+5q • where p and q are natural numbers • Prove this result using strong induction • Basis Step ○ 8=3×1+5×1 • Inductive Step ○ Inductive hypothesis: The statement is true for any n for k≥n≥8 ○ In particular it is true for k+1−3 (assuming k+1−3≥8) ○ So k+1−3=3p+5q and so k+1=3(p+1)+5q • What happens if k+1=9 or k+1=10? ○ Add those cases into the basis step ○ 9=3×3+5×0 ○ 10=3×0+5×2 Which Form of Induction Should Be Used? • We can always use strong induction instead of mathematical induction. • But there is no reason to use it if it is simpler to use mathematical induction • In fact, the principles of mathematical induction, strong induction, and the well-ordering property are all equivalent. • Sometimes it is clear how to proceed using one of the three methods, but not the other two. Proof of the Fundamental Theorem of Arithmetic • Show that if n is an integer greater than 1 • Then n can be written as the product of primes. • Let P(n) be the proposition that n can be written as a product of primes. • Basis Step ○ P(2) is true since 2 itself is prime. • Inductive Step ○ The inductive hypothesis is P(j) is true for j∈Z with 2≤j≤k ○ To show that P(k+1) must be true under this assumption ○ Two cases need to be considered: ○ If k+1 is prime, then P(k+1) is true. ○ Otherwise, k+1 is composite ○ And it can be written as the product of two positive integers ○ a and b with 2≤a≤b≤k+1 ○ By inductive hypothesis a and b can be written as product of primes ○ Therefore k + 1 can also be written as the product of those primes. • Hence, every integer greater than 1 can be written as product of primes Well-Ordering Property • Well-ordering property ○ Every nonempty set of nonnegative integers has a least element. • The well-ordering property is one of the axioms of the positive integers • The well-ordering property can be used directly in proofs. • The well-ordering property can be generalized. • Definition: A set is well ordered if every subset has a least element. ○ N is well ordered under ≤. ○ The set of finite strings over an alphabet using lexicographic ordering is well ordered. Proof of The Division Algorithm • Use the well-ordering property to prove the division algorithm ○ If a is an integer and d is a positive integer, then ○ there are unique integers q and r with 0≤r d, such that ○ a=dq+4 • Let S be the set of nonnegative integers of the form a=dq, q∈Z. • The set is nonempty since −dq can be made as large as needed • By the well-ordering property, S has a least element r=a−dq_0 • The integer r is nonnegative. • It also must be the case that r d • If it were not, then there would be a smaller nonnegative element in S ○ a−d(q_0+1)=a−dq_0−d=r−d 0 • Therefore, there are integers q and r with 0≤r d](https://shawnzhong.com/wp-content/uploads/2018/03/img_5aa936280eb33.png)
![Climbing an Infinite Ladder • Suppose we have an infinite ladder: ○ We can reach the first rung of the ladder. ○ If we can reach a particular rung of the ladder ○ then we can reach the next rung. • From (1), we can reach the first rung. • Then by applying (2), we can reach the second rung. • Applying (2) again, the third rung. And so on. • We can apply (2) any number of times to reach any particular rung, no matter how high up. Principle of Mathematical Induction • To prove that P(n) is true for all positive integers n, we complete these steps: ○ Basis Step: Show that P(1) is true. ○ Inductive Step: Show that P(k)→P(k+1) is true for all positive integers k • To complete the inductive step ○ assuming the inductive hypothesis that P(k) holds for an arbitrary integer k ○ show that must P(k+1) be true. • Climbing an Infinite Ladder Example: ○ Basis Step § By (1), we can reach rung 1. ○ Inductive Step § Assume the inductive hypothesis that we can reach rung k § Then by (2), we can reach rung k+1. ○ Hence, P(k)→P(k+1) is true for all positive integers k ○ We can reach every rung on the ladder. Important Points About Using Mathematical Induction • Mathematical induction can be expressed as the rule of inference ○ (P(1)∧∀k(P(k)→P(k+1)))→∀n P(n) ○ where the domain is the set of positive integers. • In mathematical induction, we don’t assume that P(k) is true for all positive integers! • We show that if we assume that P(k) is true, then P(k+1) must also be true. • Proofs by mathematical induction do not always start at the integer 1. • In such a case, the basis step begins at a starting point b where b is an integer. • We will see examples of this soon. Proving a Summation Formula by Mathematical Induction • Example ○ Show that:∑_(i=1)^n▒i=n(n+1)/2 • Solution ○ Basis Step § P(1) is true since 1(1+1)/2=1 ○ Inductive Step § Assume true for P(k) § The inductive hypothesis is ∑_(i=1)^k▒i=k(k+1)/2 § Under this assumption § 1+2+…+k+(k+1)=k(k+1)/2+(k+1)=(k+1)(k+2)/2 Validity of Mathematical Induction • Mathematical induction is valid because of the well ordering property, which states that • every nonempty subset of the set of positive integers has a least element • Suppose that P(1) holds and P(k)→P(k+1) is true for all positive integers k. • Assume there is at least one positive integer n for which P(n) is false. • Then the set S of positive integers for which P(n) is false is nonempty. • By the well-ordering property, S has a least element, say m. • We know that m cannot be 1 since P(1) holds. • Since m is positive and greater than 1, m−1 must be a positive integer. • Since m−1 m, it is not in S, so P(m−1) must be true. • But then, since the conditional P(k)→P(k+1) for every positive integer k holds, • P(m) must also be true. This contradicts P(m) being false. • Hence, P(n) must be true for every positive integer n. Conjecturing and Proving Correct a Summation Formula • Example ○ Conjecture and prove correct a formula for the sum of the first n positive odd integers ○ Then prove your conjecture • Solution ○ We have § 1= 1 § 1 + 3 = 4 § 1 + 3 + 5 = 9 § 1 + 3 + 5 + 7 = 16 § 1 + 3 + 5 + 7 + 9 = 25. ○ We can conjecture that the sum of the first n positive odd integers is n^2, § 1+3+5+…+(2n−1)+(2n+1)=n^2 ○ We prove the conjecture is proved correct with mathematical induction. ○ Basis Step § P(1) is true since 1^2 = 1. ○ Inductive Step: P(k)→P(k+1) for every positive integer k. § Inductive Hypothesis: 1+3+5+…+(2k−1)=k^2 § So, assuming P(k), it follows that: § 1+3+5+…+(2k−1)+(2k+1) § =[1+3+5+…+(2k−1)]+(2k+1) § =k^2+(2k+1) § =(k+1)^2 ○ Hence, we have shown that P(k+1) follows from P(k). ○ Therefore the sum of the first n positive odd integers is n^2 Proving Inequalities • Example ○ Use mathematical induction to prove that n 2^n for all positive integers n. • Solution ○ Let P(n) be the proposition that n 2^n. ○ Basis Step § P(1) is true since 1 2^1 = 2. ○ Inductive Step § Assume P(k) holds, i.e., k 2^k, for an arbitrary positive integer k. § Must show that P(k + 1) holds. § Since by the inductive hypothesis, k 2^k, it follows that: § k+1 2^k+1≤2^k+2^k=2⋅2^k=2^(k+1) § Therefore n 2^n holds for all positive integers n. • Example ○ Use mathematical induction to prove that 2^n n!, for every integer n≥4 • Solution ○ Let P(n) be the proposition that 2^n n! ○ Basis Step § P(4) is true since 24 = 16 4! = 24 ○ Inductive Step § Assume P(k) holds, i.e., 2^k k! for an arbitrary integer k ≥ 4. § To show that P(k+1) holds § 2^(k+1)=2∙2^k 2⋅2^k (k+1)k!=(k+1)! § Therefore, 2^n n! holds, for every integer n ≥ 4 Proving Divisibility Results • Example ○ Use mathematical induction to prove that n^3−n is divisible by 3 ○ for every positive integer n. • Solution ○ Let P(n) be the proposition that n^3−n is divisible by 3. ○ Basis Step § P(1) is true since 13 − 1 = 0, which is divisible by 3 ○ Inductive Step § Assume P(k) holds, i.e., k^3−k is divisible by 3, for an arbitrary positive integer k § To show that P(k + 1) follows § (k+1)^3−(k+1)=(k^3+3k^2+3k+1)−(k+1)=(k^3−k)+3(k^2+k) § By the inductive hypothesis, the first term (k^3−k) is divisible by 3 § and the second term is divisible by 3 since it is an integer multiplied by 3. § So by part (i) of Theorem 1 in Section 4.1, (k+1)^3−(k+1) is divisible by 3 ○ Therefore, n^3−n is divisible by 3, for every integer positive integer n. Number of Subsets of a Finite Set • Example ○ Use mathematical induction to show that if S is a finite set with n elements ○ where n is a nonnegative integer, then S has 2^n subsets. ○ (Chapter 6 uses combinatorial methods to prove this result.) • Solution ○ Let P(n) be the proposition that a set with n elements has 2^n subsets. ○ Basis Step § P(0) is true, because the empty set has only itself as a subset and 2^0 = 1. ○ Inductive Step § Assume P(k) is true for an arbitrary nonnegative integer k. § Inductive Hypothesis § For an arbitrary nonnegative integer k, every set with k elements has 2^k subsets § Let T be a set with k+1 elements § Then T=S∪{a}, where a∈T and S=T−{a}. Hence |S| = k. § For each subset X of S, there are exactly two subsets of T, i.e., X and X∪{a}. § By the inductive hypothesis S has 2^k subsets. § Since there are two subsets of T for each subset of S, § the number of subsets of T is 2⋅2^k=2^(k+1) An Incorrect “Proof” by Mathematical Induction • P(n)≔every set of n lines in the plane, no two of which are parallel, meet in a point • Here is a “proof” that P(n) is true for all positive integers n≥2. • Basis Step ○ The statement P(2) is true ○ because any two lines in the plane that are not parallel meet in a common point. • Inductive hypothesis ○ P(k) is true for the positive integer k ≥ 2 ○ every set of k lines in the plane, no two of which are parallel, meet in a common point. • Inductive Step ○ We must show that if P(k) holds, then P(k+1) holds ○ If every set of k lines in the plane, no two of which are parallel, k ≥ 2, meet in a point ○ Then every set of k+1 lines in the plane, no two of which are parallel, meet in a point. ○ Consider a set of k + 1 distinct lines in the plane, no two parallel. ○ By the inductive hypothesis, the first k of these lines must meet in a common point p_1. ○ By the inductive hypothesis, the last k of these lines meet in a common point p_2. ○ If p_1 and p_2 are different points, all lines containing both of them must be the same line since two points determine a line. ○ This contradicts the assumption that the lines are distinct. ○ Hence, p_1=p_2 lies on all k + 1 distinct lines, and therefore P(k+1) holds. ○ Assuming that k ≥ 2, distinct lines meet in a common point ○ Then every k + 1 lines meet in a common point. • There must be an error in this proof since the conclusion is absurd. But where is the error? ○ P(k)→P(k+1) only holds for k≥3 ○ It is not the case that P(2) implies P(3) ○ The first two lines must meet in a common point p_1 and the second two must meet in a common point p_2 ○ They do not have to be the same point since only the second line is common to both sets of lines.](https://shawnzhong.com/wp-content/uploads/2018/03/img_5aa92c5c396c9.png)
