# Mathematical Induction

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Mathematical Induction

Example Consider the predicate
Wednesday, April 19, 2017 Example Consider the predicate P(n): `` ··· + (2n-1) is equal n2’’ Let’s verify the truth value of P(n) for some n: P(1): 1 (1) is equal 12 P(2): (4) is equal 22 P(3): (9) is equal 32 .... P(9): is equal 92 …Let us write “is equal” by the symbol =. Thus, = 92 © by A-H. Esfahanian. All Rights Reserved.

Example… (no calculators!)
Wednesday, April 19, 2017 Example… (no calculators!) Suppose, without verifying, that P(21): … = 212 Given above, can we prove that P(22): … = 222 ? P(22): … = 222 ? P(22) : (by P(21)) = 222 P(22) : = 222 ? P(22): (22 – 1)2 + (22×2 – 1) = 222 ? P(22): 222 – 22× ×2 – 1 = 222 ✔ © by A-H. Esfahanian. All Rights Reserved.

Example (no calculators!)
Wednesday, April 19, 2017 Example (no calculators!) How general is our methods? Let’s replace 21 with a place holder k. Suppose, without verifying, that P(k) = … + (2k-1)= k2 Given this, can we prove that P(k+1) = … + (2(k+1)-1)= (k+1)2 ? P(k+1) = … + (2k-1) + (2(k+1)-1) = (k+1)2 ? P(k+1) = k2 (by P(k)) (2(k+1)-1) = (k+1)2 ? P(k+1) = k2 + (2(k+1)-1) = (k+1)2 ? P(k+1) = ((k+1) – 1)2 + (2(k+1)-1) = (k+1)2 ? P(k+1) = (k+1)2 – 2(k+1) (k+1) – 1 = (k+1)2 ✔ © by A-H. Esfahanian. All Rights Reserved.

Example… What have we accomplished so far?
Wednesday, April 19, 2017 Example… What have we accomplished so far? We have shown that P(k) ⇒ P(k+1), for any k. For instance: P(1) ⇒ P(2), P(2) ⇒ P(3), P(6) ⇒ P(7), P(19) ⇒ P(20), P(2000) ⇒ P(2001), …. Note that, we do NOT know if, for example, P(6) or P(19) … is true. All we know is that IF, for example, P(6) is true, then P(7) is also true. What do we need to show that P(n) is true for all n? All we need is to PROVE that P(1) is indeed true. © by A-H. Esfahanian. All Rights Reserved.

Mathematical Induction
Wednesday, April 19, 2017 Mathematical Induction Let P(n) be some propositional function involving integer n. P(n) = “n (n+3) is an even number ” P(n) = “ ··· + (2n-1) = n2 ” ….. To prove that P(n) is true for all positive integers n, we can do the following: Give a proof (usually a straight verification) that P(1) is true. Give a proof that for an arbitrary k, IF P(k) is true THEN P(k + 1) is true. That is, we validate the logical implication: P(k) ⇒ P(k+1) © by A-H. Esfahanian. All Rights Reserved.

Mathematical Induction
Wednesday, April 19, 2017 Mathematical Induction Mathematical induction amounts to the following rule of inference: [P(1)  k ((k ≥ 1)  P(k))  P(k +1))] ⇒ n P(n) where our universe is the set of positive integers A proof using mathematical induction involves: The Basis: prove P(1) is true. The Hypothesis: Assume P(k) is true for an arbitrary k ≥ 1. The Induction Step: Establish that P(k +1) A (direct) proof that ((k ≥ 1)  P(k)) ⇒ P(k +1) © by A-H. Esfahanian. All Rights Reserved.

Mathematical Induction….
Wednesday, April 19, 2017 Mathematical Induction…. In proof by mathematical induction it is not given that P(k) is true for all positive integers! It is only shown that IF P(k) is true, THEN P(k+1) is also true. Otherwise the proof by mathematical induction was a case of begging the question or circular reasoning. © by A-H. Esfahanian. All Rights Reserved.

Wednesday, April 19, 2017 Example Let P(n) = “ ··· + (2n-1) = n2 ”. We want to prove by induction that P(n) is true for all n. Basis: Prove that when n =1, the proposition is correct. P(1) is “1= 12”; thus, it is correct. Hypothesis: We will assume that P(k) is true for some arbitrary k. Step: Using the hypothesis, we want to prove that P(k +1) is true. © by A-H. Esfahanian. All Rights Reserved.

Example Wednesday, April 19, 2017

Example Wednesday, April 19, 2017

Induction can start anywhere
Wednesday, April 19, 2017 Induction can start anywhere To show P(n) for any n ≥ b : The Basis: prove P(b) is true. The Hypothesis: Assume P(k) is true and k ≥ b. The Induction Step: Establish that P(k +1) A (direct) proof that ((k ≥ b)  P(k)) ⇒ P(k +1) © by A-H. Esfahanian. All Rights Reserved. 12

Wednesday, April 19, 2017 Example Suppose we have coins of two different denominations, namely, 3 cents and 5 cents. We want to show that it is possible to pay exactly for any purchase of 8 cents or higher. In other words, that we can make up 8, 9, 10, …. cents using just these coins. Here is a proof by induction: Basis: 8 = 3+5 ✓ Hypothesis: Suppose we can pay exactly with just 3-cent coins and 5-cent coins for a purchase amount of k cents. © by A-H. Esfahanian. All Rights Reserved.

Wednesday, April 19, 2017 Example Want to show that it is we can pay exact for (k + 1) cents using just 3-cent coins and 5-cent coins. Proof: If in making up k cents, we used at least one 5- cent coin, then replace that coin with two 3-cent coins to make up (k + 1) cents. Otherwise, we must have used at least three 3-cent coins to make up k since we are considering k > 8. In the case, we will replace three 3-cent coins with two 5-cent coins to make up (k + 1) cents. © by A-H. Esfahanian. All Rights Reserved.

Wednesday, April 19, 2017 Strong induction To prove that P(k+1) holds, we may use any of the following assumptions: P(k) is true P(k – 1) is true P(k – 2) is true …. P(1) is true We must make sure that all the “assumptions” are in fact true for the induction “engine” to start. © by A-H. Esfahanian. All Rights Reserved.

Wednesday, April 19, 2017 Induction, recap Let P(n) be some propositional function about an integer n. To prove that P(n) is true for all positive integers n, we do the following: Give a proof that P(1) is true Give a proof that if all of P(1), P(2), …, P(k) are true and k ≥ 1 then P(k+1) is also true. As before, the induction can start at any integer b: Give a proof that P(b) is true Give a proof that if all of P(b), P(b+1), …, P(k) are true and k ≥ b then P(k+1) is also true. © by A-H. Esfahanian. All Rights Reserved.

Wednesday, April 19, 2017 Example Show that any positive integer n > 1 is either a prime or a product of primes. Proof: Basis: n = 2. Since 2 is prime.✓ Hypothesis: Assume true for n ≤ k Step: Need to show that, given the hypothesis, the proposition is true for k +1. Proof: If n = k +1 is a prime, then the proposition is true. If k +1 is not prime, then k +1=pq, where p ≤ k and q ≤ k. By the hypothesis, p and q are either primes or product of primes, and thus so is pq. © by A-H. Esfahanian. All Rights Reserved.

Example Show that for every nonnegative integer n
Wednesday, April 19, 2017 Example Show that for every nonnegative integer n ∙∙∙ + 2n = 2n+1 –1. Let P(n) be the corresponding predicate Basis: P(0) is true since 20 =1= 21-1. Hypothesis: Assume that P(k) is true and that k ≥ 0. Step: Need to show that P(k +1) is true given that the hypothesis is true. © by A-H. Esfahanian. All Rights Reserved.

Example P(k+1) = 20+21+22+23+ ∙∙∙ +2k + 2k+1
Wednesday, April 19, 2017 Example P(k+1) = ∙∙∙ +2k + 2k+1 = 2k+1 –1 + 2k+1 by the induction hypothesis = 2k+2 – by arithmetic = 2(k+1)+1 – by arithmetic Thus P(k) ⇒ P(k +1) © by A-H. Esfahanian. All Rights Reserved.

Example Wednesday, April 19, 2017

Hypothesis is used here

Proof of Correctness of Mathematical Induction
Wednesday, April 19, 2017 Proof of Correctness of Mathematical Induction We want to show that [P(1)  k ((k ≥ 1)  P(k))  P(k +1))] ⇒ n P(n) where the universe is the positive integers Proof is done by contradiction ¬ n P(n) Contrary assumption x ¬ P(x) , Demogan’s ¬ P(r) , EI, select smallest r P(r -1) , choice of r P(1) Premise r > 1 r is positive (by choice in 4) & 5 k ((k ≥ 1)  P(k))  P(k +1)) Premise ((r-1 ≥ 1)  P(r-1))  P(r) UI, k = r-1 P(r) 4, 6 & 8, modus ponens Contradiction & 9 © by A-H. Esfahanian. All Rights Reserved.

Example Wednesday, April 19, 2017

Example Wednesday, April 19, 2017

Proof: Wednesday, April 19, 2017

Wednesday, April 19, 2017 Example Prove by induction that if S is a finite set with n elements then S has 2n subsets. Basis: Assume n = 0. There is one set with cardinality zero: the empty set. Thus, S = . The empty set has only one subset: itself. Moreover, 20 = 1. Thus the basis is verified. Hypothesis. Assume true for any set S with cardinality 1,2,…, k Step. Want to show that if S has cardinality k+1, the proposition still holds. Let S be such a set. Let r be an arbitrary element of S. Write S = X ⋃ {r }. Now, the subsets of S either contain r or they don’t. For the ones that do not contain r , those are also subsets of X, and there are 2k of them. To form the ones that do contain r, we take each subset of X and inset r in it. That would give us another 2k subsets. Thus, S has a total of 2k+1 subsets. © by A-H. Esfahanian. All Rights Reserved.

“Induction” and “Mathematical Inductions”
Wednesday, April 19, 2017 “Induction” and “Mathematical Inductions” Let P(n) denote the number of different ways to write n as a sum of positive integers when order is not important. For instance, integer 5 can be written in the following 7 ways: = =2+2+1=3+1+1=3+2=4+1=5 Verify that: P(2) = 2 P(3) = 3 P(4) = 5 P(5) = 7 P(6) = ? P(7) ….. P(6) = 11, P(7) = 15 © by A-H. Esfahanian. All Rights Reserved.

What’s wrong with the following proof?
Wednesday, April 19, 2017 What’s wrong with the following proof? It fails when n = 1 (that is k = 0). Problem 54, page 255 © by A-H. Esfahanian. All Rights Reserved.