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Mathematical Induction

Mathematical Induction. Section 4.1. Example. Consider the predicate P ( n ): ``1 + 3 + ··· + (2 n -1) is equal n 2 ’’ Let ’ s verify the truth value of P ( n ) for some n : P ( 1 ): 1 (1) is equal 1 2 P ( 2 ): 1 + 3 (4) is equal 2 2

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Mathematical Induction

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  1. Mathematical Induction Section 4.1

  2. Example • Consider the predicate P(n): ``1 + 3 + ··· + (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): 1 + 3 (4) is equal 22 • P(3): 1 + 3 + 5 (9) is equal 32 • .... • P(9): 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 is equal 92 • …Let us write “is equal” by the symbol =. • Thus,1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 = 92

  3. Example… (no calculators!) • Suppose, without verifying, that P(21): 1 + 3 + 5 + … + 39 + 41= 212 • Given above, can we prove thatP(22): 1 + 3 + 5 + … + 39 + 41 + 43 = 222 ? • P(22): 1 + 3 + 5 + … + 39 + 41 + 43 = 222 ? • P(22) : 212 (by P(21)) +43= 222 • P(22) : 212 + 43 = 222 ? • P(22): (22– 1)2 + (22×2 –1) = 222 ? • P(22): 222–22×2 + 1 + 22×2 –1 = 222✔

  4. Example (no calculators!) • How general is our methods? • Let’s replace 21 with a place holder k. • Suppose, without verifying, that P(k) = 1 + 3 + 5 + … + (2k-1)= k2 • Given this, can we prove thatP(k+1) = 1 + 3 + 5 + … + (2(k+1)-1)= (k+1)2 ? • P(k+1) = 1 + 3 + 5 + … + (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) + 1 + 2(k+1)–1 = (k+1)2✔

  5. 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.

  6. Mathematical Induction • Let P(n) be some propositional function involving integer n. • P(n) = “n (n+3) is an even number ” • P(n) = “1 + 3 + ··· + (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 arbitraryk, IFP(k) is true THENP(k+ 1) is true. That is, we validate the logical implication: P(k) ⇒P(k+1)

  7. Mathematical Induction • Mathematical induction amounts to the following rule of inference:[P(1)  k ((k ≥ 1)  P(k))  P(k +1))] ⇒ nP(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 InductionStep: Establish that P(k +1) A (direct) proof that ((k ≥ 1)  P(k)) ⇒P(k +1)

  8. 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, THENP(k+1)is also true. Otherwise the proof by mathematical induction was a case of begging the question or circular reasoning.

  9. Example • Let P(n) = “1 + 3 + ··· + (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.

  10. Example

  11. Example

  12. 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 InductionStep: Establish that P(k +1) A (direct) proof that ((k ≥ b)  P(k)) ⇒P(k +1)

  13. 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.

  14. 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 ksince 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.

  15. 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.

  16. 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 thenP(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 ≥ bthenP(k+1) is also true.

  17. 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.

  18. Example • Show that for every nonnegative integer n 20+ 21 + 22 + 23 + ∙∙∙ + 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.

  19. Example P(k+1) = 20+21+22+23+ ∙∙∙+2k + 2k+1 = 2k+1–1 + 2k+1 by the induction hypothesis = 2k+2–1 by arithmetic = 2(k+1)+1–1 by arithmetic Thus P(k) ⇒ P(k +1)

  20. Example

  21. Example Hypothesis is used here

  22. Proof of Correctness of Mathematical Induction We want to show that [P(1)  k ((k ≥ 1) P(k))  P(k +1))] ⇒ nP(n) where the universe is the positive integers Proof is done by contradiction • ¬ nP(n) Contrary assumption • x¬P(x) 1, Demogan’s • ¬ P(r)2, EI, select smallest r • P(r -1) 3, 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 3 & 9

  23. Example

  24. Example

  25. Proof:

  26. Example • Prove by induction thatif 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.

  27. “Induction” and “Mathematical Inductions” • Let P(n) denote the number of different ways to write nas a sum of positive integers when order is not important. For instance, integer 5 can be written in the following 7 ways:1+1+1+1+1=2+1+1+1=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) …..

  28. What’s wrong with the following proof?

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