1. 1 Inductive Proofs Rosen 6 th ed., §4.1-4.4

2. 2 Mathematical Induction A powerful, rigorous technique for proving that a predicate P(n) is true for every natural number n, no matter how large.A powerful, rigorous technique for proving that a predicate P(n) is true for every natural number n, no matter how large. Based on a predicate-logic inference rule:Based on a predicate-logic inference rule: P(0)  n  0 (P(n)  P(n+1))  n  0 P(n)P(0)  n  0 (P(n)  P(n+1))  n  0 P(n)

3. 3 Outline of an Inductive Proof Want to prove  n P(n) …Want to prove  n P(n) … Base case (or basis step): Prove P(0).Base case (or basis step): Prove P(0). Inductive step: Prove  n P(n)  P(n+1).Inductive step: Prove  n P(n)  P(n+1). –e.g. use a direct proof: Let n  N, assume P(n). (inductive hypothesis)Let n  N, assume P(n). (inductive hypothesis) Under this assumption, prove P(n+1).Under this assumption, prove P(n+1). Inductive inference rule then gives  n P(n).Inductive inference rule then gives  n P(n).

4. 4 Induction Example Prove thatProve that

5. 5 Another Induction Example Prove that , n<2 n. Let P(n)=(n<2 n )Prove that , n<2 n. Let P(n)=(n<2 n ) –Base case: P(0)=(0<2 0 )=(0<1)=T. –Inductive step: For prove P(n)  P(n+1). Assuming n<2 n, prove n+1 < 2 n+1.Assuming n<2 n, prove n+1 < 2 n+1. Note n + 1 < 2 n + 1 (by inductive hypothesis) < 2 n + 2 n (because 1<2=2  2   2  2 n-1 = 2 n ) = 2 n+1Note n + 1 < 2 n + 1 (by inductive hypothesis) < 2 n + 2 n (because 1<2=2  2   2  2 n-1 = 2 n ) = 2 n+1 So n + 1 < 2 n+1, and we’re done.So n + 1 < 2 n+1, and we’re done.

6. 6 Validity of Induction Prove: if  n  0 (P(n)  P(n+1)) and P(0), then  k  0 P(k) Prove: if  n  0 (P(n)  P(n+1)) and P(0), then  k  0 P(k) (a) Given any k  0,  n  0 (P(n)  P(n+1)) implies (P(0)  P(1))  (P(1)  P(2))  …  (P(k  1)  P(k)) (P(0)  P(1))  (P(1)  P(2))  …  (P(k  1)  P(k)) Using hypothetical syllogism k-1 times we have (b) Using hypothetical syllogism k-1 times we have P(0)  P(k) P(0)  P(k) (c) P(0) and modus ponens gives P(k). Thus  k  0 P(k). Thus  k  0 P(k).

7. 7 Generalizing Induction Can also be used to prove  n  c P(n) for a given constant c  Z, where maybe c  0, then:Can also be used to prove  n  c P(n) for a given constant c  Z, where maybe c  0, then: –Base case: prove P(c) rather than P(0) –The inductive step is to prove:  n  c (P(n)  P(n+1)).  n  c (P(n)  P(n+1)).

8. 8 Induction Example Prove that the sum of the first n odd positive integers is n 2. That is, prove:Prove that the sum of the first n odd positive integers is n 2. That is, prove: Proof by induction.Proof by induction. –Base case: Let n=1. The sum of the first 1 odd positive integer is 1 which equals 1 2. (Cont…) P(n)P(n)

9. 9 Example cont. Inductive step: Prove  n  1: P(n)  P(n+1).Inductive step: Prove  n  1: P(n)  P(n+1). –Let n  1, assume P(n), and prove P(n+1). By inductive hypothesis P(n)

10. 10 Strong Induction Characterized by another inference rule: P(0)  n  0: (  0  k  n P(k))  P(n+1)  n  0: P(n)Characterized by another inference rule: P(0)  n  0: (  0  k  n P(k))  P(n+1)  n  0: P(n) Difference with previous version is that the inductive step uses the fact that P(k) is true for all smaller, not just for k=n.Difference with previous version is that the inductive step uses the fact that P(k) is true for all smaller, not just for k=n. P is true in all previous cases

11. 11 Example 1 Show that every n>1 can be written as a product p 1 p 2 …p s of some series of s prime numbers. Let P(n)=“n has that property”Show that every n>1 can be written as a product p 1 p 2 …p s of some series of s prime numbers. Let P(n)=“n has that property” Base case: n=2, let s=1, p 1 =2.Base case: n=2, let s=1, p 1 =2. Inductive step: Let n  2. Assume  2  k  n: P(k). Consider n+1. If prime, let s=1, p 1 =n+1. Else n+1=ab, where 1  a  n and 1  b  n. Then a=p 1 p 2 …p t and b=q 1 q 2 …q u. Then n+1= p 1 p 2 …p t q 1 q 2 …q u, a product of s=t+u primes.Inductive step: Let n  2. Assume  2  k  n: P(k). Consider n+1. If prime, let s=1, p 1 =n+1. Else n+1=ab, where 1  a  n and 1  b  n. Then a=p 1 p 2 …p t and b=q 1 q 2 …q u. Then n+1= p 1 p 2 …p t q 1 q 2 …q u, a product of s=t+u primes.

12. 12 Example 2 Prove that every amount of postage of 12 cents or more can be formed using just 4- cent and 5-cent stamps.Prove that every amount of postage of 12 cents or more can be formed using just 4- cent and 5-cent stamps. Base case: 12=3  4), 13=2  4)+1(5), 14=1(4)+2(5), 15=3(5), so  12  n  15, P(n).Base case: 12=3  4), 13=2  4)+1(5), 14=1(4)+2(5), 15=3(5), so  12  n  15, P(n). Inductive step: Let n  15, assume  12  k  n P(k). Note 12  n  3  n, so P(n  3), so add a 4-cent stamp to get postage for n+1.Inductive step: Let n  15, assume  12  k  n P(k). Note 12  n  3  n, so P(n  3), so add a 4-cent stamp to get postage for n+1.

13. Recursive Definitions Recursion – defining an object (or function, algorithm, etc.) in terms of itselfRecursion – defining an object (or function, algorithm, etc.) in terms of itself Recursion can be used to define sequencesRecursion can be used to define sequences – Previously sequences were defined using a specific formula, e.g., a n = 2 n for n = 0,1,2,... – This sequence can also be defined by giving the first term of the sequence, namely a 0 = 1, and a rule for finding a term of the sequence for the previous one, namely, a n+1 = 2a n for n = 0,1,2,... 13

14. Recursive Definitions When defining a set recursively, weWhen defining a set recursively, we –(1) specify the initial elements in a basis step and –(2) provide a rule for constructing new elements from those we already have in the recursive step Examples… 14

15. Recursive Algorithms Sometimes we can reduce the solution to the problem with a particular set of input to the solution of the same problem with smaller input values Sometimes we can reduce the solution to the problem with a particular set of input to the solution of the same problem with smaller input values When such a reduction can be done, the solution to the original problem can be found with a sequence of reductions, until the problem has been reduced to the initial case for which the solution is knownWhen such a reduction can be done, the solution to the original problem can be found with a sequence of reductions, until the problem has been reduced to the initial case for which the solution is known 15

16. Recursive Algorithms An algorithm is called recursive if it solves a problem by reducing it to an instance of the same problem with smaller input An algorithm is called recursive if it solves a problem by reducing it to an instance of the same problem with smaller input Examples …..Examples ….. 16