1. Inductive Proofs Rosen 6th ed., §4.1-4.3

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. • Based on a predicate-logic inference rule: • P(0)n0(P(n)P(n+1))n0 P(n)

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

4. Induction Example • Prove that

5. Another Induction Example • Prove that  , n<2n. Let P(n)=(n<2n) • Base case: P(0)=(0<20)=(0<1)=T. • Inductive step: For prove P(n)P(n+1). • Assuming n<2n, prove n+1 < 2n+1. • Note n + 1 < 2n + 1 (by inductive hypothesis) < 2n + 2n(because 1<2=22022n-1= 2n) = 2n+1 • So n + 1 < 2n+1, and we’re done.

6. Validity of Induction Prove: if n0(P(n)P(n+1)), then k0P(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)) (b) Using hypothetical syllogismk-1 times we have P(0)P(k) (c) P(0) and modus ponens gives P(k). Thus k0P(k).

7. The Well-Ordering Property • The validity of the inductive inference rule can also be proved using the well-ordering property, which says: • Every non-empty set of non-negative integers has a minimum (smallest) element. •  SN : mS : nS : mn

8. Use Well-Ordering Property and Proof by Contradiction • Suppose P(0) is true and that P(k)P(k+1) is true for all positive k. • Assume P(n) is false for some positive n. • Implies S={n|P(n)} is non-empty and has a minimum element m (i.e., P(m)=false) • But then, P(m-1)P((m-1)+1)=P(m) which is a contradiction!

9. Generalizing Induction • Can also be used to prove ncP(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)).

10. Induction Example • Prove that the sum of the first n odd positive integers is n2. That is, prove: • Proof by induction. • Base case: Let n=1. The sum of the first 1 odd positive integer is 1 which equals 12.(Cont…) P(n)

11. Example cont. • Inductive step: Prove n1: P(n)P(n+1). • Let n1, assume P(n), and prove P(n+1). By inductivehypothesis P(n)

12. Strong Induction • Characterized by another inference rule:P(0)n0: (0knP(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. P is true in all previous cases

13. Example 1 • Show that every n>1 can be written as a product p1p2…ps of some series of s prime numbers. Let P(n)=“n has that property” • Base case: n=2, let s=1, p1=2. • Inductive step: Let n2. Assume 2kn: P(k).Consider n+1. If prime, let s=1, p1=n+1.Else n+1=ab, where 1<an and 1<bn.Then a=p1p2…pt and b=q1q2…qu. Then n+1= p1p2…pt q1q2…qu, a product of s=t+u primes.

14. Example 2 • 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). • Inductive step: Let n15, assume 12knP(k). Note 12n3n, so P(n3), so add a 4-cent stamp to get postage for n+1.

15. Example 3 Suppose a0, a1, a2, … is defined as follows: a0=1, a1=2, a2=3, ak = ak-1+ak-2+ak-3for all integers k≥3. Then an ≤ 2n for all integers n≥0. P(n) • Proof (by strong induction): 1) Basis step: The statement is truefor n=0: a0=1≤1=20 P(0) for n=1: a1=2≤2=21 P(1) for n=2: a2=3≤4=22 P(2)

16. Example 3 (cont’d) 2) Inductive hypothesis: For any k>2, Assume P(i) is true for all i with 0≤i<k: ai≤ 2i for all0≤i<k . 3) Inductive step: Show that P(k) is true: ak≤ 2k ak= ak-1+ak-2+ak-3 ≤ 2k-1+2k-2+2k-3 (using inductive hypothesis) ≤ 20+21+…+2k-3+2k-2+2k-1 = 2k-1 (as a sum of geometric sequence) ≤ 2k Thus, P(n) is true by strong induction.