1 / 16

Inductive Proofs

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

zola
Télécharger la présentation

Inductive Proofs

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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.

More Related