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§3.3: Mathematical Induction

§3.3: Mathematical Induction. 1=1, 1+3=4, 1+3+5=9, .. , Will 1+3+…+(2 k- 1)= k 2 ? An important proof technique to prove that the guess is correct. Essentially a “Domino effect” principle. Based on a predicate-logic inference rule: P (0)  n 0 ( P ( n ) P ( n +1))  n 0 P ( n ).

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§3.3: Mathematical Induction

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  1. §3.3: Mathematical Induction • 1=1, 1+3=4, 1+3+5=9, .. , • Will 1+3+…+(2k-1)=k2? • An important proof technique to prove that the guess is correct. • Essentially a “Domino effect” principle. • Based on a predicate-logic inference rule: P(0)n0(P(n)P(n+1))n0 P(n) “The First Principleof MathematicalInduction” (c)2001-2003, Michael P. Frank

  2. Outline of an Inductive Proof • Basis step: show that P(0) holds. • Inductive step: Prove nP(n)P(n+1). • Let kN, assume P(k). (inductive hypothesis) • Under this assumption, prove P(k+1). • Inductive inference rule then gives nP(n). (c)2001-2003, Michael P. Frank

  3. Induction Example (1st princ.) • Prove that the sum of the first n odd positive integers is n2. That is, prove: • Proof by induction. • Basis step: Let n=1. The sum of the first 1 odd positive integer is 1 which equals 12.(Cont…) P(n) (c)2001-2003, Michael P. Frank

  4. Example cont. • Inductive step: Prove n1: P(n)P(n+1). • Let k1, assume P(k) holds, and prove P(k+1) is true. By inductivehypothesis P(n) (c)2001-2003, Michael P. Frank

  5. Another Induction Example • Prove that n>0, n<2n. Let P(n)=(n<2n) • Basis step: P(1) is T. • Inductive step: [For k>0, prove P(k)P(k+1)] • Assuming P(k) holds for k>1. • Note k + 1 < 2k + 1 (by inductive hypothesis) < 2k + 2k(because 1<2=22022n-1= 2n) = 2k+1 • So n>0, n + 1 < 2n+1, and we’re done. (c)2001-2003, Michael P. Frank

  6. More Induction Examples • n3–n is divisible by 3 • 1+2+22+…+2n=2n+1-1 • Sums of Geometric Progressions • An Inequality for Harmonic Numbers , where • DeMorgan’s law • Number of Subsets of a finite set • 2n<n! (c)2001-2003, Michael P. Frank

  7. More Induction Example • Example 11: Any chessboard with 2n×2n squares but with one removed can be tiled using L-shaped pieces (which cover 3 squares at a time). • Example 12: The greedy algorithm (selects talk with earliest ending time) schedules the most talks in a single lecture halls. (c)2001-2003, Michael P. Frank

  8. Strong Induction(2nd Principle of Mathematical Induction) 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) . If k+1 is a prime, let s=1, p1=k+1.else k+1=ab, where 1<an and 1<bn.Then a=p1p2…pt and b=q1q2…qu. Then k+1= p1p2…pt q1q2…qu, a product of s=t+u primes. (c)2001-2003, Michael P. Frank

  9. Another 2nd Principle Example • 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. (c)2001-2003, Michael P. Frank

  10. Well-Ordering Property The well-ordering property: Every nonempty set of nonnegative integer has a least element. • Example 16: If a is an integer and d is a positive integer. There are unique integers q and r with 0≦r<d and a=d×q+r. (c)2001-2003, Michael P. Frank

  11. Well-Ordering Property • Example 17: If there is a cycle of length m (m≧3) among the players in a round-robin tournament, there must be a cycle of three. • Infinite Descent： For a propositional function P(n), P(k) is false for all positive integers. • Example 18: is irrational (c)2001-2003, Michael P. Frank

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