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Discrete Structures Chapter 2 Part B Mathematical Induction

Discrete Structures Chapter 2 Part B Mathematical Induction. Nurul Amelina Nasharuddin Multimedia Department. Mathematical Induction. Use to verify a property of a sequence Used to check the conclusions about the outcomes of processes that occur repeatedly and according to definite patterns.

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Discrete Structures Chapter 2 Part B Mathematical Induction

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  1. Discrete StructuresChapter 2 Part BMathematical Induction Nurul Amelina Nasharuddin Multimedia Department

  2. Mathematical Induction • Use to verify a property of a sequence • Used to check the conclusions about the outcomes of processes that occur repeatedly and according to definite patterns

  3. Mathematical Induction • Eg: Any whole number of cents of at least 8 cents can be obtained using 3 cents and 5 cents coins • Formally: For all integers n  8, P(n) is true where P(n) = “n cents can be obtained using 3 cents and 5 cents coins” • (1) If k cents is obtained using at least one 5 cents coin, then replace the coin with two 3 cents coins = (k+1) cents • (2) If k cents is obtained without using a 5 cents coin, then replace at least three 3 cents coins with two 5 cents coins = (k+1) cents • To show that P(n) is true for all integers n  8, (1) show P(8) is true (2) show that the truth of P(k+1) follows necessarily from the truth of P(k) for each k  8

  4. Principle of Mathematical Induction Let P(n) be a predicate that is defined for integers n and let a be some integer. If the following two premises are true: • P(a) is a true • k  a, P(k)  P(k + 1) then the following conclusion is true as well P(n) is true for all n  a

  5. Method of Proof by Mathematical Induction • Consider a statement of the form, “For all integers n  a, a property P(n) is true” To prove such statement, perform these 2 steps: Step 1 (basis step): Show that when n = a, statement is true Step 2 (inductive step): Show that for all integers k  a, if the property is true for n = k then it is true for n = k+1. To perform this step, Suppose that the property is true for n = k, where k  a [This supposition is called the induction hypothesis] Then Show that the property is true for n = k+1

  6. Coins Proofing • P(n): n cents can be obtained using 3c and 5c coins.” P(n) is true for all integers n  8. • Proof: Step 1: The property is true for n = 8 because 8c = 3c + 5c. Step 2: Suppose kc can be obtained using 3c and 5c for some integer k  8. [Induction hypothesis]. We must show that (k+1)c can be obtained using 3c and 5c. (1) In case there is a 5c coin among the kc, replace it by two 3c coins = (k+1) cents (2) In case there is no 5c coin, replace three 3c coins by two 5c coins = (k+1) cents Thus in either case (k+1)c can be obtained using 3c and 5c coins (PROVED!!)

  7. Applications of Mathematical Induction • Show that 1 + 2 + … + n = (n * (n + 1)) / 2 • Sum of geometric series: r0 + r1 + … + rn = (rn+1 – 1) / (r – 1)

  8. Sum of the First n Integers • Show that 1 + 2 + … + n = (n * (n + 1)) / 2 for all integers n  1 P(n): 1 + 2 + … + n = (n * (n + 1)) / 2 Step 1: Show that the property is true for n=1 (LHS) 1 = (1 * (1 + 1)) /2 = 1 (RHS) The property true for n=1 Step 2 : Show that for all integers k  1, if the property is true for n = k then it is true for n = k+1 Suppose 1 + 2 + … + k = (k * (k + 1)) / 2, for some integer k  1 (induction hypothesis)

  9. Sum of the First n Integers (2) We must show that 1 + 2 + … + (k+1) = (k+1)((k + 1)+1) /2, or equivalently that 1 + 2 + … + (k+1) = (k+1)(k + 2) /2 (show that LHS equals the RHS) 1 + 2 + … + (k+1) = 1 + 2 + … + k +(k+1) = (k * (k + 1)) / 2 + (k+1) (by substitution from the inductive hypothesis) = (k * (k + 1)) / 2 + 2(k + 1) / 2 = [(k+1)(k+2)] / 2 LHS = RHS Consequently, by the Principle of Mathematical Induction, P(n) is true for all n  1

  10. Sum of a Geometric Sequence • For any real number r except 1, and any integer n  0, P(n): Step 1: Show that the property is true for n=0: (LHS) (RHS) The property true for n=0

  11. Sum of a Geometric Sequence (2) Step 2 : Show that for all integers k  0, if the property is true for n = k then it is true for n = k+1 Suppose , for k  0 (induction hypothesis) We must show that the property is true for (show that LHS equals the RHS) LHS = RHS Consequently, the theorem is true

  12. Proving the Divisibility Property • For all integers n  1, 22n – 1 is divisible by 3 P(n): 22n – 1 is divisible by 3 Step 1: Show that the property is true for n=1 22(1) – 1 = 22 – 1 = 3 is divisible by 3 The property true for n=1 Step 2 : Show that for all integers k  1, if the property is true for n = k then it is true for n = k+1 Suppose 22k – 1 is divisible by 3, for some integer k  1 (induction hypothesis)

  13. Proving the Divisibility Property (2) By definition of divisibility, this means that 22k – 1 = 3r, for some integer r. We must show that the 22(k+1) – 1 is divisible by 3. 22(k+1) – 1 = 22k+2 – 1 = 22k . 22 – 1 = 22k . 4 – 1 = 22k . (3 + 1) – 1 = 22k . 3 + 22k – 1 = 22k . 3 + 3r = 3(22k + r) 22k + r is an integer because it is a sum of products of integers. Consequently, the theorem is true

  14. Proving an Inequality • For all integers n  3, 2n + 1  2n P(n): 2n + 1  2n Step 1: Show that the property is true for n=3 2n + 1 = 2(3) + 1 = 7  2n = 23 = 8 The property true for n=3 Step 2 : Show that for all integers k  3, if the property is true for n = k then it is true for n = k+1 Suppose 2k + 1  2k, for some integer k  3 (induction hypothesis)

  15. Proving an Inequality (2) We must show that 2(k+1) + 1  2(k+1) or equivalently, 2k+3  2k+1 2k+3 = (2k+1) +2  2k + 2 2k+3 = (2k+1)+2  2k + 2k 2  2k for all integers k 2k + 3  2(2k) = 2k+1 Consequently, the 2n + 1  2n for all integers n  3

  16. Proving a Property of A Sequence • Define a sequence a1, a2, a3, …, as follows a1 = 2 ak = 5ak-1 • Write the first four sequence a1 = 2 a2 = 5a1 = 5.2 = 10 a3 = 5a2 = 5.10 = 50 a4 = 5a3 = 5.50 = 250 • Proof an = 2 . 5n-1 for all integers n  1 Step 1: Show that the property is true for n=1 2 . 51-1 = 2 . 1 = 2 = a1 The property true for n=1

  17. Proving a Property of A Sequence (2) Step 2 : Show that for all integers k  1, if the property is true for n = k then it is true for n = k+1 Suppose ak = 2 . 5k-1 for all integers k  1(induction hypothesis) We must show that ak+1 = 2 . 5(k+1)-1 = 2 . 5k ak+1 = 5a(k+1)-1= 5ak = 5 . (2 . 5k-1) = 2 . (5 . 5k-1) = 2 . 5k Consequently, the an = 2 . 5n-1 for all integers n  1

  18. Exercises • Show that 22n – 1 is divisible by 3 • Show that for n > 2: 2n + 1 < 2n • Show that xn – yn is divisible by x – y • Show that n3 – n is divisible by 6 • On the outside rim of a circular disk the integers from 1 to 30 are painted in random. Show that there must be three successive integers whose sum is at least 45

  19. Principle of Strong Mathematical Induction • Basis step may contains proofs for several initial values • In inductive step the truth of the predicate P(n) is assumed for all values through k-1, then the truth of P(k) is proved. Let P(n) be a predicate that is defined for integers n and let a and b be fixed integers with a  b. Suppose the following two premises are true: • P(a), P(a+1), P(a+2), …, and P(b) are true (basis step) • For any integer k  b, if P(i) is true for all integers i with a  i  k, then P(k) is true (inductive step) then the following conclusion is true as well P(n) is true for all n  a

  20. Applying Strong Mathematical Induction • Prove that any integer greater than 1 is divisible by a prime number P(n): n is divisible by a prime number where n  2 Step 1: Show that the property is true for n=2 The property is true for n=2 because 2 is a prime number and 2 is divisible by 2 Step 2: Show that for all integers k  2, if the property true for all i with 2  i  k, then it is true for k For all integers i with 2  i  k, i is divisible by a prime number. (induction hypothesis)

  21. Applying Strong Mathematical Induction (2) Either k is a prime or not. If k is a prime, k is divisible by a prime number, namely itself. If k is not a prime, then k = ab, where a and b are integers with 2  a  k and 2  b  k. By inductive hypothesis, a is divisible by a prime number p, and so by transitivity of divisibility (pg151), k is also divisible by p.  Hence regardless of whether k is a prime or not, k is divisible by a prime number

  22. Divisibility (pg 148) • Integer n is divisible by an integer d, when k  Z, n = d * k, for some integer k • Notation: d | n • Synonymous statements: • n is a multiple of d • d is a factor of n • d is a divisor of n • d divides n • Eg: If a and b are integers, is 3a + 3b divisible by 3? Yes. By distributive law, 3a + 3b = 3(a+b) and a+b is an integer because it is a sum of 2 integers

  23. Divisibility (pg 148) • Divisibility is transitive: for all integers a, b, c, if a divides b and b divides c, then a divides c We need to show a | c or in other words c = a*k (k is some integer) a | b or b = a*r; and b | c or c = b*s (r and s are some integers) c = b*s =(a*r)*s =a(rs) Let k = rs, therefore a divides c by definition of divisibility

  24. Exercises • Prove or provide counterexample: • For integers a, b, c: (a | b)  (a | bc) • For integers a, b, c: (a | (b + c))  (a | b  a | c) • If 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * m = 151 * 150 * 149 * 148 * 147 * 146 * 145 * 144 * 143, does 151 | m? • Show that an integer is divisible by 9 iff the sum of its digits is divisible by 9. Prove the same for divisibility by 3. • Show that an integer is divisible by 11 iff the alternate sum of its digits is divisible by 11

  25. Fundamental Theorem of Arithmetic (pg 153) • Given any integer n  1, the standard factored form of n, is an expression of the form where k is a positive integer; p1, p2, …, pk are prime numbers; e1, e2, …, ek are positive integers; and p1  p2  pk • Eg: Write 3300 in standard factored form 3300= 2(1650) =22(825) =2231(275) = 223152(11) = 223152111

  26. Fundamental Theorem of Arithmetic (pg 153) • Number of positive divisors of n are (e1 + 1)(e2 + 1)…(ek + 1) • Eg: 29, 338, 848,000 = 28 35 53 73 11. ∴ 29,338,848,000 has (8 + 1)(5 + 1)(3 + 1)(3 + 1)(1 + 1) = (9) (6) (4) (4) (2) = 1728 positive divisors.

  27. The Division Algorithm (pg 190) • For an integer a and a positive integer d, then there exists integers q and r such that a = dq + r and 0  r  d • In quotient-remainder theorem, q=quotient and r=remainder • Eg: Find the q and r of the division of 32 by 9 The q represents how many number 9’s contained in 32 The r is the number left over when all possible groups of 9 are subtracted 32 – 9 = 23  9, and 23 – 9 = 14  9, and 14 – 9 = 5  9 This shows that 3 groups of 9 can be subtracted from 32 with 5 leftover. Thus, q=3 and r=5

  28. The Greatest Common DivisorThe Euclidean Algorithm (pg 192) • gcd of 2 integers a and b is the largest integers that divides both a and b (gcd of 12 and 30 is 6) • Euclidean algorithm: An efficient way to compute the gcd for 2 integers • Let a and b be integers that are not 0. The gcd(a, b) is that integer d with the following properties: • d is a common divisor for both a and b or d|a and d|b • For all integers c, if c|a and c|b, then c  d.

  29. The Greatest Common DivisorThe Euclidean Algorithm (pg 192) • Lemma 1: if r is a positive integer, then gcd(r, 0) = r • Lemma 2: if a = bq + r, then gcd(a, b) = gcd(b, r) • So if a, b are 2 positive integers, division algorithm is apply as a = q1b + r1, 0  r1  b b = q2r1 + r2, 0  r2  r1 r1 = q3r2 + r3, 0  r3  r2 … ri = qi+2ri+1 + ri+2, 0  ri+2 + ri+1 ... rk-3 = qk-1rk-2 + rk-1, 0 < rk-1 < rk-2 rk-2 = qkrk-1 + rk, 0 < rk < rk-1 rk-1 = qk+1rk Then, rk , the last nonzero remainder , equals gcd(a, b)

  30. Examples • Using Euclidean algo, find gcd(330, 156) 330 = 156*2 + 18 156 = 18*8 +12 18 = 12*1 + 6 12 = 6*2 gcd(330, 156) = 6 • Find gcd(250, 111)

  31. Least Common Multiple For a, b, c Z+, c is called a common multiple of a, b if c is a multiple of both a and b. Furthermore, c is the least common multiple of a, b if it is the smallest of all positive integers. We denote c by lcm(a, b).

  32. Examples • Eg: 12 = 34 lcm(3, 4) = lcm (4, 3) = 12 • Eg: lcm(6, 15) = 30 not 90 • For all n Z+, we find that lcm(1, n) = lcm(n, 1) = n. • When a, n  Z+, we have lcm(a, na) = na. • If a, m, n  Z+ with m ≤ n, then lcm(am, an) = an. [And gcd(am, an) = am].

  33. Least Common Multiple • Theorem: Let a, b, c  Z+, with c = lcm(a, b). If d is a common multiple of a and b, then c|d. • Theorem: For all a, b  Z+, ab = lcm(a, b)  gcd(a, b). • Eg: gcd(250, 111) = 1. As a result we find that lcm(250, 111) = (250)(111) / 1 = 27,750 • Eg: For all a, b  Z+, if a, b are relatively prime, then lcm(a, b) = ab. • Eg: gcd(456, 624) = 24. lcm(456, 624) = (456)(624) / 24 = 11,856.

  34. GCD and LCM • If m, n  Z+, let and with each piprime and 0 ≤ eiand 0 ≤ fifor all 1 ≤ i ≤ t. • Then if ai= min{ei, fi}, the minimum (or smaller) of eiand, fi and bi = max { ei, fi}, the maximum (or larger) of eiand fi, for all 1 ≤ i ≤ t, we have

  35. GCD and LCM • Eg: Let m = 491,891,400 = 23 33 52 72 111 132 and n = 1,138,845,708 = 22 32 71 112 1333 171. • Then with p1= 2, p2= 3, p3= 5, p4= 7, p5 = 11, p6= 13, and p7= 17, we find a1= 2, a2= 2, a3= 0 (the exponent of 5 in the prime factorization of n must be 0, because 5 does not appear in the prime factorization), a4= 1, a5= 1, a6= 2, and a7= 0. So gcd(m, n) = 22 32 50 71 111 132 170 = 468,468. We also have lcm(m, n) = 23 33 52 72 112 133 171 = 1,195,787,993,400.

  36. Correctness of Algorithms • Assertions • Pre-condition is a predicate describing initial state before an algorithm is executed • Post-condition is a predicate describing final state after an algorithm is executed • Loop guard • Loop is defined as correct with respect to its pre- and post- conditions, if whenever the algorithm variables satisfy the pre-conditions and the loop is executed, then the algorithm satisfies the post-conditions as well

  37. Loop Invariant Theorem • Let a while loop with guard G be given together with its pre- and post- conditions. Let predicate I(n) describing loop invariant be given. If the following 4 properties hold, then the loop is correct: • Basis Property: I(0) is true before the first iteration of the loop • Inductive Property: If G and I(k) is true, then I(k + 1) is true • Eventual Falsity of the Guard: After finite number of iterations, G becomes false • Correctness of the Post-condition: If N is the least number of iterations after which G becomes false and I(N) is true, then post-conditions are true as well

  38. Correctness of Some Algorithms • Product Algorithm: pre-conditions: m  0, i = 0, product = 0 while (i < m) { product += x; i++; } post-condition: product = m * x

  39. Correctness of Some Algorithms • Division Algorithm pre-conditions: a  0, d > 0, r = a, q = 0 while (r  d) { r -= d; q++; } post-conditions: a = q * d + r, 0  r < d

  40. Correctness of Some Algorithms • Euclidean Algorithm pre-conditions: a > b  0, r = b while (b > 0) { r = a mod b; a = b; b = r; } post-condition: a = gcd(a, b)

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