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Page 1: Discrete Structures Chapter 2 Part B Mathematical Induction

Discrete StructuresChapter 2 Part B

Mathematical InductionNurul Amelina Nasharuddin

Multimedia Department

Page 2: Discrete Structures Chapter 2 Part B Mathematical Induction

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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• 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

Mathematical Induction

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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:1. P(a) is a true2. k a, P(k) P(k + 1)

then the following conclusion is true as wellP(n) is true for all n a

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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+15

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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!!)

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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)

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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)

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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

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Page 10: Discrete Structures Chapter 2 Part B Mathematical Induction

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

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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)

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LHS = RHSConsequently, the theorem is true

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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)

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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

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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)

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Page 15: Discrete Structures Chapter 2 Part B Mathematical Induction

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

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Proving a Property of A Sequence• Define a sequence a1, a2, a3, …, as follows

a1 = 2

ak = 5ak-1

a) Write the first four sequence

a1 = 2 a2 = 5a1 = 5.2 = 10

a3 = 5a2 = 5.10 = 50 a4 = 5a3 = 5.50 = 250

b) 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

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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

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Page 18: Discrete Structures Chapter 2 Part B Mathematical Induction

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

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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:1. P(a), P(a+1), P(a+2), …, and P(b) are true (basis step)2. 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

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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 2Step 1: Show that the property is true for n=2The 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)

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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

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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

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Page 23: Discrete Structures Chapter 2 Part B Mathematical Induction

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

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Page 24: Discrete Structures Chapter 2 Part B Mathematical Induction

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

Page 25: Discrete Structures Chapter 2 Part B Mathematical Induction

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

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Page 26: Discrete Structures Chapter 2 Part B Mathematical Induction

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.

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Page 27: Discrete Structures Chapter 2 Part B Mathematical Induction

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 32The r is the number left over when all possible groups of 9 are subtracted

32 – 9 = 23 9, and23 – 9 = 14 9, and14 – 9 = 5 9

This shows that 3 groups of 9 can be subtracted from 32 with 5 leftover. Thus, q=3 and r=5

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Page 28: Discrete Structures Chapter 2 Part B Mathematical Induction

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:

i. d is a common divisor for both a and b or d|a and d|b

ii. For all integers c, if c|a and c|b, then c d.

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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)

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Page 30: Discrete Structures Chapter 2 Part B Mathematical Induction

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)

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Page 31: Discrete Structures Chapter 2 Part B Mathematical Induction

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).

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Page 32: Discrete Structures Chapter 2 Part B Mathematical Induction

Examples

• Eg: 12 = 3.4lcm(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].

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Page 33: Discrete Structures Chapter 2 Part B Mathematical Induction

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.

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Page 34: Discrete Structures Chapter 2 Part B Mathematical Induction

GCD and LCM

• If m, n Z+, let and

with each pi prime and 0 ≤ ei

and 0 ≤ fi for all 1 ≤ i ≤ t.

• Then if ai = min{ei, fi}, the minimum (or smaller) of ei

and, fi and bi = max { ei, fi

}, the maximum (or larger) of ei and

fi, for all 1 ≤ i ≤ t, we have

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Page 35: Discrete Structures Chapter 2 Part B Mathematical Induction

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.

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Page 36: Discrete Structures Chapter 2 Part B Mathematical Induction

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

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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

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Page 38: Discrete Structures Chapter 2 Part B Mathematical Induction

Correctness of Some Algorithms

• Product Algorithm:

pre-conditions: m 0, i = 0, product = 0

while (i < m) {

product += x;

i++;

}

post-condition: product = m * x38

Page 39: Discrete Structures Chapter 2 Part B Mathematical Induction

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 < d39

Page 40: Discrete Structures Chapter 2 Part B Mathematical Induction

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)40


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