1. Discrete Mathematics, 1st Edition Kevin Ferland
Chapter 4
Indexed by Integers

2. A sequence is simply an ordered list of real
Sequences
A sequence is simply an ordered list of real
numbers. All sequences have an initial term,
but only finite sequences have a final term. 2
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3. S = sa + sa+1 + sa+2 + ・・ ・+sb−1 + sb , (4.10)
4.2 Sigma Notation
Given a sequence {sn }, one may be interested
in a sum of several of its terms:
S = sa + sa+1 + sa+2 + ・・ ・+sb−1 + sb , (4.10)
where a, b ∈ Z. The notation used to represent
the sum in equation (4.10) is
(4.11) 3
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4. THEOREM 4.1 4
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5. THEOREM 4.2 5
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6. THEOREM 4.3 6
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7. When b < a, it is assigned the value 1.
Product Notation
The product
P = sa ・ sa+1 ・ sa+2 ・ ・・・ ・sb
is represented by
When b < a, it is assigned the value 1. 7
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8. Let n ≥ 1. We represent a factorization of
EXAMPLE 4.16
Let n ≥ 1. We represent a factorization of
x2n − y2n in product notation. 8
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9. EXAMPLE 4.16 (Cont.)
Solution. 9
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10. 4.3 Mathematical Induction, an Introduction
Well-Ordering Principle
Each nonempty subset of the nonnegative integers has a smallest element. 10
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11. THEOREM 4.6 11
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12. Suppose it is not true that P(n) holds ∀ n ≥ a.
THEOREM 4.6
Proof. Assume conditions (i) and (ii) in the hypotheses of the theorem.
Suppose it is not true that P(n) holds ∀ n ≥ a.
Let S be the set of those integers n ≥ a for which P(n) does not hold.
By our assumptions, S is nonempty.
Hence, by the Generalized Well-Ordering Principle, S has a smallest element, say s. 12
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13. Since P(a), P(a + 1), . . . , P(b) all hold, it must be that s > b.
THEOREM 4.6 (Cont.)
Since P(a), P(a + 1), , P(b) all hold, it must be that s > b.
Therefore, s − 1 ≥ b.
Since s − 1 S, it follows that P(s − 1) holds.
However, for k = s − 1, by condition (ii), since P(k) holds, P(k + 1) must also hold. That is, P(s) holds.
This contradicts the fact that s ∈ S. 13
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14. (Proof by Mathematical Induction). To show: ∀ n ≥ a, P(n).
OUTLINE 4.1
(Proof by Mathematical Induction).
To show: ∀ n ≥ a, P(n). 14
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15. OUTLINE 4.1 Proof by induction. 1. Base cases:
Show: P(a), , P(b) are true.
2. Inductive step:
Show: ∀ k ≥ b, if P(k) is true, then P(k + 1) is true.
That is,
(a) Suppose k ≥ b and that P(k) is true.
(b) Show: P(k + 1) is true. 15
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16. EXAMPLE 4.18
Show: ∀ n ≥ 0, 2n ≥ n + 1. 16
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17. It is straightforward to see that 20 = 1 ≥ 0 + 1.
EXAMPLE 4.18
Proof.
(By Induction)
Base case: (n = 0)
It is straightforward to see that 20 = 1 ≥ 17
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18. EXAMPLE 4.18 (Cont.) Inductive step:
Suppose k ≥ 0 and that 2k ≥ k + 1.
(Goal: 2k+1 ≥ (k + 1) + 1.)
Observe that
2k+1 = 2(2k )
≥ 2(k + 1) ← By the inductive hypothesis
= (k + 1) + (k + 1)
≥ (k + 1) + 1. 18
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19. 4.4 Induction and Summations
EXAMPLE 4.26
Show: ∀ n ≥ 0, 19
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20. EXAMPLE 4.26 Proof. (By Induction) Base case: (n = 0)
It is straightforward to see that
Inductive step:
Suppose k ≥ 0 and that
(Goal: ) 20
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21. EXAMPLE 4.26 (Cont.)
Observe that 21
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22. EXAMPLE 4.26 (Cont.) 22
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23. 4.5 Strong Induction 23
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24. P(i) holds for each a ≤ i ≤ n.
Theorem 3.7
Proof.
Assume conditions (i) and (ii) in the hypotheses of the theorem.
For each n ≥ a, let Q(n) be the statement that
P(i) holds for each a ≤ i ≤ n.
Since Q(n) implies P(n), it is sufficient to show that Q(n) holds ∀ n ≥ a.
We accomplish this with a proof by regular induction. 24
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25. Theorem 3.7 (Cont.) Base cases: (n = a, . . . , b)
The truth of Q(a), , Q(b) follows from the truth of
P(a), , P(b).
Inductive step:
Suppose k ≥ a and that Q(k) holds.
That is, P(i) holds for each a ≤ i ≤ k.
From (ii), it follows that P(k + 1) holds.
By definition, Q(k + 1) also holds. 25
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26. Strong Induction
OUTLINE 4.2
To show: ∀ n ≥ a, P(n). 26
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27. OUTLINE 4.2 Proof by strong induction. 1. Base cases:
Show: P(a), , P(b) are true.
2. Inductive step:
Show: ∀ k ≥ b, if P(a), , P(k) are true, then P(k + 1) is true.
That is,
(a) Suppose k ≥ b and that P(i) is true for all a ≤ i ≤ k.
(b) Show: P(k + 1) is true. 27
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28. Every integer greater than 1 has a prime divisor.
Theorem 3.5
Every integer greater than 1 has a prime
divisor. 28
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29. Certainly, 2 is already prime and divides itself. That is, 2 | 2.
Theorem 3.5
Proof.
Base case: (n = 2)
Certainly, 2 is already prime and divides itself.
That is, 2 | 2. 29
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30. Suppose k ≥ 2 and that each integer i with
Theorem 3.5 (Cont.)
Inductive step:
Suppose k ≥ 2 and that each integer i with
2 ≤ i ≤ k has a prime divisor.
Case 1: k + 1 is prime.
Obviously, k + 1 divides itself. 30
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31. Theorem 3.5 (Cont.) Case 2: k + 1 is composite.
We can express k + 1 as a product k + 1 = rs, where 2 ≤ r ≤ k and 2 ≤ s ≤ k.
Consequently, there exists a prime p that divides r.
That is, r = pt for some integer t.
It follows that k + 1 = rs = pts.
Therefore, k + 1 is divisible by the prime p. 31
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32. Standard Factorization
DEFINITION 4.1
The expression of an integer n > 1 as a product of the
form
where m is a positive integer, p1 < p2 < ・ ・ ・ < pm
are primes, and e1, e2, , em are positive integers, is
referred to as the standard factorization of n. 32
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33. THEOREM 4.8 33
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34. Existence and uniqueness are proved
THEOREM 4.8
Proof.
Existence and uniqueness are proved
separately, but each by strong induction.
Existence is handled first.
Base case: (n = 2)
Certainly, 2 = 21 is a standard factorization. 34
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35. THEOREM 4.8 (Cont.) Inductive step:
Suppose k ≥ 2 and that each integer i with 2 ≤ i ≤ k has
a standard factorization.
(Goal: k + 1 has a standard factorization.)
Our proof naturally breaks into two cases.
Case 1: k + 1 is prime.
Here, k + 1 = (k + 1)1 is already a standard factorization. 35
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36. THEOREM 4.8 (Cont.) Case 2: k + 1 is composite.
We can express k + 1 as a product k + 1 = rs, where 2 ≤ r ≤ k and 2 ≤ s ≤ k.
By the inductive hypothesis, r and s have standard factorizations.
By appropriately grouping the primes in the product rs, we obtain a standard factorization for k + 1.
Now we handle uniqueness. 36
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37. THEOREM 4.8 (Cont.) Base case: (n = 2)
Since 2 is prime, 2 has the unique standard factorization 2 = 21. (Note that any other product of powers of primes is greater than 2.)
Inductive step:
Suppose k ≥ 2 and that each integer i with 2 ≤ i ≤ k has a unique standard factorization.
(Goal: k + 1 has a unique standard factorization.) 37
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38. THEOREM 4.8 (Cont.) Suppose k + 1 has two standard factorizations
Since p1 | (k + 1), Corollary 3.18 tells us that for some i.
Moreover, Corollary 3.19 tells us that p1 | qi .
Since p1 and qi are prime, it must be that p1 = qi .
Since the primes in a standard factorization are listed in increasing order, we have q1 ≤ qi = p1. 38
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39. THEOREM 4.8 (Cont.)
By a symmetric argument (reversing the roles of the p’s
and the q’s), we see that p1 ≤ q1. Thus, p1 = q1.
The integer now has the two standard factorizations
By the inductive hypothesis, those standard factorizations must be the same.
Therefore, the two standard factorizations for k + 1 must be the same. 39
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40. F0 = 1, F1 = 1, and ∀ n ≥ 2, Fn = Fn−2 + Fn−1.
The Fibonacci Numbers
The sequence of numbers
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . . .
is known as the Fibonacci sequence.
If we denote the Fibonacci sequence by {Fn}n≥0, then
F0 = 1, F1 = 1, and ∀ n ≥ 2, Fn = Fn−2 + Fn−1. 40
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41. Show that the Fibonacci sequence can be expressed by the formula
EXAMPLE 4.27
Show that the Fibonacci sequence can be
expressed by the formula 41
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42. It is straightforward to check that and
EXAMPLE 4.27
Proof.
Base cases: (n = 0, 1)
It is straightforward to check that
and 42
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43. EXAMPLE 4.27 (Cont.) Inductive step: Suppose k ≥ 1 and that 43
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44. EXAMPLE 4.27 (Cont.)
Observe that 44
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45. EXAMPLE 4.27 (Cont.) 45
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46. (a + b)n for some integer n ≥ 0.
4.6 The Binomial Theorem
The Binomial Theorem is a result that tells us
how to expand an expression of the form
(a + b)n for some integer n ≥ 0. 46
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47. (a + b)0 = 1 (a + b)1 = a + b (a + b)2 = a2 + 2ab + b2
(a + b)3 = a3 + 3a2b + 3ab2 + b (4.19)
(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4 … … 47
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48. This infinite array of integers is known as Pascal’s
triangle. To express this, let cn,k denote the kth entry in
the nth row. Here, both n and k start counting from 0. 48
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49. The edges of the triangle display the identity
∀ n ≥ 0, cn,0 = cn,n = 1. 49
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50. ∀ n ≥ 2 and 1 ≤ k ≤ n − 1, cn,k = cn−1,k−1 + cn−1,k . (4.20)
The internal entries of each row are determined from the previous row by Pascal’s identity
∀ n ≥ 2 and 1 ≤ k ≤ n − 1, cn,k = cn−1,k−1 + cn−1,k . (4.20) 50
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51. THEOREM 4.9 51
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52. Use the Binomial Theorem to expand each of the following.
EXAMPLE 4.29
Use the Binomial Theorem to expand each of
the following. 52
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53. EXAMPLE 4.29 (Cont.) (a) Expand (x + y)6. Solution.
We use a = x, b = y, and n = 6 in Theorem 4.9. 53
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54. EXAMPLE 4.29 (Cont.) (b) Expand (2x + 3y)5. Solution.
We use a = 2x, b = 3y, and n = 5 in Theorem 4.9. 54
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55. EXAMPLE 4.29 (Cont.) 55
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56. EXAMPLE 4.29 (Cont.) (d) Expand (x − y)n. Solution.
We use a = x and b = −y in Theorem 4.9. 56
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57. Find the coefficient of x40 in (1 + 2x)50.
EXAMPLE 4.30
Find the coefficient of x40 in (1 + 2x)50. 57
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58. EXAMPLE 4.30 Solution. By the Binomial Theorem,
Consequently, the coefficient of x40 (that is, when i = 40) is 58
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59. Verify each of the following identities. (a) Proof By Theorem 4.9,
EXAMPLE 4.31
Verify each of the following identities.
(a)
Proof
By Theorem 4.9, 59
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60. EXAMPLE 4.31 (Cont.)
(b)
Proof
By Theorem 4.9, 60
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61. EXAMPLE 4.31 (Cont.)
(c)
Proof
By Theorem 4.9, 61
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