discrete mathematics, 1st edition kevin ferland chapter 4 indexed by integers 1
TRANSCRIPT
Discrete Mathematics, 1st EditionKevin Ferland
Chapter 4
Indexed by Integers
1
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.
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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)
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THEOREM 4.1
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THEOREM 4.2
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THEOREM 4.3
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Product Notation
The product
P = sa ・ sa+1 ・ sa+2 ・ ・・・ ・ sb
is represented by
When b < a, it is assigned the value 1.
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EXAMPLE 4.16
Let n ≥ 1. We represent a factorization of
x2n − y2n in product notation.
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EXAMPLE 4.16 (Cont.)
Solution.
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4.3 Mathematical Induction, an Introduction
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Well-Ordering PrincipleEach nonempty subset of the nonnegative integers has a smallest element.
THEOREM 4.6
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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.
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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.
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OUTLINE 4.1
(Proof by Mathematical Induction).
To show: ∀ n ≥ a, P(n).
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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.
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EXAMPLE 4.18
Show: ∀ n ≥ 0, 2n ≥ n + 1.
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EXAMPLE 4.18
Proof.
(By Induction)
Base case: (n = 0)
It is straightforward to see that 20 = 1 ≥ 0 + 1.
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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.
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4.4 Induction and Summations
EXAMPLE 4.26
Show: ∀ n ≥ 0,
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EXAMPLE 4.26
Proof. (By Induction)
Base case: (n = 0)
It is straightforward to see that
Inductive step:
Suppose k ≥ 0 and that
(Goal: )
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EXAMPLE 4.26 (Cont.)
Observe that
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EXAMPLE 4.26 (Cont.)
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4.5 Strong Induction
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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
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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.
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Strong Induction
OUTLINE 4.2
To show: ∀ n ≥ a, P(n).
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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.27Ch4-p202
Theorem 3.5
Every integer greater than 1 has a prime
divisor.
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Theorem 3.5
Proof.
Base case: (n = 2)
Certainly, 2 is already prime and divides itself.
That is, 2 | 2.
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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.
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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.
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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.
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mem
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THEOREM 4.8
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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.
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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.
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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.
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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.)
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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.
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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.
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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.
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EXAMPLE 4.27
Show that the Fibonacci sequence can be
expressed by the formula
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EXAMPLE 4.27
Proof.
Base cases: (n = 0, 1)
It is straightforward to check that
and
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EXAMPLE 4.27 (Cont.)
Inductive step:
Suppose k ≥ 1 and that
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EXAMPLE 4.27 (Cont.)
Observe that
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EXAMPLE 4.27 (Cont.)
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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.
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(a + b)0 = 1
(a + b)1 = a + b
(a + b)2 = a2 + 2ab + b2
(a + b)3 = a3 + 3a2b + 3ab2 + b3
(4.19)
(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4
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… …
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.
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The edges of the triangle display the identity
∀ n ≥ 0, cn,0 = cn,n = 1.
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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)
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THEOREM 4.9
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EXAMPLE 4.29
Use the Binomial Theorem to expand each of
the following.
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EXAMPLE 4.29 (Cont.)
(a) Expand (x + y)6. Solution. We use a = x, b = y, and n = 6 in Theorem 4.9.
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EXAMPLE 4.29 (Cont.)
(b) Expand (2x + 3y)5.
Solution.
We use a = 2x, b = 3y, and n = 5 in Theorem 4.9.
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EXAMPLE 4.29 (Cont.)
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EXAMPLE 4.29 (Cont.)
(d) Expand (x − y)n. Solution.
We use a = x and b = −y in Theorem 4.9.
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EXAMPLE 4.30
Find the coefficient of x40 in (1 + 2x)50.
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EXAMPLE 4.30
Solution.
By the Binomial Theorem,
Consequently, the coefficient of x40 (that is, when i = 40) is
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EXAMPLE 4.31
Verify each of the following identities.
(a)
Proof
By Theorem 4.9,
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EXAMPLE 4.31 (Cont.)
(b)
Proof
By Theorem 4.9,
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EXAMPLE 4.31 (Cont.)
(c)
Proof
By Theorem 4.9,
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