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CSC 10400: Discrete Mathematical Structures Recitation Chapter 4
CSC 10400: Discrete Mathematical Structures
Recitation
Chapter 4
Lecturer: Pavel Rytir
The City College of New York
2015
CSC 10400: Discrete Mathematical Structures Recitation Chapter 4
Chapter 4: Properties of the Integers: Mathematical Induction
Section 4.1 The well-ordering Principle: Mathematical Induction
Problem 1(a)
Prove the following formula for all n ≥ 1 by the principle of mathematical induction.
S(n) = 12 + 32 + 52 + . . . + (2n − 1)2 = n(2n − 1)(2n + 1)/3
Answer
CSC 10400: Discrete Mathematical Structures Recitation Chapter 4
Chapter 4: Properties of the Integers: Mathematical Induction
Section 4.1 The well-ordering Principle: Mathematical Induction
Problem 1(a)
Prove the following formula for all n ≥ 1 by the principle of mathematical induction.
S(n) = 12 + 32 + 52 + . . . + (2n − 1)2 = n(2n − 1)(2n + 1)/3
Answer
S(1) = 12 = (1 × 1 × 3)/3. This is true.
Assume S(k) = k(2k − 1)(2k + 1)/3 for some k ≥ 1.
S(k + 1) = S(k) + (2k + 1)2
= k(2k − 1)(2k + 1)/3 + (2k + 1)2
= (2k + 1)(2k 2 + 5k + 3)/3
= (2k + 1)(k + 1)(2k + 3)/3
= (k + 1)(2(k + 1) − 1)(2(k + 1) + 1)/3
Therefore, S(n) = n(2n − 1)(2n + 1)/3 holds for all n ≥ 1.
CSC 10400: Discrete Mathematical Structures Recitation Chapter 4
Chapter 4: Properties of the Integers: Mathematical Induction
Section 4.1 The well-ordering Principle: Mathematical Induction
Problem 1(b)
Prove the following formula for all n ≥ 1 by the principle of mathematical induction.
S(n) = 1 · 3 + 2 · 4 + 3 · 5 + . . . + n(n + 2) = n(n + 1)(2n + 7)/6
Answer
CSC 10400: Discrete Mathematical Structures Recitation Chapter 4
Chapter 4: Properties of the Integers: Mathematical Induction
Section 4.1 The well-ordering Principle: Mathematical Induction
Problem 1(b)
Prove the following formula for all n ≥ 1 by the principle of mathematical induction.
S(n) = 1 · 3 + 2 · 4 + 3 · 5 + . . . + n(n + 2) = n(n + 1)(2n + 7)/6
Answer
S(1) = 1 · 3 = (1 × 2 × 9)/6. This is true.
Assume S(k) = k(k + 1)(2k + 7)/6 for some k ≥ 1.
S(k + 1) = S(k) + (k + 1)(k + 3)
= k(k + 1)(2k + 7)/6 + (k + 1)(k + 3)
= (k + 1)(2k 2 + 13k + 18)/6
= (k + 1)(k + 2)(2k + 9)/6
= (k + 1)((k + 1) + 1)(2(k + 1) + 7)/6
Therefore, S(n) = n(n + 1)(2n + 7)/6 holds for all n ≥ 1.
CSC 10400: Discrete Mathematical Structures Recitation Chapter 4
Chapter 4: Properties of the Integers: Mathematical Induction
Section 4.1 The well-ordering Principle: Mathematical Induction
Problem 1(c)
Prove the following formula for all n ≥ 1 by the principle of mathematical induction.
S(n) =
n
X
i=1
Answer
n
1
=
i(i + 1)
n+1
CSC 10400: Discrete Mathematical Structures Recitation Chapter 4
Chapter 4: Properties of the Integers: Mathematical Induction
Section 4.1 The well-ordering Principle: Mathematical Induction
Problem 1(c)
Prove the following formula for all n ≥ 1 by the principle of mathematical induction.
S(n) =
n
X
i=1
n
1
=
i(i + 1)
n+1
Answer
1
1
S(1) = 1(1+1)
= 1+1
. This is true.
Assume S(k) = k/(k + 1) for some k ≥ 1.
S(k + 1) = S(k) + 1/((k + 1)(k + 2))
= k/(k + 1) + 1/((k + 1)(k + 2))
= (k(k + 2) + 1)/((k + 1)(k + 2))
= (k + 1)/((k + 1) + 1)
Therefore, S(n) = n/(n + 1) holds for all n ≥ 1.
CSC 10400: Discrete Mathematical Structures Recitation Chapter 4
Chapter 4: Properties of the Integers: Mathematical Induction
Section 4.1 The well-ordering Principle: Mathematical Induction
Problem 1(d)
Prove the following formula for all n ≥ 1 by the principle of mathematical induction.
S(n) =
n
X
i=1
Answer
i3 =
n2 (n + 1)2
4
CSC 10400: Discrete Mathematical Structures Recitation Chapter 4
Chapter 4: Properties of the Integers: Mathematical Induction
Section 4.1 The well-ordering Principle: Mathematical Induction
Problem 1(d)
Prove the following formula for all n ≥ 1 by the principle of mathematical induction.
S(n) =
n
X
i=1
i3 =
n2 (n + 1)2
4
Answer
S(1) = 13 = (12 22 )/4. This is true.
Assume S(k) = (k 2 (k + 1)2 )/4 for some k ≥ 1.
S(k + 1) = S(k) + (k + 1)3
= (k 2 (k + 1)2 )/4 + (k + 1)3
= ((k + 1)2 (k 2 + 4k + 4))/4
= ((k + 1)2 ((k + 1) + 1)2 )/4
Therefore, S(n) = n2 (n + 1)2 /4 holds for all n ≥ 1.
CSC 10400: Discrete Mathematical Structures Recitation Chapter 4
Chapter 4: Properties of the Integers: Mathematical Induction
Section 4.1 The well-ordering Principle: Mathematical Induction
Problem 7
A lumberjack has 4n+110 logs in a pile consisting of n layers. Each
layer has two more logs than the layer directly above it. If the top
layer has six logs, how many layers are there?
Answer
CSC 10400: Discrete Mathematical Structures Recitation Chapter 4
Chapter 4: Properties of the Integers: Mathematical Induction
Section 4.1 The well-ordering Principle: Mathematical Induction
Problem 7
A lumberjack has 4n+110 logs in a pile consisting of n layers. Each
layer has two more logs than the layer directly above it. If the top
layer has six logs, how many layers are there?
Answer
4n + 110 = 6 + 8 + 10 + . . . + (6 + 2(n − 1))
= 6n + (0 + 2 + 4 + . . . + 2(n − 1)
= 6n + 2(1 + 2 + . . . + (n − 1))
= 6n + n(n − 1)
= n2 + 5n
Therefore, n2 + n − 110 = 0, i.e., n = 10 (n = −11 is discarded).
CSC 10400: Discrete Mathematical Structures Recitation Chapter 4
Chapter 4: Properties of the Integers: Mathematical Induction
Section 4.1 The well-ordering Principle: Mathematical Induction
Problem 8
Determine the positive integer n for which
2n
X
i=1
Answer
i=
n
X
i=1
i2
CSC 10400: Discrete Mathematical Structures Recitation Chapter 4
Chapter 4: Properties of the Integers: Mathematical Induction
Section 4.1 The well-ordering Principle: Mathematical Induction
Problem 8
Determine the positive integer n for which
2n
X
i=1
i=
n
X
i2
i=1
Answer
Here we have
Pn
i=1 i
2
=
n(n+1)(2n+1)
,
6
P2n
i=1 i
=
(2n)(2n+1)
.
2
n(n + 1)(2n + 1)
(2n)(2n + 1)
=
6
2
Therefore, n = 5.
Thus
CSC 10400: Discrete Mathematical Structures Recitation Chapter 4
Chapter 4: Properties of the Integers: Mathematical Induction
Section 4.2: Recursive Definitions
Problem 1
Give the recursive definition for each of the following
(a) cn = 7n.
(b) cn = 7n .
(c) cn = 3n + 7.
(d) cn = n2 .
(e) cn = 2 − (−1)n .
Answer
CSC 10400: Discrete Mathematical Structures Recitation Chapter 4
Chapter 4: Properties of the Integers: Mathematical Induction
Section 4.2: Recursive Definitions
Problem 1
Give the recursive definition for each of the following
(a) cn = 7n.
(b) cn = 7n .
(c) cn = 3n + 7.
(d) cn = n2 .
(e) cn = 2 − (−1)n .
Answer
(a) c1 = 7 and cn = cn−1 + 7 for n > 1.
CSC 10400: Discrete Mathematical Structures Recitation Chapter 4
Chapter 4: Properties of the Integers: Mathematical Induction
Section 4.2: Recursive Definitions
Problem 1
Give the recursive definition for each of the following
(a) cn = 7n.
(b) cn = 7n .
(c) cn = 3n + 7.
(d) cn = n2 .
(e) cn = 2 − (−1)n .
Answer
(a) c1 = 7 and cn = cn−1 + 7 for n > 1.
(b) c1 = 7 and cn = 7cn−1 for n > 1.
CSC 10400: Discrete Mathematical Structures Recitation Chapter 4
Chapter 4: Properties of the Integers: Mathematical Induction
Section 4.2: Recursive Definitions
Problem 1
Give the recursive definition for each of the following
(a) cn = 7n.
(b) cn = 7n .
(c) cn = 3n + 7.
(d) cn = n2 .
(e) cn = 2 − (−1)n .
Answer
(a) c1 = 7 and cn = cn−1 + 7 for n > 1.
(b) c1 = 7 and cn = 7cn−1 for n > 1.
(c) c1 = 10 and cn = cn−1 + 3 for n > 1.
CSC 10400: Discrete Mathematical Structures Recitation Chapter 4
Chapter 4: Properties of the Integers: Mathematical Induction
Section 4.2: Recursive Definitions
Problem 1
Give the recursive definition for each of the following
(a) cn = 7n.
(b) cn = 7n .
(c) cn = 3n + 7.
(d) cn = n2 .
(e) cn = 2 − (−1)n .
Answer
(a) c1 = 7 and cn = cn−1 + 7 for n > 1.
(b) c1 = 7 and cn = 7cn−1 for n > 1.
(c) c1 = 10 and cn = cn−1 + 3 for n > 1.
(d) c1 = 1 and cn = cn−1 + 2n − 1 for n > 1.
CSC 10400: Discrete Mathematical Structures Recitation Chapter 4
Chapter 4: Properties of the Integers: Mathematical Induction
Section 4.2: Recursive Definitions
Problem 1
Give the recursive definition for each of the following
(a) cn = 7n.
(b) cn = 7n .
(c) cn = 3n + 7.
(d) cn = n2 .
(e) cn = 2 − (−1)n .
Answer
(a) c1 = 7 and cn = cn−1 + 7 for n > 1.
(b) c1 = 7 and cn = 7cn−1 for n > 1.
(c) c1 = 10 and cn = cn−1 + 3 for n > 1.
(d) c1 = 1 and cn = cn−1 + 2n − 1 for n > 1.
(e) c1 = 3, c2 = 1, and cn = cn−2 for n > 2.
CSC 10400: Discrete Mathematical Structures Recitation Chapter 4
Chapter 4: Properties of the Integers: Mathematical Induction
Section 4.2: Recursive Definitions
Problem 13
Let Fn be the Fibonacci series defined by F0 = 0, F1 = 1 and
Fn = Fn−2 + Fn−1 for n > 1. Prove that
S(n) =
n
X
Fi−1
i=1
2i
Answer
Proof is shown in the next slide.
=1−
Fn+2
2n
CSC 10400: Discrete Mathematical Structures Recitation Chapter 4
Chapter 4: Properties of the Integers: Mathematical Induction
Section 4.2: Recursive Definitions
Answer
CSC 10400: Discrete Mathematical Structures Recitation Chapter 4
Chapter 4: Properties of the Integers: Mathematical Induction
Section 4.2: Recursive Definitions
Answer
Proof.
First, when n = 1, S(n) = F210 = 0, and 1 − F21+2
= 0. Thus, the
1
statement holds for n = 1.
Suppose the statement holds for n = k, then when n = k + 1, we
have
k
S(k + 1) = S(k) + 2Fk+1
2Fk+2 −Fk
= 1 − 2k+1
+Fk
= 1 − 2Fk+1
2k+1
+Fk+2
= 1 − Fk+12k+1
=1−
=1−
=1−
=1−
Fk+2
k
+ 2Fk+1
2k
2(Fk+1 +Fk )−Fk
2k+1
Fk+1 +(Fk+1 +Fk )
2k+1
Fk+3
2k+1
By mathematical induction, the statement holds for all
n ∈ Z+ .
CSC 10400: Discrete Mathematical Structures Recitation Chapter 4
Chapter 4: Properties of the Integers: Mathematical Induction
Section 4.3: The Divison Algorithm: Prime Numbers
Problem 5
For all a, b, c ∈ Z, prove that if a 6 |bc, then a 6 |b and a 6 |c.
Answer
Proof.
Assume that a|b. By definition, there exists an integer k so that
b = ak. Thus bc = akc, i.e, a|bc, contradicting that a 6 |bc. Thus
a 6 |b. Similary, we can prove that a 6 |c.
CSC 10400: Discrete Mathematical Structures Recitation Chapter 4
Chapter 4: Properties of the Integers: Mathematical Induction
Section 4.3: The Divison Algorithm: Prime Numbers
Problem 15
Write each of the following (base-10) integers in base 2 and base
16.
(a) 22
(b) 527
(c) 1234
(d) 6923
Answer
CSC 10400: Discrete Mathematical Structures Recitation Chapter 4
Chapter 4: Properties of the Integers: Mathematical Induction
Section 4.3: The Divison Algorithm: Prime Numbers
Problem 15
Write each of the following (base-10) integers in base 2 and base
16.
(a) 22
(b) 527
(c) 1234
(d) 6923
Answer
(a) Base-2: 10110. Base-16: 16.
CSC 10400: Discrete Mathematical Structures Recitation Chapter 4
Chapter 4: Properties of the Integers: Mathematical Induction
Section 4.3: The Divison Algorithm: Prime Numbers
Problem 15
Write each of the following (base-10) integers in base 2 and base
16.
(a) 22
(b) 527
(c) 1234
(d) 6923
Answer
(a) Base-2: 10110. Base-16: 16.
(b) Base-2: 1000001111. Base-16: 20F.
CSC 10400: Discrete Mathematical Structures Recitation Chapter 4
Chapter 4: Properties of the Integers: Mathematical Induction
Section 4.3: The Divison Algorithm: Prime Numbers
Problem 15
Write each of the following (base-10) integers in base 2 and base
16.
(a) 22
(b) 527
(c) 1234
(d) 6923
Answer
(a) Base-2: 10110. Base-16: 16.
(b) Base-2: 1000001111. Base-16: 20F.
(c) Base-2: 10011010010. Base-16: 4D2.
CSC 10400: Discrete Mathematical Structures Recitation Chapter 4
Chapter 4: Properties of the Integers: Mathematical Induction
Section 4.3: The Divison Algorithm: Prime Numbers
Problem 15
Write each of the following (base-10) integers in base 2 and base
16.
(a) 22
(b) 527
(c) 1234
(d) 6923
Answer
(a) Base-2: 10110. Base-16: 16.
(b) Base-2: 1000001111. Base-16: 20F.
(c) Base-2: 10011010010. Base-16: 4D2.
(d) Base-2: 1101100001011. Base-16: 1B0B.
CSC 10400: Discrete Mathematical Structures Recitation Chapter 4
Chapter 4: Properties of the Integers: Mathematical Induction
Section 4.4: The Greatest Common Divisor: The Euclidean Algorithm
Problem 1
For each of the following pairs a, b ∈ Z+ , determine gcd(a, b) and
express it as a linear combination of a, b.
(a) 231,1820
(b) 1369,2597
(c) 2689,4001
Answer
CSC 10400: Discrete Mathematical Structures Recitation Chapter 4
Chapter 4: Properties of the Integers: Mathematical Induction
Section 4.4: The Greatest Common Divisor: The Euclidean Algorithm
Problem 1
For each of the following pairs a, b ∈ Z+ , determine gcd(a, b) and
express it as a linear combination of a, b.
(a) 231,1820
(b) 1369,2597
(c) 2689,4001
Answer
(a) gcd(1820, 231) = 7, 7 = 1820 × 8 + (−63) × 231.
CSC 10400: Discrete Mathematical Structures Recitation Chapter 4
Chapter 4: Properties of the Integers: Mathematical Induction
Section 4.4: The Greatest Common Divisor: The Euclidean Algorithm
Problem 1
For each of the following pairs a, b ∈ Z+ , determine gcd(a, b) and
express it as a linear combination of a, b.
(a) 231,1820
(b) 1369,2597
(c) 2689,4001
Answer
(a) gcd(1820, 231) = 7, 7 = 1820 × 8 + (−63) × 231.
(b) gcd(1369, 2597) = 1, 1 = 2597 × 534 + 1369 × (−1013).
CSC 10400: Discrete Mathematical Structures Recitation Chapter 4
Chapter 4: Properties of the Integers: Mathematical Induction
Section 4.4: The Greatest Common Divisor: The Euclidean Algorithm
Problem 1
For each of the following pairs a, b ∈ Z+ , determine gcd(a, b) and
express it as a linear combination of a, b.
(a) 231,1820
(b) 1369,2597
(c) 2689,4001
Answer
(a) gcd(1820, 231) = 7, 7 = 1820 × 8 + (−63) × 231.
(b) gcd(1369, 2597) = 1, 1 = 2597 × 534 + 1369 × (−1013).
(c) gcd(2689, 4001) = 1. 1 = 4001 × (−1117) + 2689 × 1662.
CSC 10400: Discrete Mathematical Structures Recitation Chapter 4
Chapter 4: Properties of the Integers: Mathematical Induction
Section 4.4: The Greatest Common Divisor: The Euclidean Algorithm
Problem 19
If a, b ∈ Z+ with a = 630, gcd(a, b) = 105, and lcm = 242550,
what is b?
Answer
CSC 10400: Discrete Mathematical Structures Recitation Chapter 4
Chapter 4: Properties of the Integers: Mathematical Induction
Section 4.4: The Greatest Common Divisor: The Euclidean Algorithm
Problem 19
If a, b ∈ Z+ with a = 630, gcd(a, b) = 105, and lcm = 242550,
what is b?
Answer
Since ab = gcd(a, b) × lcm(a, b), we have b =
105×242550
630
= 40425.