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Task 1. Suppose we are given set π΄ = {1,2,3,4,6,12} and a relation, π , from π΄ Γ π΄. The relation is defined as follows:
π = {(π, π)| a divides b, π€βπππ(a, b) belongs to π΄ Γ π΄}. a) List all the ordered pairs (π, π) that are elements of the relation.b) Use the results from part a) to construct the corresponding zero-one matrix.Solution. a) Relation π consists of the following pairs:
(1,1), (1,2), (1,3), (1,4), (1,6), (1,12), (2,2), (2,4), (2,6), (2,12), (3,3), (3,6), (3,12)
(4,4), (4,12), (6,6), (6,12), (12,12)
b) The matrix representing this relation has the following form:1 2 3 4 6 12
1 1 1 1 1 1 12 0 1 0 1 1 13 0 0 1 0 1 14 0 0 0 1 0 16 0 0 0 0 1 112 0 0 0 0 0 1
Task 2. Let π be the relation on the set of all people who have visited a particular Web page suchthat π₯π π¦ if and only if person π₯ and person π¦ have followed the same set of links starting atthis Web page (going from Web page to Web page until they stop using the Web). Show that π is an equivalence relation (i.e. it is reflexive, symmetric, and transitive). Solution. 1) π is reflexive, that is π₯π π₯ for all persons π₯.Indeed, π₯ and π₯ have followed the same set of links starting at this Web page.
2) π is symmetric, that is π₯π π¦ then π¦π π₯.Indeed if π₯ and π¦ have followed the same set of links starting at this Web page, then π¦ and π§ have followed the same set of links starting at this Web page.
3) π is transitive, that is if π₯π π¦ and π¦π π§ then π₯π π§.Indeed if π₯ and π¦ have followed the same set of links starting at this Web page, and π¦ and π§ have followed the same set of links starting at this Web page, then π₯ and π§ also have followed the same set of links starting at this Web page.
Task 3. For the relation π on the set π΄ = {1,2,3,4} given below, deteremine whether it isreflexive, symmetric, anti-symmetric and transitive.
π = {(1,3), (1,4), (2,3), (2,4), (3,1), (3,4)}
Solution. 1) π is not reflexive, since (1,1) β π .2) π is not symmetric, since (2,3) β π but (3,2) β π .
D O M Y A S S I G N M E N T SUBMIT
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Sample: Discrete Mathematics - Properties of Relations
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3) π is not anti-symmetric, since both (1,3) and (3,1) belong to π .4) π is not transitive, since (1,3), (3,1) β π but (1,1) β π .
Task 4 . List the ordered pairs in the relation π from π΄ = {0,1,2,3,4} to π΅ = {0,1,2,3}, where (π, π) β π if and only if π + π = 4. Solution. The relation π consists of the following pairs:
(1,3), (2,2), (3,1), (4,0).
Task 5. A department manager has 4 employees involved with 6 projects throughout the fiscal year. In how many ways can the manager assign these projects so that each employee is working on at least one project? Solution. Let π = {1,2,3,4,5,6} be the set of all projects. Each assignment is a partition of π into 4 non-empty subsets. Therefore we should compute the number of partitions of π into ordered family of 4 non-empty subsets π΄1, π΄2, π΄3, π΄4.Since every π΄π is non-empty then there possible two distinct cases:Case 1). One of these sets consists of 3 elements, and each of other three sets consists of a unique element, e.g.
π΄1 = {1,2,3}, π΄2 = {4}, π΄3 = {5}, π΄4 = {6}. Each such partition is determined by three one-element sets, i.e. by choice of ordered 3-tuple from π, and then by 4 positions of the set of remained three elements of π. Then the number of partitions in this case is equal to
6 β 5 β 4 β 4 = 480. Case 2). Two sets are two-elements and other two sets are one-elements, e.g.
π΄1 = {1,2}, π΄2 = {3,4}, π΄3 = {5}, π΄4 = {6}.First let us compute the number of partitions of π into sets of 1, 1, 2 and 2 elements. Indeed, the first 1-elements set can be chosen from 6 elements. To each choice of that elemetns set correspond 5 choices of the second 1-elements set. Then third 2-elements set is chosen from the remained 4 elements, so we can choose that set into πΆ42 =
4!2!β(4β2)!
= 6 ways. The fourth2-elements set is then also determined. Hence the number partitions of π into sets of 1, 1, 2 and 2 elements is equal to
6 β 5 β πΆ42 = 6 β 5 β 6 = 180Now we should take compute the number of distinct ``words'' obtained by permutations of 1122. This number is equal to the number of choices of two elements (say 1's) from 4-elements set, and so it is 6. Hence the total number of functions in the case 2 is
180 β 6 = 1080. Therefore the total number of all ways that the manager can assign these projects so that each employee is working on at least one project is equal to
480 + 1080 = 1560.
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Task 6. A survey of households in the U.S. reveals that 96% have at least one television set, 98% have cell phone service and 95% have a cell phone and at least one television set. What percentage of households in the U.S. has neither cell phone nor a television set? Solution. Let π» be the set of all households, π be the set of all households having at least one television set, and πΆ is the set of all households having cell phone service. For a subset π΄ β π denote by |π΄| the number of elements in π΄. Then by assumption |π| = 0.96|π|, |πΆ| = 0.98|π|, |π β© πΆ| = 0.95|π|. We should find the percentage of the set π\(π βͺ πΆ) in π. Notice that πΆ\π = πΆ\(π β© πΆ) = |πΆ| β |π β© πΆ| = (0.98 β 0.95)|π| = 0.03|π|. Thus the number of households having cell phone service but not television set constitutes 3% over all households. Hence π βͺ πΆ = π βͺ (πΆ\π) = (0.96 + 0.03)|π| = 0.99|π|, so the number of households having at least one television set or cell phone service constitute 99% over all households. Therefore percentage of households in the U.S. has neither cell phone nor a television set is 100 β 99 = 1%. Task 7. Let π΄ = {π, π, π, π} and π΅ = {1,2,3,4}. a) How many functions π: π΄ β π΅ are there?
b) How many functions π: π΄ β π΅ satisfy π(π) = 2? Solution. a) Notice that a function π: π΄ β π΅ can associate to each of 4 elements π΄ one of the 4 elements of π΅. To each of 4 choices of π β¦ π(π), correspond 4 choices of π β¦ π(π), so we get 4 β 4 = 42 choices of values for π and π. To each choice of (π(π), π(π)) correspond 4 choices of π(π), so we get 4 β 4 β 4 = 43 choices of values for π, π, and π. Similarly, there are 44 choices of values π(π), π(π), π(π), π(π). Thus the number of functions π: π΄ β π΅ is equal to 44 = 256. b) A function π: π΄ β π΅ satisfying π(π) = 2, is uniquely determined by its values at points π, π, π. In other words, the number of functions π: π΄ β π΅ with π(π) = 2 is equal to the number of functions π: {π, π, π} β π΅. Similarly to a) the number of such functions π: {π, π, π} β π΅ is equal to 43 = 64.
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Task 8. Use mathematical induction to prove the following proposition: π(π): 3 + 5 + 7 + β― + 2π + 1 = π(2 + π) where π = 1,2,3, β¦ Proof. Let π = 1. Then π(1) = 3, and π(2 + π) = 1 β (2 + 1) = 3 = π(1). Suppose that we proved that 3 + 5 + 7 + β― + 2π + 1 = π(2 + π) for all π β€ π. Let us prove this for π = π + 1, that is π(π): 3 + 5 + 7 + β― + 2(π + 1) + 1 = (π + 1)(2 + π + 1) = (π + 1)(π + 3) = π2 + 4π + 3. We have that
π(π + 1) = 3 + 5 + 7 + β― + 2π + 1 + 2(π + 1) + 1 = (3 + 5 + 7 + β― + 2π + 1) + 2(π + 1) + 1 = π(2 + π) + 2(π + 1) + 1 = 2π + π2 + 2π + 2 + 1 = π2 + 4π + 3.
Now by induction relation π(π): 3 + 5 + 7 + β― + 2π + 1 = π(2 + π) holds for all π β₯ 1.
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Task 9. Express the greatest common divisor of the following pair of integers as a linear combination of the integers: 117, 213 Solution. First we find prime decompositions of 117 and 213:
117 = 3 β 39 = 3 β 3 β 13 = 32 β 13,
213 = 3 β 71.
Thus πΊπΆπ·(117,213) = 3. We should find numbers π and π such that 117π + 213π = 3. Contracting by 3 we obtain 39π + 71π = 1. Notice that this identity modulo 39 and 71 means that 71π β‘ 1mod39, 39π β‘ 1mod71. Let Ο(π) be the Euler function which is equal to the number of numbers π such that 1 β€ π <π and πΊπΆπ·(π, π) = 1. Then by Euler theorem if πΊπΆπ·(π, π) = 1, then πΟ(π) β‘ 1modπ. It is known that Ο(π) = π β 1 for any prime π, and if πΊπΆπ·(π, π) = 1, then Ο(π β π) = Ο(π) β Ο(π). Hence Ο(71) = 71 β 1 = 70, and Ο(39) = Ο(3 β 13) = Ο(3) β Ο(13) = (3 β 1) β (13 β 1) = 2 β 12 = 24. Since πΊπΆπ·(71,39) = 1, it follows from Euler theorem that 71Ο(39) = 7124 = 1mod39 39Ο(71) = 3970 = 1mod71. Thus for solving equations 71π β‘ 1mod39, 39π β‘ 1mod71. we can put π = 7124β1 = 7123mod39 π = 3970β1 = 3969mod71. Then 71π = 71 β 7123 = 7124 = 1mod39 and similarly, 39π = 39 β 3969 = 3970 = 1mod71. Let us compute 7123mod39:
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π = 7123 β‘ (2 β 39 β 7)23 β‘ (β7)23 β‘ β7 β (72)11 β‘ β7 β 4911
β‘ β7 β (39 + 10)11 β‘ β7 β 1011 = β7 β 10 β 1005 = β70 β 1005
= β(2 β 39 β 8) β (3 β 39 β 17)5 β‘ β8 β 175 = β8 β 17 β (172)2 = β136 β 2892
= β(3 β 39 + 19) β (7 β 39 + 16)2 β‘ β19 β 16 β 2 = β19 β 256 = β19 β (39 β 6 + 22) β‘ β19 β 22 = β418 = β418 + 39 β 11 = 11mod39
Then 71π = 71 β 11 = 781 = 1 + 20 β 39 β‘ 1mod39. Hence 71 β 11 β 39 β 20 = 1. Multiplying by 3 both parts of this identity we obtain: 213 β 11 β 117 β 20 = 3 which gives the required expression of πΊπΆπ·(117,213) = 3 as a linear combination of 117 and 213.
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