Let R be the relation on the set of ordered pairs of positive integers such that ((a, b), (c, d)) E R if and only if ad = bc. Arrange the proof of the given statement in correct order to show that R is an equivalence relation. (Prove the given relation is reflexive first, and then symmetric and transitive.) Rank the options below. Hence, R is reflexive. If ((a, b), (c, d)) E Rand ((c, d), (e, f)) Є R, then ad = bc and cf=de. Hence, Ris transitive. 6 7 3 Multiplying these equations gives acdf = bcde, and since all these numbers are nonzero, we have af= be, so ((a, b), (e, f)) Є R. 2 If ((a, b), (c, d)) E R then ad = bc, which also means that cb = da, so ((c, d), (a, b)) E R. Hence, R is symmetric. ((a, b), (a, b)) E R because ab = ba. Since R is reflexive, symmetric, and transitive, it is an equivalence relation. 1 ☑ × X ☑ 00 8 ☑ 5 4 ☑ Identify whether the following collections of subsets are partitions of S = {−3,-2,-1, 0, 1, 2, 3] and the correct reason for it. {−3, 3}, {−2, 2}, {−1, 1}, {0} Multiple Choice The given collection of sets forms a partition of S as these sets are not mutually disjoint and their union is S. The given collection of sets does not form a partition of S as these sets are not mutually disjoint. The given collection of sets does not form a partition of S as the union of these sets is not S. The given collection of sets forms a partition of S as these sets are mutually disjoint and their union is S.

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Chapter2: Second-order Linear Odes
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Let R be the relation on the set of ordered pairs of positive integers such that ((a, b), (c, d)) E R if and only if ad = bc. Arrange the proof of the given statement in correct order to show that R is an
equivalence relation.
(Prove the given relation is reflexive first, and then symmetric and transitive.)
Rank the options below.
Hence, R is reflexive.
If ((a, b), (c, d)) E Rand ((c, d), (e, f)) Є R, then ad = bc and cf=de.
Hence, Ris transitive.
6
7
3
Multiplying these equations gives acdf = bcde, and since all these numbers are nonzero, we have af= be, so ((a, b), (e, f)) Є R.
2
If ((a, b), (c, d)) E R then ad = bc, which also means that cb = da, so ((c, d), (a, b)) E R.
Hence, R is symmetric.
((a, b), (a, b)) E R because ab = ba.
Since R is reflexive, symmetric, and transitive, it is an equivalence relation.
1
☑
×
X
☑
00
8
☑
5
4
☑
Transcribed Image Text:Let R be the relation on the set of ordered pairs of positive integers such that ((a, b), (c, d)) E R if and only if ad = bc. Arrange the proof of the given statement in correct order to show that R is an equivalence relation. (Prove the given relation is reflexive first, and then symmetric and transitive.) Rank the options below. Hence, R is reflexive. If ((a, b), (c, d)) E Rand ((c, d), (e, f)) Є R, then ad = bc and cf=de. Hence, Ris transitive. 6 7 3 Multiplying these equations gives acdf = bcde, and since all these numbers are nonzero, we have af= be, so ((a, b), (e, f)) Є R. 2 If ((a, b), (c, d)) E R then ad = bc, which also means that cb = da, so ((c, d), (a, b)) E R. Hence, R is symmetric. ((a, b), (a, b)) E R because ab = ba. Since R is reflexive, symmetric, and transitive, it is an equivalence relation. 1 ☑ × X ☑ 00 8 ☑ 5 4 ☑
Identify whether the following collections of subsets are partitions of S = {−3,-2,-1, 0, 1, 2, 3] and the correct reason for it.
{−3, 3}, {−2, 2}, {−1, 1}, {0}
Multiple Choice
The given collection of sets forms a partition of S as these sets are not mutually disjoint and their union is S.
The given collection of sets does not form a partition of S as these sets are not mutually disjoint.
The given collection of sets does not form a partition of S as the union of these sets is not S.
The given collection of sets forms a partition of S as these sets are mutually disjoint and their union is S.
Transcribed Image Text:Identify whether the following collections of subsets are partitions of S = {−3,-2,-1, 0, 1, 2, 3] and the correct reason for it. {−3, 3}, {−2, 2}, {−1, 1}, {0} Multiple Choice The given collection of sets forms a partition of S as these sets are not mutually disjoint and their union is S. The given collection of sets does not form a partition of S as these sets are not mutually disjoint. The given collection of sets does not form a partition of S as the union of these sets is not S. The given collection of sets forms a partition of S as these sets are mutually disjoint and their union is S.
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