Types of relations

Correct answer: A. Symmetric and transitive

Step 1: The empty relation has no pair, so no pair violates symmetry. Step 2: There are no linked pairs to violate transitivity. Step 3: But on a non-empty set, it is not reflexive because self-pairs are missing.

Correct answer: A. Reflexive, symmetric and transitive

Step 1: The universal relation contains all pairs of (A\times A). Step 2: Hence self-pairs, reverse pairs and pairs required for transitivity are all available. Step 3: Since all possible pairs are present, all three properties hold.

Correct answer: A. When the set has exactly one element

Step 1: A one-element set has only one possible pair ((a,a)). Step 2: The same pair appears in both the identity and universal relations. Step 3: With two or more elements, the universal relation has extra non-self pairs.

Correct answer: A. (aRc)

Step 1: An equivalence relation is transitive. Step 2: From (aRb) and (bRc), (aRc) must follow. Step 3: In an equivalence relation, linked elements belong to the same class.

Correct answer: A. Symmetry

Step 1: Symmetry is included in an equivalence relation. Step 2: Symmetry says that if (aRb), then (bRa). Step 3: When the reverse relation is asked, identify symmetry.

Correct answer: A. Reflexivity

Step 1: Reflexivity is the first required property of an equivalence relation. Step 2: Reflexivity means every element is related to itself. Step 3: When self-pairs are involved, recall reflexivity immediately.

Correct answer: A. Symmetry

Step 1: A partial order requires reflexivity. Step 2: It also requires antisymmetry and transitivity. Step 3: Symmetry belongs to equivalence relations, not as a required condition for partial order.

Correct answer: A. Antisymmetry

Step 1: An equivalence relation needs reflexivity, symmetry and transitivity. Step 2: Antisymmetry is not a required condition for equivalence. Step 3: Remember the conditions of equivalence and partial order separately.

Correct answer: A. Equivalence relation

Step 1: Every number has the same square as itself, so it is reflexive. Step 2: If (a^2=b^2), then (b^2=a^2), so it is symmetric. Step 3: Equality of squares transfers to a third number, so it is an equivalence relation.

Correct answer: A. Reflexive and transitive, but not antisymmetric

Step 1: (|a|\le |a|) is always true, so it is reflexive. Step 2: If (|a|\le |b|) and (|b|\le |c|), then (|a|\le |c|), so it is transitive. Step 3: (1) and (-1) are different but can be related both ways, so antisymmetry fails.

Correct answer: A. Because antisymmetry fails

Step 1: (a^2\le a^2) gives reflexivity. Step 2: Comparison of squares is transitive. Step 3: For (1) and (-1), relation holds both ways but (1\ne -1), so antisymmetry fails.

Correct answer: A. It is reflexive and transitive

Step 1: All self-pairs are present, so the relation is reflexive. Step 2: ((1,2)) or ((1,3)) does not create any missing linked conclusion. Step 3: Reverse pairs are absent, so it is not symmetric, but it can be transitive.

Correct answer: A. Yes

Step 1: All self-pairs are present, so it is reflexive. Step 2: No reverse pair for unequal elements appears, so it is antisymmetric. Step 3: No required linked pair is missing, so it is also transitive.

Correct answer: A. (<) on real numbers

Step 1: (a<a) is false, so (<) is not reflexive. Step 2: If (a<b), then (b<a) is false, so symmetry fails and antisymmetry is not violated. Step 3: (a<b) and (b<c) imply (a<c), so it is transitive.

Correct answer: A. Identity relation

Step 1: Identity relation has every self-pair, so it is reflexive. Step 2: Self-pairs are their own reverses, so it is symmetric. Step 3: There are no reverse pairs for unequal elements and transitivity also holds.

Correct answer: A. It will be both symmetric and antisymmetric

Step 1: The reverse of a self-pair is the same pair, so symmetry holds. Step 2: There are no two-way pairs between unequal elements, so antisymmetry also holds. Step 3: Relations with only self-pairs can satisfy both properties.

Correct answer: A. ((2,1)) and ((3,2))

Step 1: Symmetry gives the reverse of each pair. Step 2: ((1,2)) gives ((2,1)), and ((2,3)) gives ((3,2)). Step 3: ((1,3)) may come from transitivity, not symmetry.

Correct answer: A. ((1,4))

Step 1: From ((1,2)) and ((2,3)), we get ((1,3)). Step 2: Then from ((1,3)) and ((3,4)), we get ((1,4)). Step 3: Apply transitivity step by step in a chain.

Correct answer: A. ({a,b,c})

Step 1: In equivalence classes, all elements of the same class share the same group. Step 2: Since (b) is in the class of (a), the class of (b) is the same. Step 3: Equivalence classes are either identical or disjoint.

Correct answer: A. They are either identical or disjoint

Step 1: An equivalence relation divides a set into separate classes. Step 2: An element cannot belong partly to two different classes. Step 3: Hence two classes are either identical or disjoint.

Correct answer: A. Equivalence relation

Step 1: An equivalence relation divides a set into equivalence classes. Step 2: These classes together cover the set and remain disjoint. Step 3: When you see a partition, think of an equivalence relation.

Correct answer: A. Total order relation

Step 1: In a partial order, not every pair of elements must be comparable. Step 2: When every two elements are comparable, the partial order becomes a total order. Step 3: Relations like (\le) are common examples of total order.

Correct answer: A. Total order relation

Step 1: (\le) is reflexive because (a\le a). Step 2: It is also antisymmetric and transitive. Step 3: Any two natural numbers are comparable under (\le), so it is a total order.

Correct answer: A. Because ({1}) and ({2}) are not comparable

Step 1: In a total order, every two elements must be comparable. Step 2: ({1}\subseteq{2}) is false and ({2}\subseteq{1}) is also false. Step 3: One incomparable pair is enough to stop it from being a total order.

Correct answer: A. Check reflexivity, symmetry or antisymmetry, and transitivity separately

Step 1: Check reflexivity using self-pairs. Step 2: Check symmetry or antisymmetry using reverse pairs. Step 3: Check transitivity using linked pairs before deciding the final type.