Types of relations

Correct answer: A. ({(1,1),(2,2),(3,3),(1,2),(2,1)})

Step 1: The first option has all self-pairs, so it is reflexive. Step 2: ((1,2)) is paired with ((2,1)), so it is symmetric. Step 3: The required linked-pair conditions are also satisfied, so it is an equivalence relation.

Correct answer: A. Equivalence relation

Step 1: All four self-pairs are present, so the relation is reflexive. Step 2: ((1,2)) has ((2,1)), and ((3,4)) has ((4,3)), so it is symmetric. Step 3: Transitivity is satisfied inside each small group, so it is an equivalence relation.

Correct answer: A. Equivalence relation

Step 1: (a-a=0), which is divisible by (3), so reflexivity holds. Step 2: If (a-b) is divisible by (3), then (b-a) is also divisible by (3), so symmetry holds. Step 3: The sum of such divisible differences remains divisible by (3), so it is an equivalence relation.

Correct answer: A. Reflexive and transitive

Step 1: Every number is a multiple of itself, so it is reflexive. Step 2: If (a) is a multiple of (b) and (b) is a multiple of (c), then (a) is a multiple of (c). Step 3: The multiple relation is generally not symmetric, for example (6) is a multiple of (3), but (3) is not a multiple of (6).

Correct answer: A. It is an equivalence relation

Step 1: All self-pairs are present, so it is reflexive. Step 2: Both ((1,3)) and ((3,1)) are present, so it is symmetric. Step 3: ((1,3)) and ((3,1)) require ((1,1)) and ((3,3)), which are present, so it is transitive.

Correct answer: A. ({(1,2),(2,1)}) on ({1,2})

Step 1: In the first option, both ((1,2)) and ((2,1)) are present, so it is symmetric. Step 2: But ((1,2)) and ((2,1)) require ((1,1)), which is missing. Step 3: Reverse pairs give symmetry, but not automatic transitivity.

Correct answer: A. ({(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2)}) on ({1,2,3})

Step 1: The first option has all self-pairs and reverse pairs for every non-self pair. Step 2: ((1,2)) and ((2,3)) require ((1,3)), which is missing. Step 3: An equivalence relation needs all three properties, not just two.

Correct answer: A. (R) is reflexive, symmetric and transitive

Step 1: An equivalence relation is defined by three properties. Step 2: These are reflexivity, symmetry and transitivity. Step 3: Whenever a question says equivalence, immediately apply all three properties.

Correct answer: A. ((2,3)) is missing, so transitivity fails

Step 1: The relation looks reflexive and symmetric. Step 2: ((2,1)) and ((1,3)) require ((2,3)), but it is missing. Step 3: For equivalence relation, full transitivity checking is necessary.

Correct answer: B. (2)

Step 1: When divided by (2), only two remainders are possible: (0) and (1). Step 2: (2,4) form one class and (1,3) form another. Step 3: In same-remainder relations, the number of classes is linked to possible remainders.

Correct answer: A. Equivalence relation

Step 1: Every number has the same remainder as itself, so reflexivity holds. Step 2: If (a) and (b) have the same remainder, then (b) and (a) also do. Step 3: The same-remainder link passes to a third number, so it is an equivalence relation.

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

Step 1: (1) and (2) are related to each other, so they belong to the same class. Step 2: (3) is related only to itself, so it forms a separate class. Step 3: While forming equivalence classes, group elements that are mutually related.

Correct answer: A. Relation defined by (\le)

Step 1: A partial order needs reflexivity, antisymmetry and transitivity. Step 2: The relation (\le) satisfies all three. Step 3: Remember the difference: equivalence uses symmetry, partial order uses antisymmetry.

Correct answer: A. (a=b)

Step 1: In antisymmetry, two reverse pairs can both exist only when the two elements are equal. Step 2: So ((a,b)) and ((b,a)) imply (a=b). Step 3: Do not treat antisymmetry as simply the opposite of symmetry.

Correct answer: A. Yes

Step 1: Antisymmetry fails when both ((a,b)) and ((b,a)) occur for (a\ne b). Step 2: Here ((1,2)) is present but ((2,1)) is not. Step 3: If no reverse non-self pair pair exists, antisymmetry holds.

Correct answer: B. No, because ((1,2)) and ((2,1)) both exist while (1\ne2)

Step 1: In antisymmetry, reverse pairs for unequal elements must not both be present. Step 2: Here (1\ne2), and both ((1,2)) and ((2,1)) are present. Step 3: This is a direct violation of antisymmetry.

Correct answer: A. Partial order relation

Step 1: A partial order relation has three conditions. Step 2: These are reflexivity, antisymmetry and transitivity. Step 3: Equivalence uses symmetry, while partial order uses antisymmetry.

Correct answer: A. Partial order relation

Step 1: Every positive integer divides itself, so reflexivity holds. Step 2: If (a\mid b) and (b\mid a), then for positive integers (a=b), so antisymmetry holds. Step 3: If (a\mid b) and (b\mid c), then (a\mid c), so it is a partial order.

Correct answer: A. Partial order relation

Step 1: Every set is a subset of itself, so it is reflexive. Step 2: If (A\subseteq B) and (B\subseteq A), then (A=B), so it is antisymmetric. Step 3: If (A\subseteq B) and (B\subseteq C), then (A\subseteq C), so it is a partial order.

Correct answer: A. Equivalence relation

Step 1: (a^2=a^2), so reflexivity holds. Step 2: If (a^2=b^2), then (b^2=a^2), so symmetry holds. Step 3: If (a^2=b^2) and (b^2=c^2), then (a^2=c^2), so transitivity holds.

Correct answer: A. Equivalence relation

Step 1: (|a|=|a|), so it is reflexive. Step 2: If (|a|=|b|), then (|b|=|a|), so it is symmetric. Step 3: Equality of absolute values transfers through a third number, so it is an equivalence relation.

Correct answer: A. Equivalence relation

Step 1: For every (a), (a=a), so reflexivity holds. Step 2: If (a=b) or (a=-b), the reverse relation also holds. Step 3: This relation is like (|a|=|b|), so transitivity also holds and it is an equivalence relation.

Correct answer: A. Transitive

Step 1: (a-b>0) means (a>b). Step 2: If (a>b) and (b>c), then (a>c), so it is transitive. Step 3: It is not reflexive or symmetric because no number is greater than itself.

Correct answer: A. Partial order relation

Step 1: (a-b\ge0) means (a\ge b). Step 2: The relation (\ge) is reflexive, antisymmetric and transitive. Step 3: Do not call it equivalence because it is generally not symmetric.

Correct answer: A. (<) on real numbers

Step 1: (a<a) is false, so (<) is not reflexive. Step 2: (a<b) does not imply (b<a), so it is not symmetric. Step 3: (a<b) and (b<c) imply (a<c), so it is transitive.