Types of relations

Correct answer: A. Equivalence relation

Step 1: (a-a=0) is rational, so it is reflexive. Step 2: If (a-b) is rational, then (b-a) is rational, so it is symmetric. Step 3: The sum of two rational differences is rational, so transitivity holds.

Correct answer: A. Equivalence relation

Step 1: For every (a), (a^2=a^2), so the relation is reflexive. Step 2: From (a^2=b^2), we get (b^2=a^2), so it is symmetric. Step 3: If (a^2=b^2) and (b^2=c^2), then (a^2=c^2), so it is transitive.

Correct answer: A. Equivalence relation

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

Correct answer: A. Transitive but neither reflexive nor symmetric

Step 1: (a-a=0), so (aRa) does not hold and the relation is not reflexive. Step 2: If (a-b>0), then (b-a>0) cannot hold, so it is not symmetric. Step 3: From (a>b) and (b>c), we get (a>c), so it is transitive.

Correct answer: A. Symmetric but not reflexive and not transitive

Step 1: For ((2,2)), the greatest common divisor is (2), so it is not reflexive. Step 2: The greatest common divisor does not change when order is changed, so the relation is symmetric. Step 3: (2) is coprime to (3), and (3) is coprime to (4), but (2) is not coprime to (4), so transitivity fails.

Correct answer: A. Reflexive and transitive but not symmetric

Step 1: Every number divides itself, so it is reflexive. Step 2: In divisibility, (a\mid b) and (b\mid c) imply (a\mid c), so it is transitive. Step 3: (2\mid 4) is true but (4\mid 2) is false, so it is not symmetric.

Correct answer: A. Reflexive and transitive but not symmetric

Step 1: All self-pairs are given, so the relation is reflexive. Step 2: ((1,2)) exists but ((2,1)) does not, so it is not symmetric. Step 3: Required resulting pairs such as ((1,3),(2,4),(1,4)) are present, so transitivity holds.

Correct answer: A. Equivalence relation

Step 1: Every student has the same birth month as themselves, so it is reflexive. Step 2: If (a)'s month is the same as (b)'s, then (b)'s is the same as (a)'s, so it is symmetric. Step 3: Same birth month passes through a third student, so transitivity holds.

Correct answer: A. Transitive but neither reflexive nor symmetric

Step 1: A person cannot be taller than themselves, so it is not reflexive. Step 2: If (a) is taller than (b), then (b) is not taller than (a), so it is not symmetric. Step 3: If (a) is taller than (b) and (b) is taller than (c), then (a) is taller than (c), so it is transitive.

Correct answer: A. Symmetric but not reflexive and not transitive

Step 1: (a\neq a) is false, so no self-pair exists and the relation is not reflexive. Step 2: If (a\neq b), then (b\neq a), so it is symmetric. Step 3: (1\neq 2) and (2\neq 1) hold but (1\neq 1) does not, so transitivity fails.

Correct answer: A. Symmetric but neither reflexive nor transitive

Step 1: For ((3,3)), the sum is (6), so it is not reflexive. Step 2: In (a+b\le 5), changing the order does not change the sum, so it is symmetric. Step 3: ((4,1)) and ((1,4)) exist but ((4,4)) does not, so it is not transitive.

Correct answer: A. It is an equivalence relation

Step 1: (a-a=0) is divisible by (2), so it is reflexive. Step 2: If (a-b) is divisible by (2), then (b-a) is also divisible by (2), so it is symmetric. Step 3: The sum of divisible differences is divisible, so transitivity holds.

Correct answer: A. Equivalence relation

Step 1: All self-pairs are present, so it is reflexive. Step 2: ((1,2)) has ((2,1)) and ((3,4)) has ((4,3)), so it is symmetric. Step 3: Within the groups ({1,2}) and ({3,4}), required pairs are complete, so transitivity holds.

Correct answer: A. No, for example on (A={1,2}), ({(1,1)}) is symmetric and transitive but not reflexive

Step 1: Being symmetric and transitive does not automatically give all self-pairs. Step 2: On (A={1,2}), ({(1,1)}) misses ((2,2)), so it is not reflexive. Step 3: For such theory questions, a small counterexample is the fastest method.

Correct answer: A. No, on (A={1,2,3}), ({(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2)}) is a counterexample

Step 1: Reflexive and symmetric together are not sufficient. Step 2: In the given example, ((1,2)) and ((2,3)) exist but ((1,3)) is missing. Step 3: To prove an equivalence relation, all three properties must be checked separately.

Correct answer: A. No, the (\le) relation is a counterexample

Step 1: In (\le), every number is related to itself, so it is reflexive. Step 2: (a\le b) and (b\le c) imply (a\le c), so it is transitive. Step 3: (2\le 3) is true but (3\le 2) is false, so symmetry need not hold.

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

Step 1: First check all self-pairs; option A has all three. Step 2: Option A contains both ((1,2)) and ((2,1)), so symmetry holds. Step 3: The available chains do not break transitivity, so option A is an equivalence relation.

Correct answer: A. Equivalence relation

Step 1: Every number has the same remainder as itself, so it is reflexive. Step 2: Same remainder is unaffected by changing order, so it is symmetric. Step 3: If (a) and (b) have the same remainder and (b) and (c) have the same remainder, then (a) and (c) also do.

Correct answer: A. ({1,3,5}) and ({2,4,6})

Step 1: Same parity means odd numbers together and even numbers together. Step 2: (1,3,5) are odd and (2,4,6) are even. Step 3: While forming equivalence classes, each element must appear in exactly one class.

Correct answer: A. It is reflexive and symmetric but not transitive

Step 1: Because (a=b) is allowed, all self-pairs are present, so it is reflexive. Step 2: The condition (a+b=5) is unchanged by reversing order, so it is symmetric. Step 3: The relation splits into the blocks ({1,4}) and ({2,3}), so it is actually transitive and hence an equivalence relation.

Correct answer: A. It is an equivalence relation

Step 1: (a=b) gives all self-pairs, so the relation is reflexive. Step 2: If (a+b=5), then (b+a=5), so it is symmetric. Step 3: It forms closed blocks ({1,4}) and ({2,3}), so it is also transitive.

Correct answer: A. Reflexive and symmetric but not transitive

Step 1: (|a-a|=0\le 1), so it is reflexive. Step 2: (|a-b|=|b-a|), so it is symmetric. Step 3: ((1,2)) and ((2,3)) exist, but ((1,3)) does not because the difference is (2), so it is not transitive.

Correct answer: A. Equivalence relation

Step 1: (a=b) gives all self-pairs, so it is reflexive. Step 2: The condition that (a+b) is even remains the same after reversing order, so it is symmetric. Step 3: This is essentially the same-parity relation, which is also transitive.

Correct answer: A. It is an equivalence relation

Step 1: All four self-pairs are present, so it is reflexive. Step 2: Within ({1,2,3}), all directed pairs are present, and (4) has its self-pair, so symmetry holds. Step 3: ({1,2,3}) is a closed complete block and (4) is a single class, so transitivity also holds.

Correct answer: A. Reflexive and transitive but not symmetric

Step 1: All four self-pairs are present, so the relation is reflexive. Step 2: ((1,2)) is present but ((2,1)) is not, so it is not symmetric. Step 3: From ((1,2)) and ((2,3)), ((1,3)) is required and it is present, so transitivity holds.