Reflexive relation

Correct answer: A. (R) is reflexive

Step 1: For any two numbers, (|a-b|) is never negative. Step 2: For a self-pair, (|a-a|=0), which satisfies (0\ge 0). Step 3: A condition true for every self-pair gives reflexivity.

Correct answer: A. In a self-pair, (|a-a|=0)

Step 1: For reflexivity, check ((a,a)). Step 2: (|a-a|=0), but the condition requires (>0). Step 3: So no self-pair belongs to this relation.

Correct answer: A. Reflexive

Step 1: Every number has the same parity as itself. Step 2: Therefore ((1,1),(2,2),(3,3)) all belong to the relation. Step 3: In relations based on a shared property, check the self-pair first.

Correct answer: A. For every (a), (a-a=0), which is divisible by (3)

Step 1: (a\equiv b \pmod{3}) means (a-b) is divisible by (3). Step 2: For a self-pair, (a-a=0), which is divisible by (3). Step 3: Hence every ((a,a)) belongs to the relation.

Correct answer: A. Both are reflexive

Step 1: In (R), putting ((a,a)) gives (a+a=2a), divisible by (2). Step 2: In (S), the condition (a=b) gives all self-pairs. Step 3: Therefore both relations are reflexive.

Correct answer: A. It is always true

Step 1: (I_A) contains all self-pairs of (A). Step 2: Since (R) is reflexive, all these pairs are also in (R). Step 3: Thus the identity relation is a subset of any reflexive relation.

Correct answer: A. (S) will be reflexive

Step 1: (R) contains all self-pairs. Step 2: Since (R\subseteq S), every pair of (R) is also in (S). Step 3: Hence (S) also contains all self-pairs and is reflexive.

Correct answer: B. Not reflexive

Step 1: (R) had all self-pairs. Step 2: In forming (T), ((2,2)) is removed. Step 3: Removing one required self-pair makes the relation not reflexive.

Correct answer: A. Yes, because (a-a=0) and (0) is divisible by (4)

Step 1: For reflexivity, only the condition on ((a,a)) needs to be checked. Step 2: For every (a), (a-a=0), and (0) is divisible by every positive integer. Step 3: Other unequal pairs may fail, but reflexivity can still hold.

Correct answer: A. (R) is reflexive

Step 1: Check the condition on a self-pair. Step 2: Putting (b=a) gives (a^2+a^2=2a^2), which is true. Step 3: Thus every ((a,a)) is in the relation, so (R) is reflexive.

Correct answer: A. It is reflexive

Step 1: ((a-b)^2=0) happens only when (a=b). Step 2: So ((1,1),(2,2),(3,3)) are all included. Step 3: The identity relation is reflexive.

Correct answer: A. Because (a^2-a^2=0) for every (a)

Step 1: To check reflexivity, put (b=a). Step 2: Then (a^2-a^2=0), which is true for every element. Step 3: Equality-type conditions are satisfied when an element is compared with itself.

Correct answer: A. (R) is reflexive

Step 1: Every number with itself satisfies either both even or both odd. Step 2: Therefore all self-pairs of (1,2,3,4) are in (R). Step 3: A same-parity relation is reflexive.

Correct answer: A. Not reflexive

Step 1: Check the condition on self-pairs. Step 2: For ((3,3)), (3+3=6), which is not less than or equal to (5). Step 3: If one self-pair fails, the relation is not reflexive.

Correct answer: A. (R^c) is not reflexive

Step 1: (A\times A) has ((1,1),(1,2),(2,1),(2,2)). Step 2: (R^c) contains only ((2,1)), because the self-pairs are already in (R). Step 3: Hence the complement relation is not reflexive.

Correct answer: A. (R^c) is not reflexive

Step 1: The complement of (A\times A) inside (A\times A) is the empty relation. Step 2: On a non-empty (A), the empty relation has no self-pair. Step 3: Therefore it is not reflexive.

Correct answer: A. ((2,2))

Step 1: A number compared with itself leaves the same remainder. Step 2: Hence every self-pair is in the relation, including ((2,2)). Step 3: In same-remainder questions, self-pairs are guaranteed.

Correct answer: A. Not reflexive

Step 1: In a self-pair, the sum is (2a). Step 2: For (a=1), (2a=2), which is not divisible by (3). Step 3: One failed element is enough to disprove reflexivity.

Correct answer: A. (3)

Step 1: The self-pairs are ((1,1),(2,2),(3,3)). Step 2: Each has sum (2a), which is divisible by (2). Step 3: Therefore all three self-pairs belong to (R).

Correct answer: A. Reflexive

Step 1: The condition (a=b) includes every self-pair. Step 2: Thus ((1,1),(2,2),(3,3)) are all in (R). Step 3: The second condition may add pairs, but reflexivity remains safe.

Correct answer: A. Not reflexive

Step 1: For ((1,1)), (a\ne b) is false and (a=2) is also false. Step 2: So ((1,1)) does not belong to (R). Step 3: Missing one self-pair makes the relation not reflexive.

Correct answer: A. Yes

Step 1: For a self-pair, put (b=a). Step 2: The condition becomes (a\le a+1), which is true for every (a). Step 3: In such questions, substituting (b=a) is the direct method.

Correct answer: A. (R) is not reflexive

Step 1: For a self-pair, put (b=a). Step 2: The condition becomes (a+1\le a), which is false for every number. Step 3: So no self-pair is included and (R) is not reflexive.

Correct answer: A. Reflexive

Step 1: All elements of (A) are positive. Step 2: For a self-pair, (ab=a^2), which is greater than (0) for every (a\in A). Step 3: In product conditions, check (a^2) for reflexivity.

Correct answer: A. For ((0,0)), (0\cdot 0>0) is not true

Step 1: Reflexivity requires the self-pair of every element. Step 2: For (0), the pair ((0,0)) gives product (0), which is not greater than (0). Step 3: Be careful with product conditions when zero is in the set.