Reflexive relation

Step 1: The smallest reflexive relation has (5) self-pairs. Step 2: The universal relation has (5^2=25) pairs. Step 3: The difference is (25-5=20).

Step 1: A reflexive relation on a set with (3) elements must have (3) self-pairs. Step 2: Total pairs are (5), so extra pairs are (5-3=2). Step 3: Subtract compulsory self-pairs from the total.

Step 1: (A\times A) has (4^2=16) pairs. Step 2: To be not reflexive, at least one self-pair must be absent. Step 3: The maximum possible number is (16-1=15).

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.