Reflexive relation

Step 1: (R) has all self-pairs, but (T) may be a smaller part of (R). Step 2: Some self-pairs may be removed while forming (T). Step 3: Reflexivity is not automatically preserved in sub-relations.

Step 1: (R) already contains all self-pairs. Step 2: Adding an extra pair does not remove these required pairs. Step 3: Reflexivity is not destroyed by adding pairs within (A\times A).

Step 1: The identity relation contains all self-pairs. Step 2: Removing ((3,3)) means (3) is not related to itself. Step 3: Removing even one compulsory pair destroys reflexivity.

Step 1: (A\times A) contains all self-pairs. Step 2: Removing ((2,2)) removes one required pair. Step 3: If a self-pair is removed from the universal relation, reflexivity fails.

Step 1: The removed pairs ((1,2)) and ((2,3)) are not self-pairs. Step 2: ((1,1),(2,2),(3,3)) still remain in the relation. Step 3: Removing only non-diagonal pairs does not break reflexivity.

Step 1: The minimum reflexive relation has 4 pairs. Step 2: The universal relation (A\times A) has (4^2=16) pairs. Step 3: The difference is (16-4=12).

Step 1: The number of all relations is (2^{25}). Step 2: The number of reflexive relations is (2^{25-5}=2^{20}). Step 3: The ratio is (2^{25}:2^{20}=2^5:1).

Step 1: Total relations are (2^9). Step 2: Reflexive relations are (2^{9-3}=2^6). Step 3: The fraction is (\frac{2^6}{2^9}=\frac{1}{8}).

Step 1: (A) has 3 elements, so 3 self-pairs are compulsory. Step 2: Out of the 4 pairs, 3 must be ((1,1),(2,2),(3,3)). Step 3: The remaining one may be any non-diagonal pair.

Step 1: The 3 self-pairs are compulsory. Step 2: To have exactly 4 pairs, choose 1 pair from the 6 non-diagonal pairs. Step 3: The number is (\binom{6}{1}=6).

Step 1: The 3 self-pairs are fixed for reflexivity. Step 2: To have 5 total pairs, choose 2 from the 6 non-diagonal pairs. Step 3: The number is (\binom{6}{2}=15).

Step 1: The 4 self-pairs are compulsory. Step 2: For 6 total pairs, choose 2 from (16-4=12) non-diagonal pairs. Step 3: The count is (\binom{12}{2}=66).

Step 1: For reflexivity, put (b=a). Step 2: Then (a^2=b^2) becomes (a^2=a^2), which is true. Step 3: In equality-based rules, test equality of an element with itself.

Step 1: Reflexivity requires ((a,a)) for every real (a). Step 2: Taking (a=1), (1^2+1^2=2), not 0. Step 3: One counterexample is enough to show it is not reflexive.

Step 1: Check each element with itself. Step 2: ((-1)^2=(-1)^2), (0^2=0^2), and (1^2=1^2) are all true. Step 3: Extra pairs like ((-1,1)) may exist, but reflexivity depends on self-pairs.

Step 1: Reflexivity needs all self-pairs. Step 2: ((2,2)) satisfies the rule, but ((1,1)) and ((3,3)) do not. Step 3: Having some self-pairs is not enough; all are needed.

Step 1: For self-pairs, check (a+a). Step 2: The largest element in (A) is 4, so the maximum self-sum is (4+4=8), which satisfies the rule. Step 3: In finite sets, checking the largest self-sum is useful.

Step 1: A reflexive relation also needs ((4,4)). Step 2: For ((4,4)), (4+4=8), but the rule requires (8<8), which is false. Step 3: In strict inequalities, boundary self-pairs may be missing.

Step 1: When a number is compared with itself, it remains the same. Step 2: Therefore any (a) has the same parity as itself. Step 3: In relations based on the same property, self-check is usually simple.

Step 1: For reflexivity, put (b=a). Step 2: (|a-a|=0), and (0) is even. Step 3: In parity-of-difference relations, remember the zero difference.

Step 1: Reflexivity requires every ((a,a)). Step 2: (|a-a|=0), and 0 is not odd. Step 3: If the rule requires an odd difference, self-pairs are absent.

Step 1: Check (a+a) for self-pairs. Step 2: The smallest (a) here is 1, so the minimum self-sum is 2. Step 3: Since (2\ge2) is true, all self-pairs are included.

Step 1: Test the rule on ((1,1)). Step 2: (1+1=2), and (2>2) is false. Step 3: Missing one required self-pair makes the relation not reflexive.

Step 1: A reflexive relation (R) contains all self-pairs. Step 2: The identity relation (I_A) is exactly the set of all self-pairs. Step 3: Therefore (I_A\subseteq R) is always true.

Step 1: (I_A) contains ((1,1),(2,2),(3,3)). Step 2: The given (R) does not contain ((3,3)), so (I_A\subseteq R) is false. Step 3: Every pair of the identity relation must be present.