Reflexive relation

Step 1: A reflexive relation on a four-element set must contain (4) diagonal pairs. Step 2: Since the total number of pairs is (10), the remaining pairs are (10-4=6). Step 3: Always subtract compulsory diagonal pairs first.

Step 1: (A\times A) has (16) pairs and (4) diagonal pairs are compulsory. Step 2: To get exactly (10) pairs, choose (6) more from the (12) non-diagonal pairs. Step 3: The count is (\binom{12}{6}).

Step 1: For five elements, (5) diagonal pairs are compulsory. Step 2: A relation with (9) pairs needs (9-5=4) additional non-diagonal pairs. Step 3: There are (25-5=20) non-diagonal pairs, so the answer is (\binom{20}{4}).

Step 1: The (3) diagonal pairs are compulsory. Step 2: To have (5) total pairs, choose (2) additional pairs. Step 3: There are (6) non-diagonal pairs, so the number of ways is (\binom{6}{2}).

Step 1: A reflexive relation must include every element paired with itself. Step 2: For four elements, the minimum required pairs are ((1,1),(2,2),(3,3),(4,4)). Step 3: The minimum count equals the number of elements in the set.

Step 1: Reflexivity needs all five diagonal pairs. Step 2: Since the relation has no diagonal pair, all five must be added. Step 3: Having many non-diagonal pairs does not compensate for missing diagonal pairs.

Step 1: For ((a,a)), the sum is (a+a=2a). Step 2: (2a) is always even, not odd, so no diagonal pair belongs to (R). Step 3: Parity of the diagonal sum gives a quick reflexivity check.

Step 1: To test reflexivity, put (b=a). Step 2: For every (a), (a\equiv a \pmod{2}) is true. Step 3: Congruence with the same element always includes diagonal pairs.

Step 1: On the diagonal, put (b=a), giving (a\equiv a+1 \pmod{2}). Step 2: A number and the next number do not have the same parity. Step 3: If the diagonal condition fails, the relation is not reflexive.

Step 1: For reflexivity, substitute ((a,a)) into the condition. Step 2: This gives (a^2\equiv a^2 \pmod{3}), which is always true by equality. Step 3: Even in power and modulo relations, test the same element first.

Step 1: On the diagonal, (a^2+b^2=2a^2). Step 2: (2a^2) is always even, so every ((a,a)) belongs to the relation. Step 3: Simplifying the expression on the diagonal is the fastest method.

Step 1: For ((a,a)), (a^2-a^2=0). Step 2: (0) is not odd, so no diagonal pair satisfies the condition. Step 3: When the diagonal difference is zero, an oddness condition fails immediately.

Step 1: Simplifying (a+b=2a) gives (b=a). Step 2: Thus (R) contains exactly the diagonal pairs ((a,a)). Step 3: The identity relation is reflexive.

Step 1: For a diagonal pair, (a+b=2a). Step 2: The equation (2a=5) has no integer solution in (A), so no diagonal pair is present. Step 3: All four diagonal pairs must be added.

Step 1: A diagonal pair must satisfy (2a=6). Step 2: This gives (a=3), so only ((3,3)) is a diagonal pair in (R). Step 3: To make it reflexive, the remaining four diagonal pairs would be needed.

Step 1: A reflexive relation on five elements needs five diagonal pairs. Step 2: The given relation contains only the diagonal pair ((3,3)). Step 3: Hence (5-1=4) diagonal pairs must be added.

Step 1: For ((a,a)), the condition is (a^2=0\Rightarrow a=a). Step 2: If (a=0), the conclusion (0=0) is true; if (a\neq0), the implication is still true because its premise is false. Step 3: Read implication-based conditions carefully.

Step 1: On the diagonal, (ab=a^2). Step 2: For (a=1,2), (a^2>0) is true, but for (a=0), (0>0) is false. Step 3: Only ((0,0)) must be added.

Step 1: For reflexivity, put (a=b), giving (ab=a^2). Step 2: The square of any real number is non-negative, so (a^2\geq0) is true. Step 3: Product-based diagonal checks often reduce to squares.

Step 1: On the diagonal, (a+b=2a). Step 2: (2a>0) only when (a>0), so (a=1,2). Step 3: Therefore, the diagonal pairs are ((1,1)) and ((2,2)).