Reflexive relation

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)).

Step 1: Five diagonal pairs are required. Step 2: ((1,1)) and ((2,2)) are already present because their sums are positive. Step 3: The remaining three diagonal pairs must be added.

Step 1: On the diagonal, (\min(a,a)=a). Step 2: This is true for every (a\in A), so all ((a,a)) are in the relation. Step 3: For minimum and maximum conditions, substitute equal elements first.

Step 1: For ((a,a)), (\max(a,a)=a). Step 2: Since (b=a) on the diagonal, (\max(a,a)=b) is true. Step 3: Reflexivity only needs the same-element cases to hold.

Step 1: On the diagonal, (\gcd(a,a)=a). Step 2: Hence ((a,a)\in R) for every (a\in A). Step 3: For greatest common divisor relations, remember the value for equal numbers.

Step 1: On the diagonal, (\gcd(a,a)=a). Step 2: This equals (1) only for (a=1), so only ((1,1)) is present. Step 3: The other four diagonal pairs must be added.

Step 1: For reflexivity, check ((a,a)). Step 2: The least common multiple of a number with itself is the number itself, so (\operatorname{lcm}(a,a)=a). Step 3: Evaluating the expression at equal numbers is the simplest method.

Step 1: Reflexivity requires (X\subseteq X) for every (X\in A). Step 2: Every set is a subset of itself. Step 3: The subset relation is reflexive on any collection of sets.

Step 1: (X\subset Y) means proper subset. Step 2: No set is a proper subset of itself, so (X\subset X) is false. Step 3: The difference between (\subseteq) and (\subset) is very important in exams.

Step 1: For ((X,X)), the condition becomes (X\cap X=X). Step 2: The intersection of a set with itself is the same set. Step 3: In set-operation relations, substitute the same set on the diagonal.

Step 1: (A=\mathcal{P}(S)) has (4) elements. Step 2: On the diagonal, (X\cup X=X), which equals (S) only when (X=S). Step 3: One diagonal pair is already present, so (3) must be added.

Step 1: On the diagonal, (X\triangle X=\varnothing). Step 2: The symmetric difference of a set with itself is the empty set. Step 3: For symmetric-difference relations, checking equal sets quickly proves reflexivity.

Step 1: Reflexivity requires ((1,1)) as well. Step 2: The number (1) has no prime factor, so ((1,1)\notin R). Step 3: If even one diagonal pair is missing, the relation is not reflexive.

Step 1: Every element of (A) is greater than (1), so each has at least one prime factor. Step 2: A number shares its own prime factor with itself. Step 3: Therefore, every ((a,a)) belongs to the relation.

Step 1: For reflexivity, test ((a,a)). Step 2: The condition (a=b) is true for every diagonal pair. Step 3: In an or condition, one true part is enough to include the pair.

Step 1: On the diagonal, (a=b) is true. Step 2: Also, (a+b=2a), which is even, so both conditions hold for every (a). Step 3: In an and condition, verify both parts.