Equivalence relation

Step 1: If no two distinct elements are in the same class, every class is a singleton. Step 2: This gives only the identity relation. Step 3: Hence there is exactly (1) such equivalence relation.

Step 1: If all elements are in one class, every element is related to every element. Step 2: This is only the universal relation. Step 3: Therefore there is exactly (1) such equivalence relation.

Step 1: (A\times A) has (5^2=25) ordered pairs. Step 2: (R) contains (2^2+1^2+2^2=9) pairs. Step 3: The complement has (25-9=16) pairs.

Step 1: Since (R) is an equivalence relation, every ((a,a)) is in (R). Step 2: Therefore no ((a,a)) remains in the complement. Step 3: Without reflexivity, the complement cannot be an equivalence relation.

Step 1: (1,2,3) are mutually connected and form one class. Step 2: (4) is related only to itself through ((4,4)). Step 3: Hence the class of (4) is ({4}).

Step 1: If one relation is contained in the other, the union is just the larger relation. Step 2: The larger relation is already an equivalence relation. Step 3: Therefore the union is definitely an equivalence relation in this case.

Step 1: The two class sizes must add to (7). Step 2: The sum of squares is minimized when the sizes are as balanced as possible, (3) and (4). Step 3: The number of pairs is (3^2+4^2=25).

Step 1: The two non-empty class sizes must add to (7). Step 2: The sum of squares is maximum when the sizes are most unequal, (6) and (1). Step 3: The pair count is (6^2+1^2=37).

Step 1: Elements with the same remainder modulo (3) belong to one class. Step 2: The possible remainders are (0,1,2), so there are three classes. Step 3: In exams, first group the elements by their remainders.

Step 1: For every (a), (a+a=2a) is even, so the relation is reflexive. Step 2: If (a+b) is even, then (b+a) is even, and parity also gives transitivity. Step 3: For such questions, separate even and odd integers first.

Step 1: Look for all elements related to (1). Step 2: Since ((1,1)) and ((1,2)) are in (R), the class of (1) is ({1,2}). Step 3: In exams, fix the first element and list all related second elements.

Step 1: Reflexivity holds because all three ((a,a)) pairs are present. Step 2: Symmetry also holds for the given non-diagonal pairs. Step 3: Since ((1,2)) and ((2,3)) are present but ((1,3)) is missing, transitivity fails.

Step 1: (x-x=0\in \mathbb{Z}), so reflexivity holds. Step 2: If (x-y\in \mathbb{Z}), then (y-x=-(x-y)\in \mathbb{Z}), so symmetry holds. Step 3: The sum of integers is an integer, so transitivity holds.

Step 1: (5) leaves remainder (1) when divided by (4). Step 2: In (A), elements with the same remainder are (1) and (5). Step 3: For modulo relations, use remainders as the fastest method.

Step 1: In an equivalence relation, two equivalence classes are either disjoint or identical. Step 2: If they have a common element, they must be the same class. Step 3: In exams, a non-empty intersection of classes directly implies equality.

Step 1: Elements inside the same block are related to each other. Step 2: The block ({1,3,5}) gives (3^2=9) pairs and ({2,4}) gives (2^2=4) pairs. Step 3: Add the squares of block sizes to get the total pairs.

Step 1: Within one equivalence class, every element is related to every element of that class. Step 2: Since (1) and (4) are in the same class, ((4,1)) must be present. Step 3: Do not relate elements from different classes unless the class says so.

Step 1: Every number has the same count of prime factors as itself, so reflexivity holds. Step 2: Equality of counts remains true when the order is reversed. Step 3: If two counts are equal through a third number, then the first and third counts are also equal.

Step 1: (\gcd(4,6)=2). Step 2: In (A), the elements whose gcd with (6) is (2) are (2) and (4). Step 3: First compute the assigned value, then collect elements with the same value.

Step 1: Both relations contain every ((a,a)), so the intersection does too. Step 2: Symmetry and transitivity are preserved in the intersection. Step 3: The intersection of two equivalence relations is again an equivalence relation.

Step 1: (a^2\equiv a^2 \pmod{5}), so reflexivity holds. Step 2: If (a^2\equiv b^2 \pmod{5}), then (b^2\equiv a^2 \pmod{5}), so symmetry holds. Step 3: Congruence is transitive, so the third condition also holds.

Step 1: (\frac{x}{y}>0) means (x) and (y) have the same sign. Step 2: All positive numbers form one class and all negative numbers form another. Step 3: For sign-based relations, split the set into positive and negative groups.

Step 1: The odd class is ({1,3,5}) and the even class is ({2,4,6}). Step 2: Each class gives (3^2=9) ordered pairs. Step 3: Total pairs are (9+9=18).

Step 1: An equivalence relation needs all three properties. Step 2: Here transitivity is missing, so a necessary condition fails. Step 3: Even if two properties hold, always test the third one.

Step 1: All pairs inside ({1,2,5}) must be in the relation. Step 2: If ((1,3)) were in the relation, then (3) would belong to the same class. Step 3: A pair with an outside element changes the class, so it is not possible here.