Equivalence relation

Step 1: Symmetry is necessary for an equivalence relation. Step 2: (2\leq 3) is true, but (3\leq 2) is false. Step 3: Order relations often fail symmetry.

Step 1: For (|1-b|) to be even, (b) must have the same parity as (1). Step 2: (1,3,5) are odd, so they are related to (1). Step 3: An even difference relation groups numbers by parity.

Step 1: An equivalence relation is reflexive. Step 2: Therefore (aRa), so (a\in [a]). Step 3: Equivalence classes are never empty.

Step 1: Every student is in the same class as himself or herself. Step 2: If one student is in the same class as another, the reverse is also true. Step 3: The same-class relation is transitive, so it is an equivalence relation.

Step 1: The pairs ((2,1)) and ((1,3)) are present. Step 2: Transitivity would require ((2,3)). Step 3: In transitivity checks, identify the middle element and demand the connecting pair.

Step 1: (a^2=a^2), so the relation is reflexive. Step 2: If (a^2=b^2), then (b^2=a^2), so it is symmetric. Step 3: If (a^2=b^2) and (b^2=c^2), then (a^2=c^2), so it is transitive.

Step 1: The square of (-3) is (9). Step 2: The integers whose square is (9) are (-3) and (3). Step 3: Equal-square classes usually contain both a number and its negative.

Step 1: Same remainder modulo (2) means same parity. Step 2: (2) is even and (5) is odd, so their remainders are different. Step 3: In such options, compare remainders first.

Step 1: An equivalence relation divides a set into classes. Step 2: If two classes share an element, they become equal; otherwise they are disjoint. Step 3: This idea is central in partition-based questions.

Step 1: The condition (a=b) ensures reflexivity. Step 2: The condition (a+b=5) is symmetric. Step 3: The pairs form the classes ({1,4}) and ({2,3}), so the relation is an equivalence relation.

Step 1: (\frac{a}{a}=1>0), so the relation is reflexive. Step 2: If (\frac{a}{b}>0), then (a) and (b) have the same sign, so (\frac{b}{a}>0). Step 3: Having the same sign is transitive, so the relation is an equivalence relation.

Step 1: (\frac{-5}{b}>0) only when (b) is also negative. Step 2: Hence the class of (-5) is the set of all negative real numbers. Step 3: Same-sign relations split numbers into positive and negative classes.

Step 1: Inside one equivalence class, every element is related to every element. Step 2: The class ({a,b,c}) gives (3^2=9) pairs and ({d,e}) gives (2^2=4) pairs. Step 3: Total pairs are (9+4=13).

Step 1: Each block contributes all ordered pairs within itself. Step 2: The count is (2^2+1^2+2^2=4+1+4). Step 3: Therefore there are (9) pairs, with no pairs between different blocks.

Step 1: In a partition, different blocks must not share elements. Step 2: ({1,2}) and ({2,3}) both contain (2), so they overlap. Step 3: Equivalence classes cannot overlap unless they are equal.

Step 1: (a-a=0) is rational, so the relation is reflexive. Step 2: If (a-b) is rational, then (b-a=-(a-b)) is rational. Step 3: The sum of two rational differences is rational, so transitivity holds.

Step 1: The difference between (\sqrt{2}+3) and (\sqrt{2}) is (3). Step 2: Since (3) is rational, they are in the same class. Step 3: In such questions, subtract from the given representative.

Step 1: (a\leq a) is always true, so (\leq) is reflexive. Step 2: But (2\leq 3) does not imply (3\leq 2). Step 3: Since symmetry is missing, it is not an equivalence relation.

Step 1: A sum is even when both numbers have the same parity. Step 2: (1) and (3) are both odd, so (1+3=4) is even. Step 3: Check parity quickly in such pair-selection questions.

Step 1: Equivalence classes are either disjoint or identical. Step 2: A non-empty intersection means they share an element. Step 3: Therefore the two classes must be equal.

Step 1: (2) gives remainder (2) on division by (4). Step 2: In the given set, (2) and (6) have remainder (2). Step 3: In modulo relations, classes are formed by equal remainders.

Step 1: In the identity relation, each element is related only to itself. Step 2: Hence every element forms its own equivalence class. Step 3: For (n) elements, there are (n) singleton classes.

Step 1: In the universal relation, every element is related to every element. Step 2: Thus all elements lie in one large class. Step 3: For a non-empty set (A), there is exactly (1) equivalence class.

Step 1: A relation from a partition contains only pairs from within the same block. Step 2: (2) and (3) are in different blocks. Step 3: Pairs between different blocks are not included.

Step 1: Both (1) and (4) give remainder (1) when divided by (3). Step 2: Hence they belong to the same equivalence class. Step 3: In same-remainder questions, compare remainders directly.