Equivalence relation

Step 1: (3^2-3=6). Step 2: (x^2-x=6) gives (x^2-x-6=0), that is ((x-3)(x+2)=0). Step 3: Hence (x=3) or (x=-2), so the class is ({-2,3}).

Step 1: ((-1)^2+2(-1)=1-2=-1). Step 2: (x^2+2x=-1) gives (x^2+2x+1=0), or ((x+1)^2=0). Step 3: Only (x=-1) is obtained, so the equivalence class is singleton.

Step 1: (\sin^2 0=0). Step 2: (\sin^2 \pi=0), so (\pi) is in the same class as (0). Step 3: The class contains angles whose squared sine value is (0).

Step 1: (\cos\frac{\pi}{3}=\frac{1}{2}), so the squared value is (\frac{1}{4}). Step 2: (\cos\frac{2\pi}{3}=-\frac{1}{2}), whose square is also (\frac{1}{4}). Step 3: After squaring, positive and negative values may become equal.

Step 1: A number (x) is related to (3) when (\frac{x}{3}=2^n), where (n\in Z). Step 2: This gives (x=3\cdot 2^n). Step 3: For ratio-based relations, write the class in multiplicative form.

Step 1: (12=2^2\cdot 3), so it has exactly power (2) of (2). Step 2: The relation checks only this power of (2). Step 3: Hence the class contains numbers divisible by (4) but not by (8).

Step 1: (R\cap S) keeps only pairs that are present in both relations. Step 2: Thus two elements remain together only when they are together in both (R) and (S). Step 3: Therefore its classes are common refined blocks of the two partitions.

Step 1: ((1,2)) belongs to (R\cup S) because it is in (R). Step 2: ((2,3)) belongs to (R\cup S) because it is in (S). Step 3: Transitivity would require ((1,3)), but it is in neither (R) nor (S).

Step 1: The largest relation contains all possible ordered pairs of (A). Step 2: (A\times A) is reflexive, symmetric, and transitive. Step 3: Therefore the universal relation is the largest equivalence relation.

Step 1: In an equivalence relation, every element must be related to itself. Step 2: Taking only the pairs ((a,a)) preserves symmetry and transitivity. Step 3: Hence the identity relation is the smallest equivalence relation.

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.