Equivalence relation

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

Step 1: (\gcd(6,8)=2). Step 2: In the given set, (\gcd(2,8)=2) and (\gcd(6,8)=2). Step 3: Elements with the same greatest common divisor form one equivalence class.

Step 1: The classes are ({1,3,5,7},{2,6},{4},{8}). Step 2: The number of pairs is (4^2+2^2+1^2+1^2). Step 3: The total is (16+4+1+1=22).

Step 1: In an equivalence relation, only elements inside the same class are related. Step 2: (4) is in the first class and (2) is in the second class. Step 3: A pair from different classes cannot belong to the relation.

Step 1: A class of size (m) contributes (m^2) ordered pairs. Step 2: The total is (3^2+2^2+1^2=9+4+1). Step 3: Hence the relation contains (14) pairs.

Step 1: Treat (1,2,3) as one combined block. Step 2: Then there are three objects: ({1,2,3},4,5). Three objects have (5) partitions. Step 3: Combine elements that must stay together before counting.

Step 1: This is the number of ways to partition (5) elements into exactly (3) non-empty blocks. Step 2: The block-size patterns are (3,1,1) and (2,2,1). Step 3: The count is (10+15=25), so there are (25) equivalence relations.

Step 1: A (4)-element set has (15) equivalence relations. Step 2: The number where (1) and (2) are in the same class is (5). Step 3: Hence the number where they are in different classes is (15-5=10).

Step 1: The classes are ({1,6},{2},{3},{4},{5}). Step 2: The pair count is (2^2+1^2+1^2+1^2+1^2). Step 3: The total is (4+1+1+1+1=8).

Step 1: The difference must be divisible by both (6) and (10). Step 2: Such a difference is divisible by (\operatorname{lcm}(6,10)=30). Step 3: When two modulo conditions must hold together, use the least common multiple.

Step 1: (0-4=-4) is divisible by (4), and (4-10=-6) is divisible by (6). Step 2: But (0-10=-10) is divisible by neither (4) nor (6). Step 3: Thus transitivity fails.

Step 1: (\sin a=\sin a), so reflexivity holds. Step 2: If (\sin a=\sin b), then (\sin b=\sin a), so symmetry holds. Step 3: Equality of function values is transitive, so this is an equivalence relation.

Step 1: For (0), (\sin 0=0). Step 2: The solutions of (\sin x=0) are (x=n\pi), where (n\in\mathbb{Z}). Step 3: For trigonometric function classes, include all angles with the same function value.

Step 1: (6=2\cdot3), so it has two prime factors with repetition. Step 2: (4=2^2) and (9=3^2) also have two prime factors. Step 3: Count repeated prime factors when the condition says so.

Step 1: The two numbers must have the same remainders modulo (2) and modulo (3). Step 2: This is the same as having the same remainder modulo (6). Step 3: In the given set, only (5) has remainder (5) modulo (6).

Step 1: The possible values of (\gcd(a,10)) are (1,2,5,10). Step 2: These four distinct values create four equivalence classes. Step 3: The number of distinct function values gives the number of classes.

Step 1: (\gcd(8,10)=2). Step 2: (2,4,6,8) all have greatest common divisor (2) with (10). Step 3: Only elements with the same greatest common divisor belong to the class.

Step 1: (\lfloor -1.2\rfloor=-2). Step 2: Real numbers with floor value (-2) lie in ([-2,-1)). Step 3: For negative numbers, be careful that the floor goes to the smaller integer.

Step 1: From (aRb) and (bRc), transitivity gives (aRc). Step 2: From (aRc) and (cRd), transitivity gives (aRd). Step 3: In an equivalence relation, all elements connected in a chain lie in the same class.

Step 1: The three classes share at least one common element. Step 2: If two equivalence classes have non-empty intersection, they are equal. Step 3: Applying this to all three classes gives ([a]=[b]=[c]).