Equivalence relation

Answer: A. An equivalence relation must be reflexive, symmetric, and transitive. The given relation already contains all diagonal pairs, so it is reflexive. It also contains each reverse pair: (1,2) with (2,1), and (2,3) with (3,2), so it is symmetric. Transitivity now requires (1,3), because (1,2) and (2,3) are present. Once (1,3) is added, symmetry requires its reverse (3,1) as well. Adding both pairs makes every pair in A×A present, which certainly gives reflexivity, symmetry, and transitivity. Option B breaks symmetry, option C also breaks symmetry, and option D removes pairs rather than completing the needed connection. Memory cue: equivalence classes cannot overlap through a chain without joining their endpoints.

Answer: C, (1,3). In an equivalence relation, two elements are related exactly when they belong to the same equivalence class. The classes here are {1,2} and {3,4}. Thus all ordered pairs whose two entries come from {1,2} are present, including (1,2) and (2,1). Similarly, both entries of (4,3) lie in {3,4}, so (4,3) must be present. However, 1 belongs to the first class and 3 belongs to the second class, so (1,3) joins different classes and is definitely absent. Option C is therefore the only correct choice. Memory cue: equivalence classes create complete blocks inside the relation, with no cross-block pairs.

Answer: C, 8. An equivalence relation contains every ordered pair whose two elements belong to the same equivalence class. The class {1,2} contributes 2×2=4 pairs: (1,1), (1,2), (2,1), and (2,2). The class {3,4} contributes another 2×2=4 pairs: (3,3), (3,4), (4,3), and (4,4). Pairs connecting the classes, such as (1,3) or (4,2), are excluded. Therefore the total is 4+4=8. Option A counts only one block, B does not follow the square rule, and D incorrectly counts all 4×4=16 pairs in A×A, including cross-class pairs. Memory cue: for equivalence classes, add the squares of their sizes: 2^2+2^2=8.

The correct answer is A, (1,3). An equivalence relation must be reflexive, symmetric, and transitive. Since (1,2) and (2,3) are already present, transitivity forces (1,3). Symmetry also forces (2,1) and (3,2), and then the reverse of (1,3), namely (3,1). Reflexivity requires all diagonal pairs, including (1,1), (2,2), and (3,3); (4,4) is already present. The smallest closure therefore makes {1,2,3} one equivalence class and leaves {4} as a separate class. No pair joining 4 to 1, 2, or 3 is forced. Hence B, C, and D are not necessary. A useful method is to connect elements through chains: 1 is connected to 2, and 2 to 3, so 1 must connect to 3; element 4 has no such connection.

The correct answer is D, (1,4). The three given pairs already connect 1, 2, and 3. In the smallest equivalence relation, reflexivity adds (1,1), (2,2), (3,3), and (4,4); symmetry adds the reverse pairs; and transitivity fills in every ordered pair among 1, 2, and 3. Thus {1,2,3} becomes one equivalence class, while 4 remains alone. Since no given pair connects 4 to the other elements, the equivalence closure does not add (1,4), (4,1), or any other pair between 4 and the first class. Options A, B, and C are explicitly present in the original relation, so they certainly remain present in every relation containing it. Option D is therefore the only pair that is absent. Common confusion: an equivalence closure adds what the three properties force, not arbitrary pairs between unrelated classes.

Correct answer: A. The equivalence class [a] is the set of all elements x in A such that xRa. If aRb, symmetry gives bRa. To show [a] is contained in [b], take any x in [a], so xRa. Since aRb, transitivity gives xRb, which means x belongs to [b]. The reverse inclusion is obtained similarly: if xRb, then bRa by symmetry, and transitivity gives xRa, so x belongs to [a]. Thus [a]=[b]. Option B is false because distinct equivalence classes are disjoint, but aRb means these two classes are the same class, not distinct. Option C is false because an equivalence class may contain many elements; it contains only a only in a special relation, not in general. Option D is the opposite of what equivalence means: b itself is related to a, and every member of the common class is related to a. Memory cue: in an equivalence relation, related elements are simply different representatives of the same class.

The correct answer is C. The definition separates A into two groups: multiples of 3, {3,6}, and nonmultiples of 3, {1,2,4,5}. Two elements are related exactly when they lie in the same group. This makes the relation reflexive, symmetric, and transitive, so it is an equivalence relation. It is not antisymmetric: for example, 1 and 2 are distinct, but both 1R2 and 2R1 are true because neither is a multiple of 3. A is therefore false. B is false because the definition is unchanged when a and b are interchanged. D is false because the two groups produce two equivalence classes, not six. A memory cue is that equivalence classes collect elements with the same stated property; here the property is ‘multiple of 3’ versus ‘not a multiple of 3’.

Direct answer: C, four equivalence classes. Compute the square residues modulo 6: 0² gives 0, 1² gives 1, 2² gives 4, 3² gives 3, 4² gives 16≡4, and 5² gives 25≡1. Thus the elements with the same square residue group together: residue 0 gives {0}; residue 1 gives {1,5}; residue 4 gives {2,4}; and residue 3 gives {3}. These four disjoint groups cover A, so the equivalence classes are {0}, {1,5}, {2,4}, and {3}. Option A, 2, misses two groups; B, 3, also misses one; D, 6, incorrectly treats every element as separate. The relation is an equivalence relation because equality of the values a² modulo 6 is reflexive, symmetric, and transitive. Memory cue: first calculate the invariant—in this case the square residue—then group equal values.

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.