Equivalence relation

The correct answer is A, (2,1). The relation already contains (1,1), (2,2), and (3,3), so it is reflexive. The pair (1,2) is present, and symmetry therefore requires its reverse, (2,1). After adding that pair, the relation contains all ordered pairs within the class {1,2}, together with (3,3). It is transitive: links between 1 and 2 stay inside {1,2}, and 3 is related only to itself. Thus the equivalence classes are {1,2} and {3}. B, C, and D would connect 3 with another element, but adding only one such pair would either violate symmetry or create further requirements, so they cannot produce an equivalence relation by themselves. Memory cue: equivalence classes behave like complete internally connected groups; an existing one-way link needs its reverse.

The correct answer is A. By definition, the equivalence class of √2 consists of all real numbers x that are related to √2. The condition xR√2 means x−√2 is rational. If x−√2=q, where q∈Q, then x=√2+q. Conversely, every number of the form √2+q has difference q from √2, which is rational, so it belongs to the class. Therefore [√2]={√2+q:q∈Q}. Option B is only the set of rational numbers and does not contain √2+1, for example. Option C is too small because √2+1 and √2−2 also qualify. Option D contains all irrational numbers, but not every irrational number has a rational difference from √2. The key step is always to solve the relation condition for the unknown member x.

The correct answer is C, {3,4}. The equivalence class of 3 is [3]={x∈A:3Rx}. Test each element using the definition. For x=3, the equality condition gives 3R3. For x=4, both 3 and 4 are greater than 2, so 3R4. For x=1 and x=2, the numbers are not equal to 3, and they are not both greater than 2; hence neither is related to 3. Therefore [3]={3,4}. A omits 4, even though the second condition relates it to 3. B includes 1 and 2 without justification. D incorrectly includes 2. The relation divides A into the blocks {1}, {2}, and {3,4}; an equivalence class contains exactly the elements related to the chosen element, not every element in the set.

The correct answer is C, {-3,3}. The equivalence class of -3 contains every real number x such that xR(-3), which means |x|=|-3|=3. Solving |x|=3 gives two possibilities: x=3 or x=-3. Thus [-3]={-3,3}. A is incomplete because -3 and 3 have the same absolute value. B is also incomplete because -3 itself belongs to its class. D is false because most real numbers do not have absolute value 3; for example, |0|=0 and |5|=5. The relation is an equivalence relation: absolute-value equality is reflexive, symmetric, and transitive because equality itself has these properties. Memory cue: an absolute value forgets the sign, so nonzero values usually occur in opposite-sign pairs in the same class; zero is the only value whose class has one element.

Step 1: Every element is related to itself, so the relation is reflexive. Step 2: The pair ((1,2)) has its reverse ((2,1)), so symmetry holds. Step 3: The classes are ({1,2}) and ({3}), so no extra pair is needed.

Step 1: Integers related to (7) must differ from (7) by a multiple of (4). Step 2: (-1,3,7,11) all have the same remainder (3) on division by (4). Step 3: For such questions, group all elements with the same remainder.

Step 1: The relation groups numbers by the same parity. Step 2: (1,3) are odd and (2,4) are even, so the two equivalence classes are ({1,3}) and ({2,4}). Step 3: Equivalence classes must form non-overlapping groups.

Step 1: The statement connects two related pairs to a third pair. Step 2: This is exactly transitivity. Step 3: An equivalence relation must be reflexive, symmetric, and transitive.

Step 1: The condition (a-b=0) means (a=b). Step 2: Equality is reflexive, symmetric, and transitive. Step 3: Equality is the standard example of an equivalence relation.

Step 1: Division by (3) gives possible remainders (0,1,2). Step 2: The classes are ({3,6},{1,4},{2,5}). Step 3: In modulo relations, equivalence classes are based on remainders.

Step 1: An equivalence relation is defined by three properties. Step 2: One or two properties are not enough. Step 3: Always test reflexivity, symmetry, and transitivity separately.

Step 1: The universal relation contains all ordered pairs from (A). Step 2: Hence the required pairs for reflexivity, symmetry, and transitivity are present. Step 3: On any set, (A\times A) is an equivalence relation.

Step 1: Reflexivity requires ((1,1),(2,2),(3,3)). Step 2: The empty relation has no pairs, so reflexivity fails. Step 3: On a non-empty set, the empty relation cannot be an equivalence relation.

Step 1: (a+a=2a) is always even, so the relation is reflexive. Step 2: If (a+b) is even, then (b+a) is even, so it is symmetric. Step 3: If (a+b) and (b+c) are even, then (a) and (c) have the same parity, so it is transitive.

Step 1: Reflexivity needs (aRa) for every (a). Step 2: (a+a=2a) is always even, not odd. Step 3: Once reflexivity fails, the relation cannot be an equivalence relation.

Step 1: Elements related to (2) must have absolute value (2). Step 2: Both (-2) and (2) have absolute value (2). Step 3: To find an equivalence class, apply the relation condition to the chosen element.

Step 1: Every point has the same (x)-coordinate as itself. Step 2: If two points have the same (x)-coordinate, the order does not matter. Step 3: Equality of (x)-coordinates also passes through a third point, so transitivity holds.

Step 1: An equivalence class is the block that contains the given element. Step 2: (4) lies in ({1,4}). Step 3: When a partition is given, choose the block containing the element.

Step 1: A partition has non-empty, non-overlapping blocks whose union is the whole set. Step 2: ({1,2}) and ({3,4}) are disjoint and cover (A). Step 3: Every equivalence relation creates a partition.

Step 1: In an equivalence relation, related elements belong to the same class. Step 2: Therefore (aRb) implies ([a]=[b]). Step 3: If two equivalence classes overlap, they are actually equal.

Step 1: (1) and (3) are related to each other, so they form one class. Step 2: (2) and (4) are related only to themselves. Step 3: Connected related elements form groups, while isolated elements form singleton classes.

Step 1: Every number has the same last digit as itself. Step 2: Having the same last digit is symmetric. Step 3: If two numbers share the same last digit through a middle number, the first and third also share it.

Step 1: The last digit of (23) is (3). Step 2: So numbers like (3,13,23,33) belong to the same class. Step 3: The class is built using the defining property of the relation.

Step 1: The identity relation contains all pairs ((a,a)). Step 2: Since only identical elements are paired, symmetry and transitivity also hold. Step 3: The identity relation is an equivalence relation on any set.

Step 1: Division by (5) gives only remainders (0,1,2,3,4). Step 2: Integers with the same remainder form one class. Step 3: Although integers are infinite, there are only (5) classes.