Equivalence relation

Step 1: The remainder of (3) on division by (4) is (3). Step 2: In the given set, the elements with remainder (3) are (3) and (7). Step 3: While writing an equivalence class, include only elements from the given base set.

Step 1: (3^2-5^2=9-25=-16), which is divisible by (8). Step 2: In the other options, the difference of squares is not divisible by (8). Step 3: In such questions, directly checking the difference of squares is quick and reliable.

Step 1: A number (x) is related to (\sqrt{2}) when (x-\sqrt{2}) is rational. Step 2: So (x=\sqrt{2}+q), where (q\in Q). Step 3: Classes based on rational difference contain infinitely many numbers, not just one number.

Step 1: The odd class ({1,3,5}) has (3) elements and the even class ({2,4}) has (2) elements. Step 2: The number of ordered pairs is (3^2+2^2=9+4). Step 3: Pairs between different classes are not counted, so the total is (13).

Step 1: Elements in the same block are related to each other. Step 2: (2) and (6) are together in the first block, so ((2,6)\in R). Step 3: (3) and (4) are in different blocks, so ((3,4)\notin R).

Step 1: ((-5)^2=25). Step 2: The real numbers whose square is (25) are (-5) and (5). Step 3: In equality of squares, the sign may change, so both numbers must be included.

Step 1: (\frac{a}{a}=1), which is rational, so reflexivity holds. Step 2: If (\frac{a}{b}) is rational and non-zero, then (\frac{b}{a}) is also rational. Step 3: The product of two rational ratios is rational, so transitivity also holds.

Step 1: For ((2,5)), the sum is (2+5=7). Step 2: Every pair related to it must also have sum (7). Step 3: In such relations, an equivalence class may appear as a set of points satisfying a fixed value condition.

Step 1: (1) is related to (2), and (2) is related to (3). Step 2: Transitivity requires (1) to be related to (3), and symmetry requires the reverse pair too. Step 3: (4) is only related to itself, so no new pair involving (4) is needed.

Step 1: The number of equivalence relations on a set equals the number of partitions of that set. Step 2: A set with (4) elements has (15) partitions. Step 3: In exams, remember the direct link between equivalence relations and partitions.

Step 1: The greatest integer part of (-1.2) is (-2), not (-1). Step 2: Numbers whose greatest integer part is (-2) lie in ([-2,-1)). Step 3: For negative decimals, take the lower integer while using the floor function.

Step 1: The trace of a matrix is equal to its own trace. Step 2: Equality of trace works in both directions. Step 3: If the traces are equal pairwise through a middle matrix, then the first and third traces are also equal.

Step 1: The degree of (x^2+1) is (2). Step 2: In this relation, two polynomials are related exactly when their degrees are equal. Step 3: So the class contains all non-zero polynomials of degree (2), not only the given polynomial.

Step 1: Reflexivity requires (\gcd(a,a)>1) for every (a). Step 2: For (a=1), (\gcd(1,1)=1), which is not greater than (1). Step 3: If reflexivity fails for even one element, the relation cannot be an equivalence relation.

Step 1: The relation is reflexive because (|a-a|=0<1). Step 2: It is also symmetric because distance is the same in both directions. Step 3: But (0R0.6) and (0.6R1.2), while (0R1.2) is false, so transitivity fails.

Step 1: Since (a\le a), reflexivity holds. Step 2: If (1\le 2), it does not imply (2\le 1). Step 3: Hence symmetry fails, so the relation is not an equivalence relation.

Step 1: (|3+4i|=\sqrt{3^2+4^2}=5). Step 2: All complex numbers related to it must have modulus (5). Step 3: In the complex plane, this represents the circle with distance (5) from the origin.

Step 1: (4) is not prime. Step 2: So the class of (4) contains all elements that are not prime. Step 3: (1) is also not considered prime, so it is included in this class.

Step 1: Every equation represents the same line as itself, so reflexivity holds. Step 2: If the first equation represents the same line as the second, the reverse is also true. Step 3: Representing the same line is transitive, so this is an equivalence relation.

Step 1: (125) has (3) digits. Step 2: The relation checks only the number of digits, not the actual value. Step 3: Therefore the class of (125) consists of all three-digit natural numbers.

Step 1: For reflexivity, (\max(a,a)) must be even for every (a). Step 2: For (a=1), (\max(1,1)=1), which is not even. Step 3: Therefore not every element is related to itself, so the relation is not equivalence.

Step 1: In (A\times A), every element is related to every other element. Step 2: Hence all elements fall into one equivalence class. Step 3: This is the universal relation, so the number of equivalence classes is (1).

Step 1: In the identity relation, each element is related only to itself. Step 2: Therefore each element forms its own singleton class. Step 3: Since there are (4) elements, there are (4) equivalence classes.

Step 1: Equivalence classes are either disjoint or equal. Step 2: The intersection is not empty, so they cannot be disjoint. Step 3: Therefore the two classes must be equal, not merely partially common.

Step 1: (\gcd(5,6)=1). Step 2: In the given set, (\gcd(a,6)=1) occurs only for (1) and (5). Step 3: Elements with the same greatest common divisor form one equivalence class.