Equivalence relation

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.

Step 1: On real numbers, (e^x) is one-one. Step 2: Therefore (e^a=e^b) implies (a=b). Step 3: Hence each number forms only its own equivalence class.

Step 1: (\cos 0=1). Step 2: (\cos 2\pi=1), so (2\pi) has the same cosine value as (0). Step 3: In a relation based on equal function value, all numbers with the same value lie in the same class.

Step 1: The first class is ({1,2,3}), and the second class is ({4,5,6,7,8}). Step 2: The number of ordered pairs is (3^2+5^2=9+25). Step 3: Therefore the total number of pairs is (34).

Step 1: The norm of the zero vector is (0). Step 2: A vector has norm (0) only when it is the zero vector. Step 3: Therefore the class of the zero vector contains only the zero vector itself.

Step 1: (27) leaves remainder (7) on division by (10). Step 2: For (-13), the remainder modulo (10) is (7), since (-13=-20+7). Step 3: Since the remainders are the same, the two integers are related.