Equivalence relation

Step 1: For every element, (a-a=0), which is divisible by (2), so the relation is reflexive. Step 2: If (a-b) is divisible by (2), then (b-a) is also divisible by (2), so it is symmetric. Step 3: If (a-b) and (b-c) are divisible by (2), then (a-c) is also divisible by (2). In exams, test all three properties separately.

Step 1: When (7) is divided by (5), the remainder is (2). Step 2: So all integers related to (7) must have the same remainder (2). Step 3: Since the base set is (Z), the equivalence class is not finite; write the complete residue class.

Step 1: Numbers related to (-3) must have absolute value (3). Step 2: In real numbers, (|x|=3) gives (x=3) or (x=-3). Step 3: In absolute value relations, a sign change may still keep the value equal, so include both numbers.

Step 1: (\gcd(4,6)=2). Step 2: In (A), the numbers whose greatest common divisor with (6) is (2) are (2) and (4). Step 3: Hence the equivalence class of (4) is ({2,4}), so it has (2) elements.

Step 1: Every line has the same slope as itself, so the relation is reflexive. Step 2: If the slope of one line equals the slope of another, the reverse is also true. Step 3: Equality is transitive, so the same-slope relation is an equivalence relation.

Step 1: The determinant of any matrix is equal to itself. Step 2: Equality of determinants works in both directions. Step 3: If the first and second determinants are equal, and the second and third are equal, then the first and third are equal too.

Step 1: For every function, (f(1)=f(1)), so the relation is reflexive. Step 2: If (f(1)=g(1)), then (g(1)=f(1)) is also true. Step 3: If (f(1)=g(1)) and (g(1)=h(1)), then (f(1)=h(1)), so it is an equivalence relation.

Step 1: The relation separates numbers into two types: even and odd. Step 2: In (A), the odd block is ({1,3,5}) and the even block is ({2,4}). Step 3: An equivalence relation always forms a partition, so there are (2) blocks here.

Step 1: For any integer (a), (a+a=2a), which is even, so the relation is reflexive. Step 2: If (a+b) is even, then (b+a) is also even, so it is symmetric. Step 3: If (a) has the same parity as (b), and (b) has the same parity as (c), then (a) has the same parity as (c).

Step 1: (\frac{a}{a}=1>0), so the relation is reflexive. Step 2: (\frac{a}{b}>0) means (a) and (b) have the same sign, so the reverse ratio is also positive. Step 3: Having the same sign is transitive, so it forms two equivalence classes: positive and negative.

Step 1: ((1,1),(2,2),(3,3)) are present, so the relation is reflexive. Step 2: ((1,2)) and ((2,1)) are both present, so symmetry is satisfied. Step 3: ({1,2}) is one class and ({3}) is another; transitivity is not broken, so no pair is needed.

Step 1: The relation is reflexive and appears symmetric, but transitivity must be checked. Step 2: From ((1,2)) and ((2,3)), ((1,3)) is needed; from ((3,2)) and ((2,1)), ((3,1)) is needed. Step 3: In an equivalence relation, all elements of the same class must be mutually related.

Step 1: Inside each block, every element is related to every element of that block. Step 2: ({1,4}) gives (2^2=4) pairs, ({2}) gives (1), and ({3}) gives (1). Step 3: Total pairs are (4+1+1=6).

Step 1: If an equivalence class has (n) elements, it contributes (n^2) ordered pairs. Step 2: Here the number of pairs is (3^2+2^2+1^2=9+4+1). Step 3: Total pairs are (14); pairs between different classes are not included.

Step 1: Look at the remainders of squares on division by (3). The squares (1^2,2^2,4^2) leave remainder (1). Step 2: (3^2) leaves remainder (0), so (3) is not in this class. Step 3: Numbers with the same square remainder belong to the same equivalence class.

Step 1: (a-a=0) is rational, so the relation is reflexive. Step 2: Changing the sign of a rational difference keeps it rational. Step 3: The class of a chosen number (a) is ({x\in R:x-a\in Q}); this is the key exam idea.

Step 1: Numbers with the same remainder on division by (3) belong to the same class. Step 2: Remainder (1) gives ({1,4}), remainder (2) gives ({2,5}), and remainder (0) gives ({3,6}). Step 3: Equivalence classes are disjoint and together cover the whole set.

Step 1: The relation divides letters into two groups: vowels and consonants. Step 2: Letters in the same group are related, while letters in different groups are not related. Step 3: Hence exactly (2) equivalence classes are formed; (5) is only the number of vowels, not the number of classes.

Step 1: ((1,3)) and ((3,1)) show that (1) and (3) are in the same class. Step 2: (2) and (4) are related only to themselves, so they form singleton classes. Step 3: To convert a relation into a partition, put mutually related elements in the same block.

Step 1: Every number has the same unit digit as itself, so the relation is reflexive. Step 2: Having the same unit digit works in both directions. Step 3: If two numbers have the same unit digit as a third number, they also have the same unit digit as each other.