Equivalence relation

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.

Step 1: Each equivalence class contributes all ordered pairs within itself. Step 2: Hence the total number is (1^2+2^2+4^2=1+4+16). Step 3: The total is (21); the full (A\times A) would occur only with one class.

Step 1: In the given (R), every element of (A) is related to every element of (A). Step 2: It is equal to (A\times A). Step 3: The universal relation is reflexive, symmetric, and transitive, so it is an equivalence relation.

Step 1: ((1,3)) belongs to the relation. Step 2: For symmetry, ((3,1)) must also belong to it. Step 3: Since ((3,1)) is missing, symmetry fails and the relation is not an equivalence relation.

Step 1: (3.25-0.25=3), which is an integer. Step 2: Therefore both numbers lie in the same equivalence class. Step 3: If the decimal part is the same, a relation based on integer difference can be identified quickly.

Step 1: For (1), (\log 1=0). Step 2: A number (x) is related to (1) when (\log x-0) is an integer. Step 3: Hence the class contains positive numbers whose logarithm is an integer.

Step 1: On division by (2), only two remainders are possible: (0) and (1). Step 2: The odd class is ({1,3,5}), and the even class is ({2,4,6}). Step 3: The quotient set is the set of all equivalence classes.

Step 1: Every rectangle has the same area as itself. Step 2: Having equal area works in both directions. Step 3: If the first and second have equal area, and the second and third have equal area, then the first and third have equal area.

Step 1: Reflexivity would require (a+a=6) for every (a). Step 2: For (a=1), (1+1=2), so (1R1) is false. Step 3: Since reflexivity fails, it cannot be an equivalence relation.

Step 1: Both relations are reflexive, so every ((a,a)) lies in both and also in the intersection. Step 2: Symmetry is preserved in the intersection. Step 3: Transitivity also works because it holds in both relations, so (R\cap S) is an equivalence relation.

Step 1: Reflexivity and symmetry may often remain in the union. Step 2: But transitivity can fail because one pair may come from (R) and another from (S). Step 3: Therefore (R\cup S) is not assumed to be an equivalence relation without checking.

Step 1: (5) is not a multiple of (2). Step 2: Its class will contain elements that are also not multiples of (2). Step 3: In the given set, these are (1,3,5).

Step 1: The sum of digits of (29) is (2+9=11). Step 2: The sum of digits of (56) is (5+6=11). Step 3: Since the relation is based on digit sum, (56) lies in the same equivalence class.

Step 1: (\tan 0=0). Step 2: (\tan \pi=0), so (\pi) has the same tangent value as (0). Step 3: But tangent is undefined at (\frac{\pi}{2}), so the domain must be checked carefully.

Step 1: The condition (a=b) gives reflexivity. Step 2: The condition (a+b=6) is symmetric. Step 3: The relation creates the blocks ({1,5}), ({2,4}), and ({3}), so it is actually an equivalence relation.

Step 1: Since (1+5=6), (1) and (5) lie in one class. Step 2: Since (2+4=6), (2) and (4) lie in one class. Step 3: (3) is related only to itself, so the correct partition is ({{1,5},{2,4},{3}}).

Step 1: The last digit of (102) is (2). Step 2: The relation is based on equality of last digit. Step 3: So its class consists of all positive integers with last digit (2), not merely all even numbers.

Step 1: (2) and (5) are in different classes, so they are not related. Step 2: (4) and (3) are in the same class ({3,4,5}). Step 3: Therefore ((2,5)\notin R) and ((4,3)\in R).

Step 1: (|a|+|b|=0) happens only when (a=0) and (b=0). Step 2: For (1R1), (|1|+|1|=2), so (1R1) is false. Step 3: Since reflexivity fails on all real numbers, it is not an equivalence relation.

The relation \(5\mid(a-b)\) is reflexive, symmetric and transitive: from \(a-b=5m\) and \(b-c=5n\), we get \(a-c=5(m+n)\). Option B fails reflexivity since \(1R1\) is false. Exam tip: test all three properties.

Step 1: (2^2=4), so the square of (2) leaves remainder (4) modulo (5). Step 2: (3^2=9) also leaves remainder (4) modulo (5). Step 3: The squares of (1) and (4) leave remainder (1), so the class of (2) is ({2,3}).