Introduction to Relations

Step 1: (a+a=2a) is always even, so it is reflexive. Step 2: (a+b) and (b+a) are the same, so it is symmetric. Step 3: This relation is also transitive because it connects elements of the same parity.

Step 1: Reflexivity requires ((a,a)) to be in the relation. Step 2: But (a+a=2a) is always even, not odd. Step 3: Therefore, no self-pair is formed by this rule.

Step 1: Every number divides itself, so the relation is reflexive. Step 2: If (a) divides (b) and (b) divides (c), then (a) divides (c). Step 3: (1) divides (2), but (2) does not divide (1), so it is not symmetric.

Step 1: For every (a), (a\ge a) is true, so it is reflexive. Step 2: (a\ge b) and (b\ge c) imply (a\ge c), so it is transitive. Step 3: (2\ge1) is true but (1\ge2) is false, so it is not symmetric.

Step 1: (1) and (3) are related to each other. Step 2: (2) and (4) are related to each other. Step 3: Hence the equivalence classes are ({1,3}) and ({2,4}).

Step 1: In the inverse relation, the order of each pair is changed. Step 2: ((1,2)) and ((2,1)) are reverses, and so are ((2,3)) and ((3,2)). Step 3: Therefore, the inverse relation is (R) itself.

Step 1: Interchange the entries of every ordered pair. Step 2: ((1,3)) gives ((3,1)), ((2,3)) gives ((3,2)), and ((3,1)) gives ((1,3)). Step 3: Do not skip any pair while finding the inverse.

Step 1: (1) is related to (2), and (1) is related to (3). Step 2: By transitivity and symmetry, (2) and (3) must also be related to each other. Step 3: Hence ((2,3)) and ((3,2)) must be added.

Step 1: In the identity relation, each element has only its self-pair. Step 2: For (4) elements, there are (4) self-pairs. Step 3: The number of pairs in the identity relation equals the number of elements.

Step 1: The universal relation is equal to (A\times A). Step 2: Since (A) has (4) elements, (A\times A) has (4^2=16) pairs. Step 3: No possible pair is left out in the universal relation.

Step 1: The relation is reflexive and symmetric. Step 2: From ((1,2)) and ((2,3)), transitivity requires ((1,3)). Step 3: Since ((1,3)) is missing, it is not an equivalence relation.

Step 1: For (A={1,2,3}), (A\times A) has (9) pairs. Step 2: The given relation contains all (9) pairs. Step 3: A relation containing all pairs is the universal relation.

Step 1: The empty relation has no pair, so symmetry and transitivity conditions are not violated. Step 2: A non-empty set needs self-pairs for reflexivity, but they are absent. Step 3: Hence it is not reflexive.

Step 1: In a symmetric relation, the reverse of any pair is also present. Step 2: The reverse of ((2,3)) is ((3,2)). Step 3: Symmetry requires reverse pairs, not necessarily self-pairs.

Step 1: In a reflexive relation, every element is related to itself. Step 2: Since (6) is an element of (A), ((6,6)) must be present. Step 3: Pairs with different elements are not compulsory for reflexivity.

Step 1: Transitivity gives ((a,c)) from ((a,b)) and ((b,c)). Step 2: Here (a=4), (b=5), and (c=8). Step 3: Therefore, ((4,8)) must be present.

Since R is symmetric, the presence of (1,2) forces (2,1) to be present as well. The remaining independent choices are the three diagonal pairs (1,1), (2,2), (3,3), and the two unordered off-diagonal pairs represented by (1,3) and (2,3). Thus there are 5 freely selectable pairs, giving 2^5=32 relations. The other options use an incorrect number of independent choices.

Step 1: Reflexivity makes ((1,1)) and ((2,2)) compulsory. Step 2: The distinct pair group ((1,2)) and ((2,1)) is either included together or excluded together. Step 3: Therefore, two relations are possible.

Step 1: (|a-b|=0) only when (a=b). Step 2: So only pairs like ((1,1)), ((2,2)), and ((3,3)) occur. Step 3: A relation based on equality is the identity relation.

Step 1: (|a-a|=0) is even, so it is reflexive. Step 2: (|a-b|=|b-a|), so it is symmetric. Step 3: Even difference means same parity, which also gives transitivity.