Introduction to Relations

Step 1: In the inverse relation, the two entries of each pair are interchanged. Step 2: ((1,2)), ((2,4)), and ((4,1)) become ((2,1)), ((4,2)), and ((1,4)). Step 3: While finding inverse, reverse entries inside each pair.

Step 1: In (S\circ R), apply (R) first and then (S). Step 2: (1) goes to (2), and (2) goes to (4), giving ((1,4)); (2) goes to (3), and (3) goes to (5), giving ((2,5)). Step 3: The middle element must match in composition.

Step 1: In (R\circ R), ((a,c)) occurs when some (b) satisfies ((a,b)\in R) and ((b,c)\in R). Step 2: ((1,2)) and ((2,1)) give ((1,1)). Step 3: In composition, look for connected pairs.

Step 1: Every pair has its reverse, so it is symmetric. Step 2: ((1,1)), ((2,2)), and ((3,3)) are absent, so reflexivity is missing. Step 3: An equivalence relation needs reflexivity, symmetry and transitivity.

Step 1: Reflexivity requires three self-pairs. Step 2: Being both symmetric and antisymmetric forbids pairs between distinct elements. Step 3: So only ((1,1)), ((2,2)), and ((3,3)) remain, giving (3) pairs.

Step 1: Reflexivity makes all self-pairs compulsory. Step 2: Symmetry and antisymmetry together do not allow non-self pairs. Step 3: Hence the relation must be the identity relation.

Step 1: A universal relation contains all self-pairs, so it is reflexive. Step 2: It also contains every reverse pair, so it is symmetric. Step 3: Since all pairs are present, transitivity also holds, making it an equivalence relation.

Step 1: The universal relation contains both ((1,2)) and ((2,1)). Step 2: Since (1\ne2), antisymmetry fails. Step 3: On a set with two or more elements, the universal relation is usually not antisymmetric.

Step 1: On a one-element set, the only pair is ((1,1)). Step 2: This pair satisfies all required conditions. Step 3: On a singleton set, many relation properties can hold together.

Step 1: ((1,2)) and ((2,3)) are in the relation. Step 2: Transitivity requires ((1,3)). Step 3: Since ((1,3)) is missing, the relation is not transitive.

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 possible pairs is the universal relation.

Step 1: The class ({1,4}) gives (2^2=4) pairs. Step 2: The singleton classes ({2}) and ({3}) give (1+1) pairs. Step 3: Total pairs are (4+1+1=6).

Step 1: One class means all elements are related to one another. Step 2: A class with (4) elements gives (4^2=16) ordered pairs. Step 3: Hence the relation is universal and has (16) pairs.

Step 1: Singleton classes mean each element is related only to itself. Step 2: Therefore, only four self-pairs are formed. Step 3: This is the identity relation and it has (4) pairs.

Step 1: Reflexivity gives self-pairs, and symmetry gives reverse pairs. Step 2: ((1,2)) and ((2,3)) require ((1,3)) by transitivity. Step 3: To keep symmetry, ((3,1)) is also needed.

Step 1: Find pairs whose sum is (5). Step 2: The pairs are ((1,4)), ((2,3)), ((3,2)), and ((4,1)). Step 3: Therefore, there are (4) pairs.

Step 1: If (a+b=5), then (b+a=5), so it is symmetric. Step 2: No self-pair exists because (2a=5) is not possible in the set. Step 3: ((1,4)) and ((4,1)) would require ((1,1)), which is absent, so it is not transitive.

Step 1: All self-pairs ((1,1)), ((2,2)), and ((3,3)) are included. Step 2: Pairs with difference (1), namely ((1,2)), ((2,1)), ((2,3)), and ((3,2)), are also included. Step 3: Total pairs are (3+4=7).

Step 1: (|a-a|=0), so reflexivity holds. Step 2: (|a-b|=|b-a|), so symmetry holds. Step 3: ((1,2)) and ((2,3)) are present but ((1,3)) is absent, so transitivity fails.

Step 1: Even difference means the two elements have the same parity. Step 2: The odd class ({1,3}) gives (4) pairs and the even class ({2,4}) gives (4) pairs. Step 3: Total pairs are (8).