Introduction to Relations

Step 1: If ((a,b)\in R-S), then ((a,b)\in R) and ((a,b)\notin S). Step 2: Since (R) is symmetric, ((b,a)\in R); since (S) is symmetric, ((b,a)\in S) would imply ((a,b)\in S). Step 3: Hence ((b,a)\notin S), so ((b,a)\in R-S).

Step 1: Since both relations are reflexive, all diagonal pairs lie in both (R) and (S). Step 2: In (R-S), pairs that also lie in (S) are removed. Step 3: Therefore no diagonal pair remains in (R-S), so it cannot be reflexive.

Step 1: In a symmetric relation, the reverse of every pair is already in the relation. Step 2: Hence (R^{-1}=R). Step 3: Therefore (R\cup R^{-1}=R\cup R=R).

Step 1: If ((a,b)\in R\cap R^{-1}), then ((a,b)\in R) and ((a,b)\in R^{-1}). Step 2: This means ((b,a)\in R) as well. Step 3: Hence ((b,a)) also lies in (R\cap R^{-1}), so the relation is symmetric.

Step 1: If ((a,b)\in R\cup R^{-1}), it lies in (R) or in (R^{-1}). Step 2: In either case, the reverse pair ((b,a)) lies in the other or same part of the union. Step 3: Therefore (R\cup R^{-1}) is always symmetric.

Step 1: If ((a,b)\in R), reflexivity gives ((b,b)\in R). Step 2: The pairs ((a,b)) and ((b,b)) imply ((a,b)\in R\circ R). Step 3: Thus every pair of (R) also belongs to (R\circ R).

Step 1: ((a,c)\in R\circ R) means there is some (b) such that ((a,b)\in R) and ((b,c)\in R). Step 2: Since (R) is transitive, ((a,c)\in R). Step 3: Hence (R\circ R\subseteq R).

Step 1: For composition, we need linked pairs of the form ((a,b)) and ((b,c)). Step 2: Here ((1,2)) and ((2,3)) combine to give ((1,3)). Step 3: No other linked chain exists, so (R\circ R={(1,3)}).

Step 1: ((1,2)) and ((2,1)) are linked pairs. Step 2: Their composition gives ((1,1)). Step 3: A pair enters (R\circ R) when a suitable middle element exists.

Step 1: The relation contains both ((1,2)) and ((2,3)). Step 2: Transitivity then requires ((1,3)). Step 3: Since ((1,3)) is absent, the relation is not transitive.

Step 1: The (5) diagonal pairs are independently optional. Step 2: For (\frac{5\cdot4}{2}=10) unordered pairs of distinct elements, each gives three choices. Step 3: Hence the total number is (2^5\cdot3^{10}).

Step 1: Irreflexivity forbids all diagonal pairs. Step 2: Symmetry leaves (6) independent off-diagonal reverse-pair groups. Step 3: Removing the empty relation from (2^6) gives (2^6-1).

Option B is correct. If a relation is symmetric, the presence of (a,b) forces (b,a). Antisymmetry forbids both of these when a and b are distinct, so no off-diagonal pair can be included. Only the four diagonal pairs (1,1),(2,2),(3,3),(4,4) may occur, and including all of them gives the greatest possible size: 4.

Step 1: Reflexivity forces all diagonal pairs to be present. Step 2: Symmetry together with antisymmetry allows no pair between distinct elements. Step 3: Only the identity relation is possible, so the number is (1).

Step 1: An equivalence relation partitions a set into disjoint classes. Step 2: The number of partitions of a (5)-element set is (52). Step 3: Therefore the number of equivalence relations is (52).

Step 1: Exactly two classes means splitting (5) elements into two non-empty parts. Step 2: There are (2^5-2=30) non-empty proper subsets, but each partition is counted twice. Step 3: Hence the number is (\frac{30}{2}=15).

Step 1: A three-element class can be chosen in (\binom{6}{3}=20) ways. Step 2: The remaining three elements automatically form the other class. Step 3: Since interchanging the two classes does not create a new partition, the count is (\frac{20}{2}=10).

Step 1: We must split six elements into three two-element classes. Step 2: The count is (\frac{6!}{(2!)^3\cdot3!}). Step 3: This equals (15), so there are (15) such equivalence relations.

Step 1: Treat (1,2,3) as one combined block. Step 2: Now there are three units: ({1,2,3},4,5). Step 3: The number of partitions of three units is (5), so the answer is (5).

Step 1: First treat (1) and (2) as one block. Step 2: Element (3) must not be in that block, while (4) and (5) may be placed freely. Step 3: Partitions with (1,2) together are (15), and those with (1,2,3) together are (5), so the answer is (15-5=10).