Types of relations

Correct answer: B. (2^{12}-1)

Step 1: The number of reflexive relations is (2^{16-4}=2^{12}). Step 2: The identity relation is one of these reflexive relations. Step 3: Removing it gives (2^{12}-1).

Correct answer: A. (2^9-2^6)

Step 1: The total number of relations is (2^9). Step 2: The number of reflexive relations is (2^{9-3}=2^6). Step 3: Therefore the number of non-reflexive relations is (2^9-2^6).

Correct answer: A. It will be equal to (R)

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

Correct answer: A. It is symmetric

Step 1: If ((a,b)) is in (R\cap R^{-1}), it is in both (R) and (R^{-1}). Step 2: This ensures ((b,a)) is also in both (R) and (R^{-1}). Step 3: Hence every pair in the intersection has its reverse, so it is symmetric.

Correct answer: A. It is symmetric

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

Correct answer: A. (R^{-1}) is also an equivalence relation and (R^{-1}=R)

Step 1: An equivalence relation is symmetric, so (R^{-1}=R). Step 2: The same relation remains reflexive, symmetric and transitive. Step 3: Hence (R^{-1}) is also an equivalence relation.

Correct answer: A. ({2,3})

Step 1: Since (2+3=5), (2) and (3) are related. Step 2: Because of (a=b), (2) is related to itself as well. Step 3: The class of (2) is ({2,3}).

Correct answer: C. (12)

Step 1: The equivalence classes are ({1,4},{2,5},{3,6}). Step 2: Each class has size (2), so each contributes (2^2=4) pairs. Step 3: Total pairs are (4+4+4=12).

Correct answer: C. (20)

Step 1: The classes are ({3,6}) and ({1,2,4,5}). Step 2: The first class gives (2^2=4) pairs and the second gives (4^2=16) pairs. Step 3: Total pairs are (4+16=20).

Correct answer: B. Equivalence relation

Step 1: The condition (a=b) includes all self-pairs, so reflexivity holds. Step 2: The condition that both numbers are prime is unchanged by reversing order, so symmetry holds. Step 3: Prime elements ({2,3,5}) form one closed class and the others stay single, so transitivity holds.

Correct answer: A. (R\cap I=I)

Step 1: A reflexive relation (R) contains all self-pairs. Step 2: The identity relation (I) contains only self-pairs. Step 3: Therefore the common part is exactly the identity relation, so (R\cap I=I).

Correct answer: A. It will be symmetric

Step 1: In (R\cup R^{-1}), every pair of (R) is accompanied by its reverse. Step 2: Therefore the reverse of any pair in the union is also present. Step 3: Even if (R) itself is not symmetric, (R\cup R^{-1}) is symmetric.