Introduction to Relations

Step 1: Reflexivity makes (4) self-pairs compulsory. Step 2: There are (\frac{4\cdot3}{2}=6) unordered distinct pairs and each pair-group may be included or excluded. Step 3: Hence the number of relations is (2^6).

Step 1: Reflexivity fixes all (4) self-pairs. Step 2: For each of the (\frac{4\cdot3}{2}=6) unordered distinct pairs, there are three choices. Step 3: Therefore, the count is (3^6).

Step 1: Symmetry asks for reverse pairs. Step 2: Antisymmetry forbids two-way pairs between distinct elements, so only self-pairs may be chosen. Step 3: There are (4) free self-pairs, so the count is (2^4).

Step 1: The original relation has three pairs. Step 2: Transitivity adds ((1,3)), ((2,4)), and ((1,4)). Step 3: Total pairs are (3+3=6).

Step 1: ((1,3)) and ((3,4)) require ((1,4)). Step 2: ((1,2)) and ((2,4)) also require ((1,4)). Step 3: This is the key missing pair, so it must be added.

Step 1: The class ({1,2}) gives (2^2=4) pairs. Step 2: The class ({3,4,5}) gives (3^2=9) pairs. Step 3: Total pairs are (4+9=13).

Step 1: The singleton class gives (1^2=1) pair. Step 2: The two classes with two elements each give (2^2+2^2=8) pairs. Step 3: Total pairs are (1+8=9).

Step 1: The even class is ({2,4,6}) and the odd class is ({1,3,5}). Step 2: Each class has (3) elements, so each gives (3^2=9) pairs. Step 3: Total pairs are (9+9=18).

Step 1: Dividing (2) by (3) gives remainder (2). Step 2: (5) and (8) also leave remainder (2). Step 3: Elements with the same remainder lie in the same class.

Step 1: Dividing (1) by (4) gives remainder (1). Step 2: Dividing (5) by (4) also gives remainder (1). Step 3: Therefore, the class of (1) is ({1,5}).

Step 1: For the sum to be even, both elements must have the same parity. Step 2: The odd class ({1,3,5}) gives (3^2=9) pairs and the even class ({2,4}) gives (2^2=4) pairs. Step 3: Total pairs are (9+4=13).

Step 1: The sum is odd when one element is odd and the other is even. Step 2: There are (3) odd and (2) even elements. Step 3: Both directions give (3\cdot2+2\cdot3=12) pairs.

Step 1: (1) divides all four elements. Step 2: (2) divides (2) and (6); (3) divides (3) and (6); (6) divides only (6). Step 3: Total pairs are (4+2+2+1=9).

Step 1: The greatest element is the one divisible by every element of the set. Step 2: (2,3,4,6,12) all divide (12). Step 3: Hence (12) is the greatest element.

Step 1: The least element divides every element in the set. Step 2: (2) divides (2,4,6,12). Step 3: Therefore, (2) is the least element.

Step 1: A least element must divide every element. Step 2: (2) does not divide (3), and (3) does not divide (2). Step 3: Hence this set has no least element.

Step 1: All self-pairs are present, so it is reflexive. Step 2: No two-way pair exists between distinct elements, so it is antisymmetric. Step 3: ((1,2)) and ((2,4)) require ((1,4)), which is present, so transitivity holds.

Option A is correct because transitivity says that if (a,b) and (b,c) are in R, then (a,c) must also be in R. Here (1,2) and (2,4) are present, but (1,4) is absent. The other statements either use a pair that is not given or demand a conclusion unrelated to transitivity. Thus the relation fails the partial-order test specifically because transitivity fails.

Step 1: All self-pairs are present, so reflexivity holds. Step 2: Both ((1,2)) and ((2,1)) are present while (1\ne2). Step 3: Therefore, antisymmetry definitely fails.

Step 1: By symmetry, ((5,3)\in R) also holds. Step 2: Taking inverse, ((5,3)\in R) gives ((3,5)\in R^{-1}). Step 3: For a symmetric relation, remember (R^{-1}=R).