Introduction to Relations

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.

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).

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.