Introduction to Relations

Step 1: A minimal element has no distinct element below it. Step 2: There is no distinct element of (A) below (2). Step 3: Each of (4,6,12,18) has (2) or another element below it, so there is only (1) minimal element.

Step 1: Two elements are comparable if one divides the other. Step 2: The comparable pairs are ({1,2},{1,3},{1,6},{2,6},{3,6}). Step 3: ({2,3}) is not comparable, so the count is (5).

Step 1: In a chain, every two elements must be comparable. Step 2: (1\mid2\mid6) and (1\mid3\mid6) are chains of three elements. Step 3: Since (2) and (3) are not comparable, a four-element chain is impossible.

Step 1: An upper bound must be divisible by both (2) and (8). Step 2: (8) and (16) are upper bounds. Step 3: Under divisibility, the least among them is (8).

Step 1: A lower bound must divide both (4) and (16). Step 2: (1,2,4) are lower bounds. Step 3: Under divisibility, the greatest among them is (4).

Step 1: Reflexivity gives (R\subseteq R\circ R). Step 2: Transitivity gives (R\circ R\subseteq R). Step 3: Combining both inclusions gives (R\circ R=R).

Step 1: ((1,2)) and ((2,3)) give ((1,3)). Step 2: ((2,3)) and ((3,4)) give ((2,4)). Step 3: There is no other two-step chain, so this is (R\circ R).

Step 1: A triple composition represents a three-step chain. Step 2: Here (1\to2\to3\to4) is the only three-step chain. Step 3: Therefore (R\circ R\circ R={(1,4)}).

Step 1: (1\to2\to1) gives ((1,1)), and (1\to2\to3) gives ((1,3)). Step 2: (3\to2\to1) gives ((3,1)). Step 3: There is no two-step chain from (1) to (2), so ((1,2)) is not in (R\circ R).

Step 1: For a symmetric relation, (R^{-1}=R). Step 2: So in (R\circ R^{-1}), we may replace (R^{-1}) by (R). Step 3: Hence (R\circ R^{-1}=R\circ R).

Step 1: Taking the inverse reverses every ordered pair. Step 2: Reversing twice brings each pair back to its original direction. Step 3: Therefore ((R^{-1})^{-1}=R).

Step 1: A pair in the union comes from (R) or from (S). Step 2: Its reverse will then lie in (R^{-1}) or in (S^{-1}). Step 3: Hence ((R\cup S)^{-1}=R^{-1}\cup S^{-1}).

Step 1: A pair is in the intersection only when it belongs to both (R) and (S). Step 2: Its reverse then belongs to both (R^{-1}) and (S^{-1}). Step 3: Therefore ((R\cap S)^{-1}=R^{-1}\cap S^{-1}).

Step 1: If ((a,b)\in R^{-1}), then ((b,a)\in R). Step 2: Since (R\subseteq S), ((b,a)\in S). Step 3: Therefore ((a,b)\in S^{-1}), so (R^{-1}\subseteq S^{-1}).

Step 1: Transitivity gives (R\circ R\subseteq R). Step 2: (R^3=R\circ R\circ R). Step 3: Combining the first two factors into (R) shows (R^3\subseteq R).

Step 1: ((1,2)) and ((2,1)) are in the relation. Step 2: Transitivity requires ((1,1)) from these two pairs. Step 3: Therefore ((1,1)) is a compulsory pair to add.

Step 1: There is a chain (1\to2), (2\to3), and (3\to4). Step 2: The transitive closure adds all reachable pairs. Step 3: Since (4) is reachable from (1), ((1,4)) will be included.

Option A is correct. In an equivalence relation, transitivity applies to (1,2) and (2,3), so (1,3) must be included. Symmetry also adds the reverse pairs, and reflexivity adds all diagonal pairs. Elements 1, 2, and 3 therefore form one equivalence class, while 4 remains separate; hence pairs joining 4 to another element are not forced.

Step 1: Elements (1,2,3) are connected into one class. Step 2: Elements (4,5) form another class. Step 3: The first class gives (3^2=9) pairs and the second gives (2^2=4), so the total is (13).

Option D is correct. The given pairs connect 1, 2, and 3, and closure under reflexivity, symmetry, and transitivity makes them one equivalence class. Element 4 is not connected to that class, so the smallest equivalence relation contains no pair from 4 to 1, 2, or 3, including (1,4). The first three options are already given explicitly.