Introduction to Relations

Step 1: For (a=1), (5) values work; for (a=2), (4); and for (a=3), (3). Step 2: Similarly, for (a=4), (2) values work, and for (a=5), (1) value works. Step 3: Total pairs are (5+4+3+2+1=15).

Step 1: For (a=1), there are (4) larger values. Step 2: For (a=2,3,4,5), the counts are (3,2,1,0). Step 3: Total pairs are (4+3+2+1=10).

Step 1: For (a=1), there are (5) choices. Step 2: For (a=2,3,4,5), the choices are (4,3,2,1). Step 3: Total pairs are (5+4+3+2+1=15).

Step 1: (1) divides all (6) elements. Step 2: (2) divides (2,4,6,12); (3) divides (3,6,12); (4) divides (4,12); (6) divides (6,12); and (12) divides (12). Step 3: The total is (6+4+3+2+2+1=18), so the correct answer is (18).

Step 1: An upper bound is an element divisible by both (3) and (4). Step 2: In (A), (12) is such an element. Step 3: No smaller upper bound exists in this order, so the least upper bound is (12).

Step 1: A lower bound must divide both (4) and (6). Step 2: Both (1) and (2) are lower bounds. Step 3: Under divisibility, (2) is greater than (1), so the greatest lower bound is (2).

Step 1: A least element must divide every element of the set. Step 2: (2) does not divide (3), and (3) does not divide (2). Step 3: No single element divides all elements, so there is no least element.

Step 1: A greatest element is one that is divisible by every element of the set. Step 2: (2,3,6,9,18) all divide (18). Step 3: Therefore (18) is the greatest element in this order.

Step 1: In the divisibility order, (2\mid4\mid8\mid16) forms a full chain. Step 2: All four elements are ordered one after another. Step 3: Hence the longest chain contains (4) elements.

Step 1: Two elements are comparable if one divides the other. Step 2: (2) does not divide (3), and (3) does not divide (2). Step 3: Therefore (2) and (3) are not comparable.

Step 1: (1) and (2) must be in the same class. Step 2: (3) remains as a singleton class. Step 3: The class ({1,2}) gives (2^2=4) pairs, and ({3}) gives (1) pair, making (5) pairs total.

Step 1: (1,2,3) get connected into one class. Step 2: (4) remains a singleton class. Step 3: The class ({1,2,3}) gives (3^2=9) pairs and ({4}) gives (1), so the total is (10).

Step 1: ((1,2)) and ((2,3)) require ((1,3)) for transitivity. Step 2: ((4,4)) only requires itself, which is already present. Step 3: Hence the minimal new pair is ((1,3)).

Step 1: ((1,2)) and ((2,1)) require ((1,1)). Step 2: ((2,1)) and ((1,2)) require ((2,2)). Step 3: No pair involving (3) creates a chain, so ((3,3)) is not required for transitivity.

Step 1: The original three pairs are already present. Step 2: Transitivity adds ((1,3)), ((2,4)), and ((1,4)). Step 3: Thus the transitive closure contains (3+3=6) pairs.

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.