Introduction to Relations

Step 1: (a+a=2a) is even, so it is not reflexive. Step 2: If (a+b) is odd, then (b+a) is odd, so it is symmetric. Step 3: (1R2) and (2R3) hold, but (1R3) fails, so it is not transitive.

Step 1: Same remainder here means same parity. Step 2: One class is ({1,3,5}) and the other is ({2,4,6}). Step 3: The number of possible remainders often gives the number of classes, but do not count empty classes.

Step 1: In (A), all three remainders (0,1,2) occur modulo (3). Step 2: The classes are ({3}), ({1,4}), and ({2,5}). Step 3: Count only remainders that actually occur in the set.

Step 1: If (a+b\equiv0\pmod3), then (b+a\equiv0\pmod3), so it is symmetric. Step 2: ((1,1)) is absent because (1+1\equiv2\pmod3), so it is not reflexive. Step 3: (1R2) and (2R1) hold, but (1R1) fails, so it is not transitive.

Step 1: The difference is even when the two numbers have the same parity. Step 2: In (A), there are (2) odd and (2) even numbers. Step 3: The number of ordered pairs is (2^2+2^2=8).

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.