Introduction to Relations

Step 1: The sum is even only when both numbers have the same parity. Step 2: The odd class ({1,3}) gives (4) pairs and the even class ({2,4}) gives (4) pairs. Step 3: Total pairs are (4+4=8).

Step 1: Since (a+b=b+a), if ((a,b)) is present, ((b,a)) is also present. Step 2: For ((a,a)), (2a) is even, so no self-pair appears. Step 3: Hence it is symmetric but not reflexive.

Step 1: ((a,b)) appears only when (a) divides (b). Step 2: (4) does not divide (2). Step 3: In divisibility relations, the order of the pair is very important.

Step 1: The least element divides every element, so (1) is least. Step 2: The greatest element is divisible by every element, so (6) is greatest. Step 3: In divisibility, least and greatest are judged by the relation, not by usual size only.

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 no element divides all elements.

Step 1: Every number divides itself, so it is reflexive. Step 2: If (a) divides (b) and (b) divides (a), then (a=b), so it is antisymmetric. Step 3: Divisibility is transitive, so it is a partial order relation.

Step 1: All self-pairs are present, so it is reflexive. Step 2: No reverse pair appears with a distinct pair, so it is antisymmetric. Step 3: ((1,2)) and ((2,3)) require ((1,3)), which is present, so it is transitive.

Step 1: A partial order needs antisymmetry. Step 2: Here (1\ne2), yet both ((1,2)) and ((2,1)) are present. Step 3: Two-way pairs between distinct elements break antisymmetry.

Step 1: In (R^{-1}), every ordered pair is reversed. Step 2: If the inverse is the same relation, every reverse pair is already present. Step 3: This is exactly the identity of a symmetric relation.

Step 1: In (R^{-1}\circ R), ((a,c)) appears when some (b) satisfies ((a,b)\in R) and ((b,c)\in R^{-1}). Step 2: For (a=1), ((1,2)\in R) and ((2,1)\in R^{-1}). Step 3: Hence ((1,1)) belongs to this composition.

Step 1: In an inverse relation, the two entries of every pair are interchanged. Step 2: ((2,4)) gives ((4,2)), and ((3,4)) gives ((4,3)). Step 3: Reverse every pair separately while finding the inverse.

Step 1: (A\times A) contains all possible pairs. Step 2: Therefore, it cannot be empty; it is the universal relation. Step 3: The universal relation is reflexive, symmetric and transitive.

Step 1: On a non-empty set, reflexivity needs self-pairs. Step 2: The empty relation has no pair, so self-pairs are absent. Step 3: The empty relation may be symmetric and transitive, but it is not reflexive on a non-empty set.

Step 1: Antisymmetry fails when both directions exist for distinct elements. Step 2: Here ((1,2)) is present but ((2,1)) is not, and ((1,3)) is present but ((3,1)) is not. Step 3: So antisymmetry is not violated.

Step 1: If ((a,b)) is present and the relation is symmetric, then ((b,a)) is also present. Step 2: Antisymmetry says both can occur only when (a=b). Step 3: For (a\ne b), this is impossible, so such a pair cannot be present.

Step 1: A partial order needs reflexivity, antisymmetry and transitivity. Step 2: ((1,2)) and ((2,3)) are present but ((1,3)) is missing. Step 3: Thus transitivity is absent, so it is not a partial order.

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

Step 1: (|a-a|=0), so the relation is reflexive. Step 2: (|a-b|=|b-a|), so it is symmetric. Step 3: ((1,2)) and ((2,3)) are present, but ((1,3)) is absent because (|1-3|=2), so transitivity fails.

Step 1: (|a-a|=0) is divisible by (2), so it is reflexive. Step 2: (|a-b|=|b-a|), so it is symmetric. Step 3: Same parity continues through a chain, so it is transitive.

Step 1: If (a+b=5), then (b+a=5), so the relation is symmetric. Step 2: For ((a,a)), we need (2a=5), which is not true for any integer in the set. Step 3: Hence the relation is symmetric but not reflexive.