Introduction to Relations

Step 1: An equivalence relation needs reflexive, symmetric, and transitive properties. Step 2: If even one of these is missing, it is not an equivalence relation. Step 3: In exams once one property fails, you can conclude it is not equivalence.

Step 1: Symmetry is necessary for an equivalence relation. Step 2: Here ((1,2)) is present but ((2,1)) is missing, so symmetry fails. Step 3: In equivalence relation questions check reverse pairs for non-diagonal pairs.

Step 1: An equivalence relation requires all three properties together. Step 2: Since transitivity is missing, the relation is not an equivalence relation. Step 3: Even if two properties hold, the third must be checked.

Step 1: Diagonal pairs are the same as their reverse. Step 2: The non-diagonal pair ((1,3)) appears with ((3,1)). Step 3: Pairs appearing with their reverses show symmetry.

Step 1: In ((1,2)) and ((2,2)), the middle element (2) matches. Step 2: Transitivity requires ((1,2)), which is already present. Step 3: Sometimes the required pair is already in the relation.

Step 1: Transitivity needs the third pair from connected pairs. Step 2: The required pairs ((1,1),(1,2),(2,2)) are already present. Step 3: A reverse pair is not required for transitivity.

Step 1: Reflexivity requires each element to be paired with itself. Step 2: Here ((1,1)) and ((2,2)) are missing, so it is not reflexive. Step 3: Being symmetric does not automatically mean being reflexive.

Step 1: Each element is paired with itself. Step 2: Therefore the relation is reflexive. Step 3: A relation with only diagonal pairs can still be reflexive.

Step 1: A diagonal pair ((a,a)) remains ((a,a)) after reversing. Step 2: All given pairs are diagonal so symmetry is satisfied. Step 3: Diagonal pairs do not create any problem for symmetry.

Step 1: In transitivity, ((a,a)) and ((a,a)) require ((a,a)). Step 2: All such required diagonal pairs are already present. Step 3: The identity relation is a simple example of transitivity.

Step 1: Elements of the domain become first components. Step 2: Elements of the range become second components, so ((1,3)) may be possible. Step 3: While forming a pair take the first position from the domain and the second from the range.

Step 1: The domain is formed from first components. Step 2: Here the first components are (1,2,3,4). Step 3: In ordered pairs choose the left component for the domain.

Step 1: The range is formed from second components. Step 2: Here the second components are (2,4,6,8). Step 3: Read the right components to find the range quickly.

Step 1: To be a relation on (A), all pairs must come from (A\times A). Step 2: Both ((1,2)) and ((2,3)) are pairs from (A\times A). Step 3: A relation does not need to contain all possible pairs.

Step 1: Both components of a pair in (A\times A) must belong to (A). Step 2: In ((3,1)), (3) is not an element of (A). Step 3: In Cartesian products check membership of every component.

Step 1: Reflexivity needs the diagonal pairs for (2,4,6). Step 2: ((2,2),(4,4),(6,6)) are present so reflexivity is visible. Step 3: Check the diagonal pair for every element of the base set.

Step 1: ((1,2)) and ((2,1)) are reverse pairs. Step 2: Diagonal pairs remain the same when reversed. Step 3: Therefore the symmetric property holds here.

Step 1: Transitivity requires ((a,c)) from ((a,b)) and ((b,c)). Step 2: From ((1,2)) and ((2,3)), ((1,3)) is needed and it is present. Step 3: In connected pairs identify the common middle element.

Step 1: A symmetric relation contains the reverse of a pair. Step 2: The reverse of ((5,7)) is ((7,5)). Step 3: For symmetry remember to interchange the two components.

Step 1: In a transitive relation, ((a,b)) and ((b,c)) require ((a,c)). Step 2: From ((2,5)) and ((5,9)), ((2,9)) must be present. Step 3: Remove the common middle element to form the new pair.