Transitive relation

Step 1: If (a^2=b^2) and (b^2=c^2), then (a^2=c^2). Step 2: Hence (aRc), so the relation is transitive. Step 3: In equality-based relations, connect the equal quantities.

Step 1: If (a=b+1) and (b=c+1), then (a=c+2). Step 2: Transitivity would require (a=c+1), which is generally false. Step 3: For algebraic relation conditions, substitute carefully.

Step 1: (a-b\ge 0) means (a\ge b). Step 2: If (a\ge b) and (b\ge c), then (a\ge c), so (a-c\ge 0). Step 3: First convert the condition into a simple inequality.

Step 1: If (A\subseteq B) and (B\subseteq C), every element of (A) is also in (C). Step 2: Hence (A\subseteq C), so the relation is transitive. Step 3: In subset questions, follow the chain of inclusion.

Step 1: Two sets can be disjoint, and the second can be disjoint from the third. Step 2: Still, the first and third need not be disjoint. For example, ({1}) and ({2}) are disjoint, ({2}) and ({1}) are disjoint, but ({1}) and ({1}) are not disjoint. Step 3: For set relations, build a small counterexample.

Step 1: If (|a|=|b|) and (|b|=|c|), then (|a|=|c|). Step 2: Therefore ((a,c)) belongs to the relation. Step 3: Treat equal absolute values like a chain of equalities.

Step 1: Take (a=0), (b=0.6), and (c=1.2). Step 2: (|0-0.6|<1) and (|0.6-1.2|<1), but (|0-1.2|<1) is false. Step 3: In distance-based relations, two small distances may combine into a larger one.

Step 1: From (a\le b+2) and (b\le c+2), we only get (a\le c+4). Step 2: Transitivity requires (a\le c+2), which need not hold. Take (a=4), (b=2), (c=0). Step 3: When adding inequalities, the bound may change.

Step 1: If (a\le b) and (b\le c), then by order (a\le c). Step 2: Hence ((a,c)) is also in the relation. Step 3: The same inequality rule applies even on a finite set.

Step 1: ((1,2)) and ((2,3)) require ((1,3)). Step 2: ((2,3)) and ((3,4)) require ((2,4)), and then ((1,3)) and ((3,4)) require ((1,4)). Step 3: In a long chain, both short and long jumps are needed.

Step 1: Extra pairs in (S) can create new chains. Step 2: If the required direct pair for a new chain is missing in (S), then (S) is not transitive. Step 3: A property does not automatically pass from a subset relation to a larger relation.

Step 1: The smaller relation (R) may lose a required direct pair from (S). Step 2: For example, (S={(1,2),(2,3),(1,3)}) is transitive, but (R={(1,2),(2,3)}) is not. Step 3: A subrelation must be checked separately.

Step 1: ((2,3)) and ((3,2)) are in the relation. Step 2: Transitivity requires ((2,2)), but it is missing. Step 3: Always check the self-pair produced by reverse pairs.

Step 1: ((3,2)) and ((2,3)) require ((3,3)), which is present. Step 2: ((1,3)) and ((3,2)) require ((1,2)), also present; other self-pair checks do not create a missing pair. Step 3: After full checking, the relation is transitive, so saying not transitive is wrong.

Step 1: The definition of transitivity applies to every linked pair. Step 2: So for every ((a,b)) and ((b,c)), ((a,c)) must be found. Step 3: Making a table reduces mistakes in hard questions.

Step 1: ((a,b)) and ((b,a)) imply ((a,a)). Step 2: ((b,a)) and ((a,b)) imply ((b,b)). Step 3: When reverse pairs exist, both self-pairs become necessary.

Step 1: ((1,2)) and ((2,1)) require ((1,1)) for transitivity. Step 2: If ((1,1)) is absent, the condition fails. Step 3: This missing pair gives an immediate counterexample.

Step 1: To break transitivity, two pairs like ((a,b)) and ((b,c)) are needed. Step 2: The empty relation has no pairs, so no counterexample exists. Step 3: The condition is considered vacuously true.

Step 1: To test transitivity, the first element of the second pair must match the second element of the first pair. Step 2: ((1,2)) and ((3,4)) do not form any linked chain. Step 3: Without a linked chain, no required pair is missing.

Step 1: ((1,3)) and ((3,4)) require ((1,4)). Step 2: ((1,4)) is not in the list, while the other main requirements are satisfied. Step 3: Also use already formed long jumps in further checking.

Step 1: Take (a=0), (b=1), and (c=2). Step 2: (0<1<2) gives (0R1), and (1<2<3) gives (1R2), but (0<2<2) is false, so (0R2) fails. Step 3: In open-bound relations, watch equality at the boundary.

Step 1: If (a-b) is rational and (b-c) is rational, their sum (a-c) is also rational. Step 2: Hence (aRc) is true. Step 3: In difference-based relations, adding differences is the key idea.

Step 1: The sum of two irrational differences is not always irrational. Step 2: Take (a=\sqrt{2}), (b=0), (c=\sqrt{2}); (a-b) and (b-c) are irrational, but (a-c=0) is rational. Step 3: Build counterexamples carefully in irrational-number relations.

Step 1: ((1,2)) and ((2,1)) require ((1,1)), which is present. Step 2: ((2,1)) and ((1,2)) require ((2,2)), also present. The other self-pairs satisfy their own requirements. Step 3: Reverse pairs can still be transitive if both self-pairs are present.

Step 1: An equivalence relation includes reflexive, symmetric, and transitive properties. Step 2: Therefore such a relation must be transitive. Step 3: When you hear equivalence relation, remember all three properties.