Transitive relation

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.

Step 1: (a\mid b) means (b=ak) for some natural number (k). Step 2: If (b\mid c), then (c=bl), so (c=a(kl)), hence (a\mid c). Step 3: In divisibility, connect the multipliers.

Step 1: If (a^3=b^3) and (b^3=c^3), then (a^3=c^3). Step 2: Therefore ((a,c)) is also in the relation. Step 3: In equality of powers, connect through the middle equality.

Step 1: ((1,2)) and ((2,3)) require ((1,3)), which is present. Step 2: ((2,3)) and ((3,2)) require ((2,2)), and ((3,2)) and ((2,3)) require ((3,3)), both present. Step 3: In hard lists, conclude only after pair-by-pair checking.

Step 1: ((1,2)) and ((2,3)) require ((1,3)). Step 2: ((1,3)) is missing, so the relation is not transitive, even though ((2,2)) is present. Step 3: One missing required pair is enough.

Step 1: Hard questions often contain many hidden chains. Step 2: Even if the first chain works, another chain may reveal a missing pair. Step 3: Checking the whole list systematically is the safest method.

Step 1: ((1,2)) and ((2,4)) require ((1,4)), which is present. Step 2: ((2,1)) and ((1,2)) require ((2,2)), and ((1,2)) and ((2,1)) require ((1,1)), both present. Step 3: Check reverse pairs and forward chains together.

Step 1: ((1,2)) and ((2,4)) require ((1,4)). Step 2: ((1,4)) is not in the list, so the relation is not transitive. Step 3: Use already present derived-looking pairs in further chains too.

Step 1: If (a-b) and (b-c) are both divisible by (6), then their sum (a-c) is also divisible by (6). Step 2: Hence (aRc), so the relation is transitive. Step 3: Adding differences is the most useful method here.

Step 1: If (a^2\le b^2) and (b^2\le c^2), then by order of inequalities (a^2\le c^2). Step 2: Hence (aRc) is true, so the relation is transitive. Step 3: Do not get confused by signs when comparing squares.

Step 1: If (a^2<b^2) and (b^2<c^2), then (a^2<c^2). Step 2: Thus ((a,c)) also belongs to the relation. Step 3: For strict inequalities, connect the middle inequality carefully.

Step 1: From (a=b^2) and (b=c^2), we get (a=c^4). Step 2: Transitivity would require (a=c^2), which is not generally true. Take (c=2), (b=4), (a=16). Step 3: For algebraic relations, substitute and compare the required condition.

Step 1: If (A\supseteq B) and (B\supseteq C), then (A) contains every element of (C). Step 2: Hence (A\supseteq C), so the relation is transitive. Step 3: Follow the chain from larger set to smaller set carefully.

Step 1: If (A\subset B) and (B\subset C), every element of (A) is in (C). Step 2: Also (A) and (C) cannot be equal, so (A\subset C). Step 3: In proper subset questions, think separately about equality.

Step 1: ((1,4)) and ((4,2)) require ((1,2)), which is present. Step 2: ((4,4)) with ((4,2)) requires ((4,2)), and requirements with ((2,2)) are also satisfied. Step 3: Do not ignore self-pairs, but their requirements are often already present.

Step 1: ((1,2)) and ((2,1)) require ((1,1)). Step 2: ((1,1)) is not in the given list, so the relation is not transitive. Step 3: In cycle-like pairs, quickly look for required self-pairs.