Which of the following Math relation is transitive but neither reflexive nor symmetric?

Question

BlackTom AI · Accepted Answer

The '<' relation on a set (typically numbers) is transitive but neither reflexive nor symmetric.

1. Relevant concepts:
- Reflexive: For all a, aRa must hold. For '<' this would require a < a for every a.
- Symmetric: For all a,b, if aRb then bRa. For '<' this would require that whenever a < b, then b < a.
- Transitive: For all a,b,c, if aRb and bRc then aRc. For '<' this requires that if a < b and b < c, then a < c.

2. Step-by-step reasoning:
- Reflexivity check: Pick any element a. Is a < a true? No — a number is never strictly less than itself. Therefore '<' is not reflexive.
- Symmetry check: Suppose a < b. Does that imply b < a? No — if 2 < 5, it is false that 5 < 2. Thus '<' is not symmetric.
- Transitivity check: Suppose a < b and b < c. By the ordered property of real numbers (or any strict total order), a < c. For example, 1 < 3 and 3 < 7 imply 1 < 7. Hence '<' is transitive.

3. Intermediate logical steps shown:
- Reflexive failure: ∃a such that not (a < a) — in fact for all a, not (a < a).
- Symmetric failure: ∃a,b with a < b and not(b < a) — e.g., a=2, b=5.
- Transitive verification: For arbitrary a,b,c, (a < b) ∧ (b < c) ⇒ a < c by order axioms.

4. Why this is the correct answer:
Because '<' satisfies the transitive property while explicitly failing reflexivity and symmetry as demonstrated by the general logical checks and concrete counterexamples.

5. Why other common relations would be incorrect:
- '≤' is reflexive (a ≤ a) so it fails the "neither reflexive" requirement.
- '=' is reflexive and symmetric, so it fails both requirements.
(If other choices were given, they can be judged similarly.)

6. Conclusion:
Therefore the relation '<' is transitive but neither reflexive nor symmetric, matching the provided answer "<".

Which of the following Math relation is transitive but neither reflexive nor symmetric?Single choice

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