Sets, Relations & Functions
General
Grade 11
Question:
<p>Among S = {(a, b) : a, b ∈R \ {0}, 2 + a/b > 0} and T = {(a, b) : a, b ∈R, a2 −b2 ∈Z}:</p>
S is transitive but T is not
T is symmetric but S is not
Neither S nor T is symmetric
Both S and T are symmetric
Step-by-Step Solution
Key Concept: For S: test whether a/b > -2 implies b/a > -2 — it fails when |b/a| is large. For T: negating an
integer gives an integer — immediate symmetry.
<p><strong>Step 1</strong>: S — symmetry fails: Take a = 0.1, b = -1. a/b = -0.1 > -2 ✓, so (0.1, -1) \in S. But b/a = -10 < -2</p><br>✗, so (-1, 0.1) /\in S.<p><strong>Step 2</strong>: T — symmetric: If a2 -b2 \in Z, then b2 -a2 = -(a2 -b2) \in Z (negation of an integer is an integer). ✓</p><p><strong>Step 3</strong>: T is in fact a full equivalence: Reflexive (a2 -a2 = 0 \in Z) ✓; Transitive ((a2 -b2) + (b2 -c2) =</p><br>a2 -c2 \in Z) ✓.
Correct Answer: 2