<p>Let R1 and R2 be defined on R by a R1 b ⇔ab ≥0 and a R2 b ⇔a ≥b. Then:</p>
R1 is an equivalence relation but not R2
R2 is an equivalence relation but not R1
Both R1 and R2 are equivalence relations
Neither R1 nor R2 is an equivalence relation
Step-by-Step Solution
Key Concept: On all of R (including 0), R1 fails transitivity via a = 1, b = 0, c = -1. Meanwhile R2 is a total
order — it lacks symmetry and is never an equivalence relation.
<p><strong>Step 1</strong>: Analyse R1: ab \geq0.</p><br>• Reflexive: a \cdot a = a2 \geq0 ✓<br>• Symmetric: ab \geq0 \Leftrightarrow ba \geq0 ✓<br>• Transitivity FAILS on R: Take a = 1, b = 0, c = -1.<br>ab = 1 \cdot 0 = 0 \geq0 ✓,<br>bc = 0 \cdot (-1) = 0 \geq0 ✓,<br>but ac = 1 \cdot (-1) = -1 < 0. ✗<br>The element b = 0 acts as a “bridge” that both signs can cross, breaking transitivity. R1 is NOT an equivalence<br>relation on R.<p><strong>Step 2</strong>: Analyse R2: a \geqb.</p><br>• Reflexive: a \geqa ✓<br>• Symmetry FAILS: 3 \geq2 but 2 ̸\geq3. ✗<br>R2 is a total order (reflexive, antisymmetric, transitive) — never an equivalence relation.
Correct Answer: 4