<p>Let R = {(a, b) : 3a −3b +
√
7 is irrational} on R. Then R is:</p>
Reflexive but neither symmetric nor transitive
Reflexive and transitive but not symmetric
Reflexive and symmetric but not transitive
An equivalence relation
Step-by-Step Solution
Key Concept: 3a -3b +
\sqrt{7 = 3(a -b) +}
\sqrt{7. Reflexivity holds trivially since}
\sqrt{7 is irrational. Both symmetry}
and transitivity fail via carefully chosen real numbers.
<p><strong>Step 1</strong>: Reflexive: 3(a -a) +</p><br>\sqrt<br>7 =<br>\sqrt<br>7, which is irrational. So (a, a) \in R for all a \in R. ✓<p><strong>Step 2</strong>: Symmetry fails: Choose a -b =</p><br>\sqrt<br>7<br>3<br>(a specific real).<br>• Forward: 3 \cdot <br>\sqrt<br>7<br>3 +<br>\sqrt<br>7 = 2<br>\sqrt<br>7 — irrational ✓, so (a, b) \in R.<br>• Reverse: 3 \cdot <br><br>-<br>\sqrt<br>7<br>3<br>!<br>+<br>\sqrt<br>7 = 0 — rational ✗, so (b, a) /\in R.<br>R is NOT symmetric.<p><strong>Step 3</strong>: Transitivity fails: Take a -b =</p><br>\sqrt<br>7<br>3<br>(so (a, b) \in R) and b -c = -2<br>\sqrt<br>7<br>3<br>(check: 3 \cdot (-2<br>\sqrt<br>7<br>3 ) +<br>\sqrt<br>7 = -<br>\sqrt<br>7<br>— irrational ✓, so (b, c) \in R). Then a -c =<br>\sqrt<br>7<br>3 -2<br>\sqrt<br>7<br>3<br>= -<br>\sqrt<br>7<br>3 , so 3(a -c) +<br>\sqrt<br>7 = -<br>\sqrt<br>7 +<br>\sqrt<br>7 = 0 — rational.<br>Thus (a, c) /\in R. ✗R is NOT transitive.
Correct Answer: 1