Sets, Relations & Functions
General
Grade 11
Question:
<p>Let R1 = {(a, b) ∈R2 : a2 + b2 ∈Q} and R2 = {(a, b) ∈R2 : a2 + b2 /∈Q}, where Q is the
set of rationals. Then:</p>
<p>R2 is transitive but R1 is not</p>
<p>R1 is transitive but R2 is not</p>
<p>R1 and R2 are both transitive</p>
<p>Neither R1 nor R2 is transitive</p>
Step-by-Step Solution
Key Concept: Key identity: a2 + c2 = (a2 + b2) + (b2 + c2) -2b2. Even if the first two sums are rational, 2b2
may be irrational — breaking transitivity for both.
<p><strong>Step 1</strong>: R1 not transitive. Need: if a2 + b2 \in Q and b2 + c2 \in Q, is a2 + c2 \in Q?</p><br>a2 + c2 = (a2 + b2)<br>|<br>{z<br>}<br>\in Q<br>+ (b2 + c2)<br>|<br>{z<br>}<br>\in Q<br>-2b2.<br>If b2 /\in Q then 2b2 /\in Q, so a2 + c2 may be irrational.<br>Counterexample: a =<br>p<br>2 +<br>\sqrt<br>3, b =<br>p<br>2 -<br>\sqrt<br>3, c = a.<br>• a2 + b2 = (2 +<br>\sqrt<br>3) + (2 -<br>\sqrt<br>3) = 4 \in Q ✓<br>• b2 + c2 = 4 \in Q ✓<br>• a2 + c2 = 2(2 +<br>\sqrt<br>3) = 4 + 2<br>\sqrt<br>3 /\in Q ✗<br>R1 is NOT transitive.<p><strong>Step 2</strong>: R2 not transitive. Using the same identity, irrational -2b2 can cause the difference to become rational,</p><br>so a2 + c2 may fall in Q even when both inputs are not. R2 also NOT transitive.
Correct Answer: 4