Question:easy

The relation \(R=\{(a,b): a\le b^2\}\) is defined in the set of real numbers. Then \(R\) is:

Show Hint

Test a=0.5 for reflexivity, (0,10) vs (10,0) for symmetry, and (10,4,2) for transitivity — all fail.
Updated On: Sep 23, 2026
  • Reflexive and symmetric but not transitive
  • Reflexive and transitive but not symmetric
  • Symmetric and transitive but not reflexive
  • Not reflexive, not symmetric and not transitive
Show Solution

The Correct Option is D

Solution and Explanation

Step 1: Reflexivity via the boundary a(a-1)\(\ge\)0:
$a\le a^2 \iff a^2-a\ge0 \iff a\in(-\infty,0]\cup[1,\infty)$. Any $a$ strictly between $0$ and $1$, e.g. $a=1/2$, breaks reflexivity since $1/2\le1/4$ is false.

Step 2: Symmetry via a large-small pair:
Pick $a$ small and $b$ large so $a\le b^2$ easily holds, e.g. $a=-5,\,b=1$: $-5\le1$ holds. Now check the reverse $b\le a^2$: $1\le25$ — this particular pair happens to hold, so try instead $a=0,b=10$: forward holds ($0\le100$), reverse fails ($10\le0$ is false). Symmetry breaks.

Step 3: Transitivity via a squeeze example:
Choose $b$ with a moderate square and $c$ with a small square so the chain breaks: $a=10\le b^2=16$ and $b=4\le c^2=4$, yet $a=10\le c^2=4$ fails.

Final Answer:
All three properties fail on at least one explicit example. \[ \boxed{\text{Not reflexive, not symmetric, not transitive}} \]
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