SELECT * FROM question_mgmt as q WHERE id=965 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=3 AND subId=6 AND chapterId=88 and ex_no='1' AND status=1 ORDER BY CAST(question_no AS UNSIGNED)
R = {(a, b): a ≤ b3}
It is observed that
\begin{align} \left(\frac{1}{2},\frac{1}{2}\right) ∉ R , as \frac{1}{2}>\left(\frac{1}{2}\right)^3 = \frac{1}{8}\end{align}
∴ R is not reflexive.
Now,
(1, 2) ∈ R (as 1 < 23 = 8)
But,
(2, 1) ∉ R (as 23 > 1)
∴ R is not symmetric.
We have
\begin{align} \left(3,\frac{3}{2}\right),\left(\frac{3}{2},\frac{6}{5}\right) ∉ R , as 3>\left(\frac{3}{2}\right)^3 and \frac{3}{2}<\left(\frac{6}{5}\right)^3 \end{align}
But
\begin{align} \left(3,\frac{6}{5}\right) ∉ R , as 3>\left(\frac{6}{5}\right)^3 \end{align}
∴ R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.
Comments
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Report a problem on Specifications: