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Question 5

Check whether the relation R in R defined as R = {(a, b): a ≤ b3} is reflexive, symmetric or transitive.

Answer

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.

 

 

 

Popular Questions of Class 12th mathematics

 

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Let A = R – {3} and B = R – {1}. Consider the function  f : A → B defined by

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