Question 2

# Show that the relation R in the set R of real numbers, defined as R = {(a, b): a ≤ b^{2}} is neither reflexive nor symmetric nor transitive.

Answer

R = {(*a*, *b*): *a* ≤ *b*^{2}}

It can be observed that

\begin{align} \left(\frac{1}{2},\frac{1}{2}\right) ∉ R , since \frac{1}{2}>\left(\frac{1}{2}\right)^2 = \frac{1}{4}\end{align}

∴R is not reflexive.

Now, (1, 4) ∈ R as 1 < 4^{2}

But, 4 is not less than 1^{2}.

∴(4, 1) ∉ R

∴R is not symmetric.

Now,

(3, 2), (2, 1.5) ∈ R

(as 3 < 2^{2} = 4 and 2 < (1.5)^{2} = 2.25)

But, 3 > (1.5)^{2} = 2.25

∴(3, 1.5) ∉ R

∴ R is not transitive.

Hence, R is neither reflexive, nor symmetric, nor transitive.

## 1 Comment(s) on this Question

Prashant Sharma

2019-03-04 13:00:43
Sir please help me?