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show that the relation r in the set r of real numb...

Question: Show that the relation R in the set R of real numbers, defined as R = {(a, b): a ≤ b²} is neither reflexive nor symmetric nor transitive.

Answer:

R = {(a, b): a ≤ b²}

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 < 4²

But, 4 is not less than 1².

∴(4, 1) ∉ R

∴R is not symmetric.

Now,

(3, 2), (2, 1.5) ∈ R

(as 3 < 2² = 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.

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Question: Show that the relation R in the set R of real numbers, defined as R = {(a, b): a ≤ b2} is neither reflexive nor symmetric nor transitive.

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Question: Show that the relation R in the set R of real numbers, defined as R = {(a, b): a ≤ b²} is neither reflexive nor symmetric nor transitive.