Question

The relation S is defined on R as follows : S={(a,b)| a ≤ b²,a,b ∈ R} Prove S is not reflexive, not symmetric and not transitive.

Answer 1

Answer:

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Answer 2

Answer:

S is not reflexive, not symmetric and not transitive.

Step-by-step explanation:

According to this question

Given that

S is defined on R

S={(a,b)| a ≤ b²,a,b ∈ R}

So, Reflective ⇒ $\frac{1}{2}$ ∈ R

                           $\frac{1}{2}$ ≥ $(\frac{1}{2})^2$

                           $\frac{1}{2}$ ≥ $\frac{1}{4}$

So, ( $\frac{1}{2}$,  $\frac{1}{2}$) ∉ S

Then S is not reflective.

Symmetric ⇒ (1, 2) ∈ S = 1 ≤ $2^2$

                                     = 1 ≤ 4

                                      = 4 ≥ 1

                                      = (2, 1) ∉ S

So, S is not symmetric.

Transitive ⇒ (3, 2)(2, ${3}{2}$) ∈ S

3 ≤ $2^2$ and 2 ≤ $(\frac{3}{2})^2$

But (3, ${3}{2}$) ∉ S

Because, 3 ≥ $(\frac{3}{2})^2$

                3 ≥ 2.25

So, S is not transitive.