Question

For the relation R₁ defined on R by the rule (a, b) ∈ R₁ ⇔ 1 + ab > 0.
Prove that: (a, b) ∈ R₁ and (b , c) ∈ R₁ ⇒ (a, c) ∈ R₁ is not true for all a, b, c ∈ R.

Answer 1

We have:
(a, b) ∈ R₁ ⇔ 1 + ab > 0
Let:
a = 1, b = $-\frac{1}{2}$ and c = $-$4
Now,

$\left(1,-\frac{1}{2}\right)\in R_{1}and\left(-\frac{1}{2},-4\right)\in R_1$, as $1+\left(-\frac{1}{2}\right)>0and1+\left(-\frac{1}{2}\right)\left(-4\right)>0$.

But $1+1\times \left(-4\right)<0$.
∴ (1,$-$4) $\notin R_1$
And,
(a, b) ∈ R₁ and (b , c) ∈ R₁
Thus, (a, c) ∈ R₁ is not true for all a, b, c ∈ R.