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.
Solution:
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 $1+\left(-\frac{1}{2}\right)>0$ and $1+\left(-\frac{1}{2}\right)(-4)>0$
But $1+1 \times(-4)<0$
$\therefore(1,-4) \notin R_{1}$
And,
(a, b) ∈ R1 and (b , c) ∈ R1
Thus, (a, c) ∈ R1 is not true for all a, b, c ∈ R.