For the relation
Question:

For the relation R1 defined on R by the rule (ab) ∈ R1 ⇔ 1 + ab > 0.

Prove that: (ab) ∈ R1 and (b , c) ∈ R1 ⇒ (ac) ∈ R1 is not true for all abc ∈ R.

Solution:

We have:

(ab) ∈ 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,

(ab) ∈ R1 and (b , c) ∈ R1

Thus, (ac) ∈ R1 is not true for all abc ∈ R.

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