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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. - Mathematics

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.

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Solution

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

Now,

\[\left( 1, - \frac{1}{2} \right) \in R_1 \text{ and }  \left( - \frac{1}{2}, - 4 \right) \in R_1 \] , as 

\[1 + \left( - \frac{1}{2} \right) > 0 \text{ and }  1 + \left( - \frac{1}{2} \right)\left( - 4 \right) > 0\] But 
\[1 + 1 \times \left( - 4 \right) < 0\] 
∴ (1, - 4) \[\not\in R_1\] And,
(ab) ∈ R1 and (b , c) ∈ R1
Thus, (ac) ∈ R1 is not true for all abc ∈ R.
 

 

Concept: Relation
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 2 Relations
Exercise 2.3 | Q 21 | Page 21
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