Question

a pair of irrational numbers whose proudct is a rational number is. (a)√16,√4 (b) √5√4 (c) √3,√27 (d) √36,√2​

Answer 1

Answer:

$\boxed{\sf \: Product \: of \: \sqrt{16}, \sqrt{4} \: and \: \sqrt{3}, \sqrt{27} \: are \: rational \: number \: }$

Step-by-step explanation:

Consider,

$\sf \: \sqrt{16} \times \sqrt{4}$

$\sf \: = \: 4 \times 2$

$\sf \: = \: 8$

Hence,

$\implies\sf \: \sqrt{16} \times \sqrt{4} = \: 8$

$\implies\sf \: \sqrt{16} \times \sqrt{4} \: is \: a \: rational \: number$

Now, Consider

$\sf \: \sqrt{5} \times \sqrt{4}$

$\sf \: = \: \sqrt{5 \times 4}$

$\sf \: = \: \sqrt{20}$

Thus,

$\implies\sf \: \sqrt{5} \times \sqrt{4} \: is \: an \: irrational \: number$

Now, Consider

$\sf \: \sqrt{3} \times \sqrt{27}$

$\sf \: = \: \sqrt{3 \times 27}$

$\sf \: = \: \sqrt{81}$

$\sf \: = \: 9$

Thus,

$\implies\sf \: \sqrt{3} \times \sqrt{27} \: is \: a \: rational \: number$

Now, Consider

$\sf \: \sqrt{36} \times \sqrt{2}$

$\sf \: = \: 6 \times \sqrt{2}$

$\sf \: = \: 6 \sqrt{2}$

Thus,

$\implies\sf \: \sqrt{36} \times \sqrt{2} \: is \: an \: irrational \: number$

Hence,

$\implies\sf \: Product \: of \: \sqrt{4}, \sqrt{16} \: and \: \sqrt{3}, \sqrt{27} \: are \: rational \: number.$