Question

give examples of two irrational numbers whose product is a rational number

Answer 1

${\bold{\sqrt{2} \text { and } \sqrt{8}}$ are the two irrational numbers, where their product 4 is the rational number

To Find:

Two irrational numbers whose product is the rational number

Solution:

A number is said to be rational number if that number is written as fraction and an irrational number cannot be written as the ratio of any two integers.  

For example, square roots.

Now, we can take $\sqrt{2} \text { and } \sqrt{8}$ as irrational number.

By multiplying them,

$\sqrt{2} \times \sqrt{8}=\sqrt{16}=4$

We know that 4 is the rational number.

Answer 2

Answer:

$i) Let \: \sqrt{3} \:and \: 5\sqrt{3} \\ are \: two \: irrational\: numbers$

$Product \: of \: these \\irrational \: numbers \\= \sqrt{3}\times 5\sqrt{3}\\=5\times \sqrt{3}\times \sqrt{3}\\=5\times 3\\=15\\(Rational \:number)$

Step-by-step explanation:

$i) Let \: \sqrt{3} \:and \: 5\sqrt{3} \: are \\ two \: irrational\: numbers$

$Product \: of \: these \\irrational \: numbers \\= \sqrt{3}\times 5\sqrt{3}\\=5\times \sqrt{3}\times \sqrt{3}\\=5\times 3\\=15\\(Rational \:number)$

$ii) Let \: 5\sqrt{7} \:and \: \sqrt{7} \: are \\two \: irrational\: numbers$

$Product \: of \: these \\irrational \: numbers = 5\sqrt{7}\times \sqrt{7}\\=5\times \sqrt{7}\times \sqrt{7}\\=5\times 7\\=35\\(Rational \:number)$

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