Question

show that the reciprocal of 3+2√2 is an irrational number

Answer 1

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Answer 2

$\underline{ \huge \bold { solution : }}$

▪️Question : Show that reciprocal of 3 + 2√2 is an irrational.

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Given : Reciprocal of 3 + 2√2 i.e., $\sf \frac{1}{3 + 2\sqrt{2}}$

To prove : Reciprocal of 3 + 2√2 is an irrational number.

Proof :

First of all, rationalise the denominator of the reciprocal of 3 + 2√2.

$\sf 3 + 2 \sqrt{2} \\ \\ \sf \frac{1}{3 + 2 \sqrt{2} } \times \frac{3 - 2 \sqrt{2} }{3 - 2 \sqrt{2} } \\ \\ \sf \frac{3 - 2 \sqrt{2} }{(3 ){}^{2} - (2 \sqrt{2} ) {}^{2} } \\ \\ \sf \frac{3 - 2 \sqrt{2} }{9 - 8} \\ \\ \bf 3 - 2 \sqrt{2}$

After rationalising its denominator, we get ( 3 - 2√2 ) as a result.

Now, let us assume that ( 3 - 2√2 ) is an irrational number. So, taking a rational number i.e., 3 and subtracting from it.

We have ;

[ 3 - 2√2 - 3 ]

⇒ - 2√2

As a result, we get ( - 2√2 ) which is an irrational number.

Hence, the reciprocal of ( 3 + 2√2) is an irrational number.