Show that reciprocal of 3+2√2 is …
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Posted by Gracy Mehar 6 years, 7 months ago
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Yogita Ingle 6 years, 7 months ago
First of all, rationalise the denominator of the reciprocal of 3 + 2√2.
{tex}\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}{/tex}
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.
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