Question

show that the reciprocal of the given number is an irrational number
$3 + 2\sqrt{2}$
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Answer 1

$3 + 2 \sqrt{2} = \frac{{3 + 2 \sqrt{2} } }{1}$

$Its \: \: reciprocal = \frac{1}{3 + 2 \sqrt{2} } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{1(3 - 2 \sqrt{2} }{ {(3)}^{2} - {(2 \sqrt{2} )}^{2} } = \frac{3 - 2 \sqrt{2} }{9 - 8} = \frac{3 - 2 \sqrt{2} }{1} = 3 - 2 \sqrt{2}$

As $3 - 2\sqrt{2}$ can't have even an approximate value because it is non-terminating & non-recurring.

Answer 2

Answer:

hey just see the attachment