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| 1. |
Show that reciprocal of 3+2√2 is an irrational number |
| Answer» 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√2As a result, we get ( - 2√2 ) which is an irrational number.Hence, the reciprocal of ( 3 + 2√2) is an irrational number.\xa0 | |