Math, asked by SanidhyaNayyar, 6 months ago

Prove that 3+2√5 is irrational number​

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Answered by Anonymous
15

Answer:

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Answered by Anonymous
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Solution:

Let  \: 3+2 \sqrt{5}  \: is \: not \: irrational, \: \\ so \: 3 + 2 \sqrt{5} \: is \: a \: rational \: number. \\  Now \:  let  \: 3 + 2 \sqrt{5}  = x, \: \\  a \:  rational \: number. \\ Therefore \:  \:  \: x {}^{2}  = (3 + 2 \sqrt{5} ) {}^{2}  \\  = 9 + 20 + 2 \times 3 \times 2 \sqrt{5}  \\  = 9 + 20 + 12 \sqrt{5}  \\  = 29 + 12 \sqrt{5}  \\  =  > 12 \sqrt{5}  = x {}^{2}  - 29 \\ and, \:so \:   \sqrt{5}  =  \frac{x {}^{2} - 29 }{12}  \\ Since, \: i t \: is \: assumed \: that \: \\  3 + 2 \sqrt{5}  \:  = x \: is \: rational \:  \\  =  > x {}^{2}  \: is \:ra tional \: ............(I) \\ x {}^{2}  - 29 \: is \: rational \\ and, \: so \:  \frac{x {}^{2} - 29 }{12} is \: rational \\  =  >  \frac{x {}^{2} - 29 }{12}  =  \sqrt{5}  \: is \: rational \\ But \:  \sqrt{5} \:  is \: irrational \: i.e. \: \\  x {}^{2}  \: is \: irrational. ...........(II) \\ From \:  I, x {}^{2}  \: is \: rational \: and \\ from \: II, \: x {}^{2}  \: i s \: irrational. \\ Therefore, \: we \: arrive \: at \: a  \\ \: contradiction. \\ So, \: our \: assumption \: that \\  \: 3 + 2 \sqrt{5} \:  is \: a \: rational \: number \\  \: is \: wrong. \\ Therefore, \: 3 +  2\sqrt{5}  \: is \: irrational.

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