Question

prove that √11 is an irrational number​

Answer 1

Answer&Explaination:

$\sf{</strong><strong>L</strong><strong>et \: us \: assume \: \sqrt{11} \: is \: rational \: by \: contradiction \: method}$

$\sf{\sqrt{11} = \frac{a}{b}(a \: and \: b \: are \: co - primes \: where \: b \: is \: not \: equal \: to \: 0}$

$\sf{b \sqrt{11} = a}$

$\sf{</strong><strong>N</strong><strong>ow \: squaring \: on \: both \: sides}$

$\sf(b \sqrt{11})^{2} = {a}^{2}$

$\sf11b = {a}^{2}\rightarrow \: equation1$

$\sf{here \: 11 \: divides \: {a}^{2} so \: 11 \: also \: divides \: a}$

$\sf \: a = 11c$

$\sf \: </strong><strong>N</strong><strong>ow \: squaring \: on \: both \: sides$

$\sf {a}^{2} = ( {11c)}^{2}$

$\sf \: {a}^{2} = {121c}^{2}$

$\sf{substitute \: {a}^{2} \: in \: equation \: 1}$

$\sf{11b}^{2} = 121 {c}^{2}$

$\sf{\cancel{11}}^{1} {b}^{2} = {\cancel{121}}^{11} {c}^{2}$

$\sf{b}^{2} = {11c}^{2}$

$\sf\frac{ {b}^{2} }{11} = c$

Here,11 divides b²,so 11 also b

Therefore 11 is a common factor of both a and b.

This results in that our assumption a and b are co-primes is incorrect.

This contradiction has arisen because of our incorrect assumption that √11 is rational.

$\sf \: Hence \: \sqrt{11} is \: irrational$

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