coukd-you-prove-please-that-11-is-not-racional-number-

Question Number 128462 by Ari last updated on 07/Jan/21

coukd you prove please that \sqrt{11} is not racional number

Answered by Dwaipayan Shikari last updated on 07/Jan/21

\sqrt{11} = 3 + \sqrt{11} - 3 = 3 + \frac{2}{\sqrt{11} + 3} = 3 + \frac{2}{3 + 3 + \frac{2}{3 + 3 + \frac{2}{3 + 3 + \frac{2}{3 + 3 \dots}}}}

= 3 + \frac{2}{6 + \frac{2}{6 + \frac{2}{6 + \frac{2}{6 + \frac{2}{6 + \dots}}}}}

I t i s a n I n f i n t e c o n t i n u e d f r a c t i o n . S o i t c a n n e v e r b e R a t i o n a l

O r

\sqrt{11} = 1 + \frac{10}{\sqrt{11} + 1} = 1 + \frac{10}{2 + \frac{10}{2 + \frac{10}{2 + \frac{10}{2 + \dots}}}}

Commented by Ari last updated on 07/Jan/21

Yes,one way !

Answered by physicstutes last updated on 07/Jan/21

{let}^{'} s do it by contradiction .

{let}^{'} s assume that \sqrt{11} was rational .

then \sqrt{11} = \frac{a}{b} where a and b are in thier simpliest term .

\Rightarrow 11 = {(\frac{a}{b})}^{2}

\Rightarrow a^{2} = 11 b^{2} so we can say a = 11 k (an assumption)

\Rightarrow a^{2} = 11^{2} k^{2}

\Rightarrow 11 b^{2} = 11^{2} k^{2} hence b^{2} = 11 k^{2} so b = 11 m

thus \gcd (a, b) = 11 which contradicts that fact that \gcd (a, b) = 1

hence \sqrt{11} is not rational .

Commented by Ari last updated on 07/Jan/21

Yes,you are right!