Math, asked by sr9, 22 days ago

prove that
 \sqrt{2 - \sqrt{3} }
prove that
irrational number ​

Answers

Answered by as3801504
8

Answer:

Step-by-step explanation:

let \: we \: assume \: that \: it \: is \: nation \: no. \\   {\implies}{ \boxed{\mathbb{\red{\frac{ \sqrt{2 -  \sqrt{3} } }{1 }  =  \frac{a}{b}  }}}}\\ now \:  \\  \sqrt{3}  =  \frac{a}{b}  +  \sqrt{2}  \\ {\implies}{ \boxed{\mathbb{\blue{squarring \: both \: side \: we \: get}}}} \\ 3 =  \frac{a {}^{2} }{b {}^{2} }  +  \frac{2a}{b}  \sqrt{2}  + 2 \\  \frac{2a}{b}  \sqrt{2}  =  \frac{a {}^{2} }{b {}^{2} }   + 1 \\  \sqrt{2}  =  \frac{a {}^{2}  \:  +  b {}^{2}  }{2ab}  \\ since \: a \: and \: b \: are \: interger \: so \\  \frac{a {}^{2}  +   \: b {}^{2} }{2ab} is \: rational \\ thus \:  \sqrt{2}  \: is \: also \: rational \\ but \: we \: know \: that \:  \sqrt{2} is \: irratinal \\ thus \: we \: arrive \: at \: a \: contration \:  \\ since \: the \: contradicton \: arises \: by \: assuming \: that \:  (\sqrt{2}  -  \sqrt{3} ) \: is \: rational \\ {\implies}{ \boxed{\mathbb{\red{hence \: it \: is \: irrational}}}} \\ \\ {\implies}{ \boxed{\mathbb{\orange{ hence \: proved}}}}

Similar questions