Question

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

Answer 1

Answer:

Step-by-step explanation:

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