Question

6. Prove that 5 – 2 √3 is an irrational number

Answer 1

$\underline{ \boxed{ \sf \red \bigstar GIVEN }}$

$\sf \rightarrow 5 - 2 \sqrt{3}$

$\underline{ \boxed{ \sf \green \bigstar PROVE }}$

$\sf \rightarrow 5 - 2 \sqrt{3} \: is \: an \: irrational \: no.$

$\underline{ \boxed{ \sf \orange \bigstar SOLUTION}}$

$\sf ☽ let \: 5 - 2 \sqrt{3} \: is \: an \: rational \: no.$

$\sf ✵ and \: 5 - 2 \sqrt{3} = \frac{p}{q} , \\ \sf where \: p,q \: any \: integer \: q \not= 0$

$\sf❂ squaring \: both \: sides$

$\sf ⇝ {(5 - 2 \sqrt{3})}^{2} = {( \frac{p}{q})}^{2}$

$\sf ⚝by \: using \: {(a - b)}^{2} = {a}^{2} - 2ab + {b}^{2}$

$\sf ⇝ {5}^{2} - 2(5)( 2\sqrt{3} ) + {(2 \sqrt{3} )}^{2} = \frac{ {p}^{2} }{ {q}^{2} }$

$\sf ⇝ 25 - 20 \sqrt{3} + 12 = \frac{ {p}^{2} }{ {q}^{2} }$

$\sf ⇝ 37 - 20 \sqrt{3} = \frac{ {p}^{2} }{ {q}^{2} }$

$\sf ⇝ 37 = \frac{ {p}^{2} }{ {q}^{2} } - 20 \sqrt{3}$

$\sf❂ 37 \: is \: rational \: but \: \frac{ {p}^{2} }{ {q}^{2} } + 20 \sqrt{3} \: is \: irrational \\ \sf bcz \: \sqrt{3} \: is \: irrational \: no.$

$\sf ⚝rational \: no. \: can't \: be \: equal \: to \: irrational \\ \sf no. \: Hence, by \: contradiction \: my \: assumption \: is \: wrong. \\ \sf i.e. \: 5 - 4 \sqrt{3} \: is \: a\:rational \: no. \: is \: wrong ❌$

$\sf Therefore, 5 - 4 \sqrt{3} \: is \: an \: irrational \: no. \\ \mathbb{ \huge HENCE \: PROVED}$

Answer 2

U HAVE ALREADY GOT UR ANS