Question

Prove that 5-√3 is irrational,given that √3 is irrational?? please give the sol in proper way... i mean in A to Z explain properly please guys please​

Answer 1

Given :-

$\sqrt{3} \: is \: irrational$

To prove :-

$5 - \sqrt{3} \: is \: irrational.$

Proof:-

As any rational number is in the form p/q, where p and q are integers and q ≠0.

$let \: consider \: 5 - \sqrt{3} to \\ \: be \: a \: rational \: number \: then$

$5 - \sqrt{3} = \frac{p}{q}$

On squaring both sides we have,

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

$(25 - 10 \sqrt{3} + 3) = \frac{ {p}^{2} }{ {q}^{2} }$

$(28 - 10 \sqrt{3} ) = \frac{ {p}^{2} }{ {q}^{2} }$

$(28 - \frac{ {p}^{2} }{ {q}^{2} } ) = 10 \sqrt{3}$

$\frac{28 {q}^{2} - {p}^{2} }{ {q}^{2} } = 10 \sqrt{3}$

$\frac{28 {q}^{2} - {p}^{2} }{10 {q}^{2} } = \sqrt{3}$

As

$\sqrt{3} \: is \: irrational$

So 28q²-p²/10q² is also irrational.

Hence proved that,

$5 - \sqrt{3} \: is \: irrational.$