Question

prove that root3/5 is irrational.​

Answer 1

answer explanation:

A rational number can be written in the form of p/q where p,q are integers. p,q are integers then (p²+2q²)/2pq is a rational number. Then √5 is also a rational number. ... Therefore, √3+√5 is an irrational number

Answer 2

Answer:

HEY MATE HERE'S YOUR ANSWER

Step-by-step explanation:

$To \: prove : 3+5 \: is \: irrational. \\ Let \: us \: assume \: it \: to \: be \: a \: rational \: number. \\ \\ Rational \: numbers \: are \: the \: ones \: that \: can \: be \: expressed \: in  \frac{p}{q} \: form \: where \: p,q \: are \: integers \: and \: q \: isn't \: equal \: to \: zero. \\ \sqrt{3} + \sqrt{5} = \frac{p}{q} \\ \\ \sqrt{3} \: \frac{p}{q} - \sqrt{5} \\ squaring \: on \: both \: sides, \\ \\ 3 = \frac{ {p}^{2} }{ {q}^{2} } - 2. \sqrt{5} ( \frac{p}{q} ) + 5 \\ ( \frac{ 2\sqrt{5} p}{q} ) = 5 - 3 + ( \frac{ {p}^{2} }{ {q}^{2} } ) \\ \\ \frac{ ( 2\sqrt{5} p)}{q} = \frac{ {2q}^{2} - {p}^{2} }{ {q}^{2} } ) \\ \sqrt{5} = \frac{ {2q}^{2} - {p}^{2} }{ {q}^{2} } . \frac{q}{2p} \\ \sqrt{5} = ( \frac{ {2q}^{2} - {p}^{2} }){2pq} \\ \\ As \: p \: and \: q are \: integers \: RHS \: is \: also \: rational. \\ \\ As \: RHS \: is \: rational \: LHS \: is \: also \: rational \: i.e  \sqrt{5} \: is \: rational. \\ \\ But \: this \: contradicts \: the \: fact \: that \sqrt{5} \: is \: irrational. \\ \\ This \: contradiction \: arose \: because \: of \: our \: false \: assumption. \\ so, \sqrt{?} + \sqrt{5} \: irrational.</p><p></p><p>$

HOPE IT WILL HELP YOU PLZ MRK AS BRAINLIST

DON'T FORGET TO FØLLØW ME TO

HV A GOOD DAY