Question

prove that √3+√5is an irrational number​

Answer 1

Answer:

let as assume

$\sqrt{3 + \sqrt{5} }$

is a rational number. we can find the co prime integer

so that

$\sqrt{3 + \sqrt{5} }$

$\sqrt{3 + \sqrt{5 = a \div b } }$

$\sqrt{5 = a \div b - \sqrt{3} }$

$\sqrt{5 = \sqrt{3 + a \div b} }$

therefore

$\sqrt{3 + a \div b}$

is a rational , so

$\sqrt{5}$

is rational

But this contradict the fact that

$\sqrt{5}$

is irrational number

hence,

$\sqrt{3 + \sqrt{5} }$

is irrational number

Answer 2

Step-by-step explanation:

Let √3+√5 be a rational number. 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. ... Therefore, √3+√5 is an irrational number.