Question

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

Answer 1

Step-by-step explanation:

To prove : V3 + V5 is irrational.

Let us assume it to be a rational number.

Rational numbers are the ones that can be expressed in р form where p, q are

integers and q isn't equal to zero.

V3 + V5 = P

q

V5

V3 =P

q

squaring on both sides,

3 = p q? 2.5 + 5

(2/5p) = 5-3+

(25p) 2q? p? q?

5 2q? – p q q? 2p

> V5 (2q? – p?) 2pq

As p and q are integers RHS is also rational.

As RHS is rational LHS is also rational i.e 5

is rational

Answer 2

Answer:

Let √3 +√5 be rational number

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

where a and b is a co-prime number

squaring both side :

${( \sqrt{3} + \sqrt{5}) }^{2} = \frac{a}{b}$

$\implies \: 3 + 5 + \sqrt{15} = \frac{ {a}^{2} }{ {b}^{2} }$

$\implies \: 8 + \sqrt{15} = \frac{ {a}^{2} }{ {b}^{2} }$

$\sqrt{15} = \frac{ {a}^{2} }{ {b}^{2} } - 8$

$\sqrt{15} = \frac{ {a}^{2} - 8 {b}^{2} }{ {b}^{2} }$

RHS is not equal to LHS as lhs is irrational and Rhs is rational

our assumption is wrong

hence , √3+√5 is irrational number