Question

shiw that root
$\sqrt{5} \times \frac{?}{?}$
5 is irrational number​

Answer 1

Answer:

(1) √5

➡️ let us assume that √5 is a rational number

rational number is of the form of a/b , a, b, E, Z and b ≠ 0 .

$\sqrt{5} = \: \frac{a}{b}$

$\frac{ \sqrt{5} }{1} = \frac{a}{b}$

cross multiplication

1 X a = √5 X b

√5b = a

squaring on both sides

(√5b) ² = (a) ²

5b² = a² —————————(1)

use Theorem 1.6 : 2 divides a² then 2 divides a

let a = 5c (c is a integer)

squaring on both sides

(a) ² = (5c) ²

a² = 25c²

From equation (1) a²=5b²

5b² = 25c²

$= 25 = \frac{25c}{5}$

b² = 5c²

5c² = b²

5 divides b² then 5 divides b

both a and b have common factors but a and are co primes so √5 is not a rational number.

the assume is False

Hence proved

√5 is a irrational number.

Hope it helps ☺️☺️

Answer 2

$\underline{\underline{\huge{\pink{\tt{\textbf Answer :-}}}}}$

Given:- √5

We need to prove that √5 is irrational

Proof:-

Let us assume that √5 is a rational number.

So it can be expressed in the form p/q where p,q are co-prime integers and q≠0

⇒ √5 = p/q

On squaring both the sides we get,

⇒5 = p²/q²

⇒5q² = p² —————–(i)

p²/5 = q²

So 5 divides p

p is a multiple of 5

⇒ p = 5m

⇒ p² = 25m² ————-(ii)

From equations (i) and (ii), we get,

5q² = 25m²

⇒ q² = 5m²

⇒ q² is a multiple of 5

⇒ q is a multiple of 5

Hence, p,q have a common factor 5. This contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number

√5 is an irrational number.

Hope it will helps you ❤️