Question

Prove that √5 is irrational.
Prove that 3 + 2√5 is irrational.​

Answer 1

Answer:

$\huge \mathrm \color{black}{Question}$

Prove that √5 is irrational.

$\large \mathrm \color{navy}{ANSWER}$

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.

Hence proved

$\huge \mathrm \color{black}{Question}$

Prove that 3 + 2√5 is irrational.

$\large \mathrm \color{navy}{ANSWER}$

Given: 3 + 2√5

To prove: 3 + 2√5 is an irrational number.

Proof:

Let us assume that 3 + 2√5 is a rational number.

So, it can be written in the form a/b

3 + 2√5 = a/b

Here a and b are coprime numbers and b ≠ 0

Solving 3 + 2√5 = a/b we get,

=>2√5 = a/b – 3

=>2√5 = (a-3b)/b

=>√5 = (a-3b)/2b

This shows (a-3b)/2b is a rational number. But we know that √5 is an irrational number.

So, it contradicts our assumption. Our assumption of 3 + 2√5 is a rational number is incorrect.

3 + 2√5 is an irrational number

Hence proved

Answer 2

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.

(2) 3+2√5

➡️ let us assume 3+2√5 is a rational number

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

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

$= 2 \sqrt{5} = \frac{a}{b} - 3$

$= 2 \sqrt{5} = \frac{a}{b} - \frac{3}{1}$

$= 2 \sqrt{5} = \frac{a}{b} - \frac{3}{1} \times \frac{b}{b}$

$= 2 \sqrt{5} = \frac{a}{b} - \frac{3b}{b}$

$= 2 \sqrt{5} = \frac{a - 3b}{b}$

$= \sqrt{5} = \frac{ a - 3b}{b} \times \frac{1}{2}$

= √5 = a-3b/2b

a-3b/2b is of the form of a/b

hence it is a rational number but √5 is a irrational number

the assume is false

3+2√5 is an irrational number

Hence proved

Hope this helps you

#Ravalika Rajula

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