# Irrational number

A number which we can't show in fractional form is known as Irrational number.

Example:- √2, √3, π, √5, etc.

For a detailed explanation on an irrational number, you may read this article.

Now we will discuss how to prove √2, √3 & √5 is an irrational number.

## Prove that √2 is an irrational number.

First, Let

**√2**is a rational number. So we may write it as below-**√2 = a/b**

Let, where

**a**&**b**are co-prime numbers, means common factor between**a**&**b**is**1.****b√2 = a**

Now squaring both sides-

**(b√2)² = a²**

**2b² = a²**

**b² = a²/2......Eq.1**

where we can see

**a**is divisible by**2,**hence**a**has**2**as a factor.Let,

**a = 2c**where c is some integer & we will put**a = 2c**in Eq.1.**b² = (2c)²/2**

**b² = 4c²/2**

**b² = 2c²**

**b²/2 = c²**

Here we can see

**b**is also divisible by**2**, hence**b**has**2**as a factor.Now here we can say that

**a**&**b**have**2**as a common factor other than**1**. So they are not co-prime.Hence our assumption

**√2**is a rational number is wrong & we conclude**√2**will be an irrational number.## Prove that √3 is an irrational number.

First, Let

**√3**is a rational number. So we may write it as below-**√3 = a/b**

Let, where

**a**&**b**are co-prime numbers, means common factor between**a**&**b**is**1.****b√3 = a**

Now squaring both sides-

**(b√3)² = a²**

**3b² = a²**

**b² = a²/3......Eq.1**

where we can see

**a**is divisible by**3,**hence**a**has**3**as a factor**.**Let,

**a = 3c**where c is some integer & we will put**a = 3c**in Eq.1.**b² = (3c)²/3**

**b² = 9c²/3**

**b² = 3c²**

**b²/3 = c²**

Here we can see

**b**is also divisible by**3**, hence**b**has**3**as a factor**.**Now here we can say that

**a**&**b**have**3**as a common factor other than**1**. So they are not co-prime.Hence our assumption

**√2**is a rational number is wrong & we conclude**√2**will be an irrational number.## Prove that √5 is an irrational number.

First, Let

**√5**is a rational number. So we may write it as below-**√5 = a/b**

Let, where

**a**&**b**are co-prime numbers, means common factor between**a**&**b**is**1.****b√5 = a**

Now squaring both sides-

**(b√5)² = a²**

**5b² = a²**

**b² = a²/5......Eq.1**

where we can see

**a**is divisible by**5,**hence**a**has**5**as a factor**.**Let,

**a = 5c**where c is some integer & we will put**a = 5c**in Eq.1.**b² = (5c)²/5**

**b² = 25c²/5**

**b² = 5c²**

**b²/5 = c²**

Here we can see

**b**is also divisible by**5**, hence**b**has**5**as a factor**.**Now here we can say that

**a**&**b**have**5**as a common factor other than**1**. So they are not co-prime.Hence our assumption

**√5**is a rational is wrong & we conclude**√5**will be an irrational number.

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