Question

Prove that √2 is irrational?​

Answer 1

Step-by-step explanation:

Let us assume that √2 is a rational number.

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

√2 = p/q

Here p and q are coprime numbers and q ≠ 0

Solving

√2 = p/q

On squaring both the side we get,

=>2 = (p/q)2

=> 2q2 = p2……………………………..(1)

p2/2 = q2

So 2 divides p and p is a multiple of 2.

⇒ p = 2m

⇒ p² = 4m² ………………………………..(2)

From equations (1) and (2), we get,

2q² = 4m²

⇒ q² = 2m²

⇒ q² is a multiple of 2

⇒ q is a multiple of 2

Hence, p, q have a common factor 2. This contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number

√2 is an irrational number.

Answer 2

Step-by-step-Explanation :

Given :

√2

To Prove :

That √2 is irrational.

According to question,

Let us assume on the contrary that   √2  is a rational number. Then, there exist positive integers a and b such that

√2  = a/b (∴ where a and b are , co-prime i.e. their HCF is 1.)

⇒ (√2)² = (a/b)²

⇒ 2 = a²/b²

⇒ 2b² = a²

⇒ 2∣a²    [∵2∣2b²  and 2b²  = a²  ]

⇒ 2∣a² ... (i)

⇒ a = 2c for some integer c

⇒ a² = 4c²  

⇒ 2b² = 4c²  [∵2b²  = a²  ]  

⇒ b²  = 2c²    

⇒ 2∣b²    [∵2∣2c² ]  

⇒ 2∣b...(ii)

∴ From (i) and (ii), we obtain that 2 is a common factor of a and b. But, this contradicts the fact that a and b have no common factor other than 1. This means that our supposition is wrong.

Hence, √2 is irrational.