Question

prove √2 is an irrational number​

Answer 1

Answer:

Lets first assume √2 to be rational.

√2 = a/b, where b is not equal to 0.

Here, a and b are co-primes whose HCF is 1.

√2 = a/b ( squaring both sides )… 2 = a^2/b^2

2b^2 = a^2 ….. Eq.1

Here, 2 divides a^2 also a ( bcz, If a prime number divides the square of a positive integer, then it divides the integer itself )

Now, let a = 2c ( squaring both sides )…

a^2 = 4c^2…Eq.2

Substituting Eq. 1 in Eq. 2,

2 b^2 = 4c^2

b^2 = 2c^2

2c^2 = b^2

2 divides b^2 as well as b.

Conclusion:

Here, a & b are divisible by 2 also. But our assumption that their HCF is 1 is being contradicted.

Therefore, our assumption that √2 is rational is wrong. Thus, it is irrational.

Answer 2

Answer:

$answer$

Given √2 is irrational number.

Let √2 = a / b wher a,b are integers b ≠ 0 we also suppose that a / b is written in the simplest form

Now √2 = a / b ⇒ 2 = a2 / b2 ⇒ 2b2 = a2 ∴ 2b2 is divisible by 2 ⇒ a2 is divisible by 2 ⇒ a is divisible by 2

let a = 2c a2 = 4c2 ⇒ 2b2 = 4c2 ⇒ b2 = 2c2 ∴ 2c2 is divisible by 2 ∴ b2 is divisible by 2 ∴ b is divisible by 2

a are b are divisible by 2 . this contradicts our supposition that a/b is written in the simplest form Hence our supposition is wrong

∴ √2 is irrational number.