Question

prove That √2 is an Irrational No​

Answer 1

Answer:

To prove that √2 is irrational by the contradiction method, we first assume that √2 is a rational number. Now, if it is a rational number, there exist two co-prime integers x and y, such that √2 = x/y, where x and y have no other common factors except 1 and y ≠ 0. So, our equation is √2 = x/y.

Answer 2

Given

→ √2

To prove: √2 is an irrational number.

Proof:

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.

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