Question

prove that root 2 is irrational​

Answer 1

Step-by-step explanation:

Answer:

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.

Answer 2

Step-by-step explanation:

let root 2 is rational no.

root2 =a by b

b root2 =a

squaring both side

2b^2 =a^2 ( put it equation 1)

b^2= a^2/2

so , 2 divides a^2

hence 2 divides a also

now,

let c be any integer such that

a=2c

putting a value in equation 1

2b^2 = 2c ka whole square

2b^2 =4c^2

b^2= 4/2 c^2

b^2= 2 c^2

b^2/2 =c^2

so 2 divides b^2 , hence 2 divides b also .

so a and b have 2 as a common factor which contradicts the fact that a and b are co prime .

so our assumption is wrong.

hence root 2 is irrational no. ( hence proved )

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