Question

Prove that
$\sqrt{2}$
is a irrational number



​

Answer 1

Let us assume that √2 is rational number.

Now,

√2 = p/q [ where p and q are co-primes and q is not equal to zero ]

★ Squaring both sides

⇒ ( √2 )² = ( p/q )²

⇒ 2 = p²/q²

⇒ p² = 2q² .......( 1 )

⇒ 2q² = p²

⇒ 2 divides p²

⇒ 2 divides p [ By concept 3 ] .......( 2 )

Let p = 2m

★ Squaring both sides

⇒ p² = 4m²

Putting the value of p² in ( 1 ) ,we get:

⇒ 2q² = 4m²

⇒ 4m² = 2q²

⇒ 2m² = q²

⇒ 2 divides q²

⇒ 2 divides q ......( 3 ) [ By concept 3 ]

Thus, 2 divides p and q [ from 2 and 3 ]

It means 2 is a common factor of p and q . This contradicts the assumption as there is no common factor of p and q .

Hence, √2 is irrational number.

Answer 2

$\huge\mathfrak{Amswer:}$

Given:

• We have been given a number √2.

To Prove:

• We need to prove that √2 is irrational.

Solution:

Let us assume that √2 is rational.

Therefore, it can be written in the form of p/q where p and q are coprime.

=> √2 = p/q

On squaring both sides, we have

(√2)² = (p/q)²

=> p²= 2q² _______(1)    

2 divides p² => 2 also divides p.

Therefore, p is a Multiple of 2.

p = 2a [where a is any integer]

Putting p = 2a in equation 1, we have

(2a)² = 2q²

4a² = 2q²

2a² = q² __________(2)

2 divides q² => 2 also divides q.

Therefore, q is a Multiple of 2.

=> q = 2b

From equation 1 and 2, we have

p and q have 2 as a common factor. But this contradicts the fact that p and q are coprime.

Therefore, our assumption was wrong.

Hence, √2 is irrational.

Hence proved!!