Question

Prove that √2 I an irrational number ?

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Answer 1

Answer:

Let √2 be a rational number 

Therefore, √2= p/q  [ p and q are in their least terms i.e., HCF of (p,q)=1 and q ≠ 0

On squaring both sides, we get 

                   p²= 2q²                                                                                    ...(1)

Clearly, 2 is a factor of 2q²

⇒ 2 is a factor of p²                                                                    [since, 2q²=p²]

⇒ 2 is a factor of p

 Let p =2 m for all m ( where  m is a positive integer)

Squaring both sides, we get 

            p²= 4 m²                                                                                          ...(2)

From (1) and (2), we get 

           2q² = 4m²      ⇒      q²= 2m²

Clearly, 2 is a factor of 2m²

⇒       2 is a factor of q²                                                      [since, q² = 2m²]

⇒       2 is a factor of q 

Thus, we see that both p and q have common factor 2 which is a contradiction that H.C.F. of (p,q)= 1

     Therefore, Our supposition is wrong

Hence √2 is not a rational number i.e., irrational number.

Answer 2

So, √2 = p/q (where p and q are co-prime number and q is not equal to 0)

Squaring Both Sides-

2 = p^2 / q^2

Cross Multiplying-

q^2 = 2 p^2 ………………………………………(1)

So, We can say that q^2 is divisible by 2 and q is also divisible by 2.

Let q = 2 r

Squaring Both Sides-

q^2 = 4 r^2 ………………………………………(2)

From Eq. 1 and 2-

2 p^2 = 4 r^2

Dividing Both Sides By 2-

p^2 = 2 r^2

So, We can say that p^2 is divisible by 2 and p is also divisible by 2.

From This we have reach a conclusion that p and q both divisible by 2.

It contradict the fact that p and q are co-prime numbers.

So, Our Supposition is wrong.

Hence, we can say that √2 is an irrational number.