Question

prove that root 2 is irrational number​

Answer 1

Answer:

Step-by-step explanation:

Let us assume √2 is rational no.

We know that the rational no's. Are in the form of p/q form where p,q are integers

So, √2=p/q

p=√2q

We know that 'p' is a rational no. So √2q must be rational no. since it equal to p

But it doesn't occurs with √2 since it's not an integer

Therefore p=/= √2q

This contradicts the fact that √2is an irrational no.

Hence our assumption is wrong and √2 is an irrational no.

Answer 2

Statement:

To prove that root 2 is irrational no.

Proof:

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.