Question

prove that √7 is not a rational no.​

Answer 1

Answer:

Step-by-step explanation:

Prove √7 is not a rational number.

Prove:- Let us suppose that √7 is a rational number if possible and it is in the simplest form of p/q. Then, p and q are integers, having no common factor other than 1, q is not equal to 0.    

p/q ≠ 0    

 

Now, √7 = p/q => p2 = (√7.q)2 => p2 = 7q2                     (i)

=> p2 is multiple of 7

=> p is multiple of 7                                                             (ii)

Let p = 7m for some integer m, then

P2 =7m => p2 =49m2

=>7q2 = 49m2 =>q2 =7m2

=>q2 is multiple of 7

=>q is multiple of 7                                                      (iii)

From (ii) and (iii), it follows that 7 is a common factor of p and q.

This contradicts the hypothesis that there is no common factor of p and q other than 1. So our supposition is wrong.

Hence, √7 is not a rational number.

                                                           Proved

Answer 2

Prove:- Let us suppose that √7 is a rational number if possible and it is in the simplest form of p/q. Then, p and q are integers, having no common factor other than 1, q is not equal to 0.

p/q ≠ 0

Now, √7 = p/q => p2 = (√7.q)2 => p2 = 7q2 (i)

=> p2 is multiple of 7

=> p is multiple of 7 (ii)

Let p = 7m for some integer m, then

P2 =7m => p2 =49m2

=>7q2 = 49m2 =>q2 =7m2

=>q2 is multiple of 7

=>q is multiple of 7 (iii)

From (ii) and (iii), it follows that 7 is a common factor of p and q.

This contradicts the hypothesis that there is no common factor of p and q other than 1. So our supposition is wrong.

Hence, √7 is not a rational number.

Proved