Prove that (root) √2 is an Irrational Number Class 9,10- Number System - Ranker Studys

Solution:

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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.

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