Prove that √2 is irrational also …

Ans: √2 Let us assume that √2 is a rational number Take a and b are co prime numbers where b is not equal to 0 Such that √2 =a/b √2b= a take it as eqn 1 Squaring both side (√2b) 2 = a2 (2b) 2 = a2 take it as eqn two b2= a2/2 2 divides a2 Also 2 divides a ( if a prime number p divides a2 then divides a also) Let a = 2c Substitute in 2 (2b)2 =( 2c) 2 2b2 = 4c2 2b2 / 4= c2 b2 /2 = c2 2 divides b2 Also 2 divides b Therefore, 2 divides both a and b also a and b are co prime number. Our assumption that √2 is a rational number is wrong. √2 is a irrational number.

1. 

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.