Lets assume that √11 is rational
Therefore, √11 = P/Q such that HCF (P,Q) =1 and Q not 0
Now, square on both sides,
11= P² / Q²
11 Q² = P²
Therefore, P² is divisible by 11 as well as P
Therefore, P = 5A (where A is any integer
Hence √11 Q = 11 A
11 Q² = 121 A²
Q² = 11A²
Q² and Q are divisible by 11
Sice both P and Q are divisible by 11, HcF is not 1 , which is contradicting
And hence by contradiction method, √11 is an irrational number
