we will use method of contradiction let root 2 be rational no. root2=p\q where q is not equal to zero and h.c.f.(p,q)=1 squaring on both sides, we get 2=p square÷q square 2q square=p square (1) p square is divisible by 2 therefore,p is divisible by 2 p=2k k is integer put in (1) 2q square=(2k) square 2q squrae=4k square q square = 2k square q square is divisible by 2 q is divisible by 2 now p and q both are divisible by 2 h.c.f.(p,q)=2 but their h.c.f. must be 1 so our supposition becomes wrong hence,root 2 is an irrational no. hope it helps uhh!!

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