<√n> =λ
λ-1/2 < √n ≤ λ+1/2
λ2-λ+1/4 < n ≤ λ2+λ+1/4
λ2-λ+1 ≤ n ≤λ2+λ
x = n + < √n > = n+λ
ie
λ2+1 ≤ x ≤ λ2+2λ
so x takes all values btw λ2 and (λ+1)2
hence proved
let 'x' be positive non-square integers such that
x1=2, x2=3, x3 = 5, x4=6 .. and so on...
and we define <m> such that if the decimal part is <=0.5 we have <m> =m
and if decimal part >0.5 , then <m>=m+1
for ex..
<1.2> = 1
<2.5> = 2
<3.8> = 4
then prove that
x = n + < √n >
<√n> =λ
λ-1/2 < √n ≤ λ+1/2
λ2-λ+1/4 < n ≤ λ2+λ+1/4
λ2-λ+1 ≤ n ≤λ2+λ
x = n + < √n > = n+λ
ie
λ2+1 ≤ x ≤ λ2+2λ
so x takes all values btw λ2 and (λ+1)2
hence proved
one thing was missed in problem
it has to be proved that xn=n+<√n>
oh ok sky dint mention that
anyway all nos of form x = n + < √n > are part of the sequence
and series goes in ascending n
given
term 1 = has n=1
and term 2 must have n=2
hence trm r must have n=r
...