ISI-2009 - SIB

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 >

4 Answers

9
Celestine preetham ·

<√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

1
Rohan Ghosh ·

one thing was missed in problem

it has to be proved that xn=n+<√n>

9
Celestine preetham ·

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

...

1
skygirl ·

oh am really vry sry :(
that was a typo .... sry for that...

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