# Elementary Number Theory and Its Application, 6th Edition by Kenneth H. Rosen Exercises 1.1.19

19. Show that $[\sqrt{[x]}]=[\sqrt{x}]$ whenever $x$ is a nonnegative real number.
Solution:  We have to prove the equation
$[\sqrt{[x]}]=[\sqrt{x}]$
Whenever $x$ is a non-negative real number
We assume that
$x=k+\epsilon$
Where $k$ is an integer and
$0<=\epsilon<1$
Further, we assume that
$k=a^2+b$
Where a is the largest integer such that
$a^2<=k$
Now,
$a^2<=k$
$=a^2+b$
$<=x$
$=a^2+b+\epsilon$
$a^2<(a+1)^2$
This implies that
$[\sqrt{x}] = a$
And
$[\sqrt{[x]}]=[\sqrt{k}]$
$=a$
Hence we have achieved the equality in the equation
$[\sqrt{[x]}]=[\sqrt{x}]$

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