**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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