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}]
Students Helping Clan
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