MHB Is Floor Function Equality Proven for All Non-Negative x?

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The discussion centers on proving the equality of the floor function for non-negative values of x, specifically that for all x ≥ 0, the equation floor(sqrt[n]{x}) equals floor(sqrt[n]{floor(x)}) holds true. Participants acknowledge the proof provided by kaliprasad, affirming its correctness. The conversation emphasizes the importance of understanding the behavior of the floor function in relation to roots and integer values. Overall, the proof is confirmed and appreciated by the community. The equality is thus established for all non-negative x.
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For all $x\ge 0$, prove that $$\left\lfloor{\sqrt[n]{x}}\right\rfloor=\left\lfloor{\sqrt[n]{\left\lfloor{x}\right\rfloor}}\right\rfloor$$.
 
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anemone said:
For all $x\ge 0$, prove that $$\left\lfloor{\sqrt[n]{x}}\right\rfloor=\left\lfloor{\sqrt[n]{\left\lfloor{x}\right\rfloor}}\right\rfloor$$.

let $l^n <= x < (l+1)^n$
then $\lfloor{\sqrt[n]{x}}\rfloor = l\cdots(1)$

from the given condition because l is integer $l^n$ is also integer and it cannot be greater than $\lfloor x \rfloor$

so $l^n <= \lfloor x \rfloor < (l+1)^n$

so $l = \lfloor{\sqrt[n]{\lfloor{x}\rfloor}\rfloor}\cdots(2)$
from (1) and (2) we get the result
 
Well done, kaliprasad! And thanks for participating!
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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