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!