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

[anemone]

For all $x\ge 0$, prove that \(\displaystyle \left\lfloor{\sqrt[n]{x}}\right\rfloor=\left\lfloor{\sqrt[n]{\left\lfloor{x}\right\rfloor}}\right\rfloor\).

 

[kaliprasad]

For all $x\ge 0$, prove that \(\displaystyle \left\lfloor{\sqrt[n]{x}}\right\rfloor=\left\lfloor{\sqrt[n]{\left\lfloor{x}\right\rfloor}}\right\rfloor\).

Spoiler

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

 

[anemone]

Well done, kaliprasad! And thanks for participating!