Can you prove that for any real number $\displaystyle x$ there exists asuniqueinteger $\displaystyle n$ such as,

$\displaystyle x-1\leq n\leq x$

thus, the function $\displaystyle f(x)=[\x x\x ]$ is well-defined.

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- Mar 25th 2006, 06:21 PMThePerfectHackerGreatest Integer Function
Can you prove that for any real number $\displaystyle x$ there exists as

**unique**integer $\displaystyle n$ such as,

$\displaystyle x-1\leq n\leq x$

thus, the function $\displaystyle f(x)=[\x x\x ]$ is well-defined. - Mar 28th 2006, 06:12 AMRebesques
Well... there certainly exists one such number, n(x); Or else, the naturals would be bounded in the reals, something the Archimedean Property denies.

So there is one, at least. The set {n(x): n(x)-1< x <n(x)} must have a least element, by the Well-Ordering Property of the naturals. This least element, is exactly [x].

Personally, I would not bother myself so much :p just take x, and kill its integer part!