If $x$ is any positive real number, then prove that there exists a non-negative integer denoted by $[x]$ such that $[x]\leq x<[x]+1$.

(A Hints is given as: Apply Archimedean Property in $\mathbb{R}$)

I know the Archimedian property as:

If $a(>0), b\in \mathbb{R}$, then there exits at least one positive integer $n$, such that $na>b$.

here, how to choose $a,b, n$.? How to use the theory to prove the result.

  • $\begingroup$ Your "Archimidean property" is actually a corollary of the Archimidean property that for every real number $x$ there exists a natural number $n$ such that $n>x$ $\endgroup$ – smcc Aug 22 '16 at 14:22
  • $\begingroup$ Don't you think this is the definition for the greatest integer function? $\endgroup$ – StubbornAtom Aug 22 '16 at 14:46
  • $\begingroup$ @StubbornAtom Yes. the definition is for the greatest integer function. But how to prove by Archimedean Property is a problem $\endgroup$ – user1942348 Aug 22 '16 at 14:55
  • $\begingroup$ @smcc Yes. I understand. Applying Archimedean Property how to prove that. $\endgroup$ – user1942348 Aug 22 '16 at 14:56
  • $\begingroup$ This appears to be math.stackexchange.com/questions/117734/… restricted to positive reals. $\endgroup$ – David K Aug 22 '16 at 17:14

AP $\implies$ there exist $n$ such that $n > x$. The set X = {$m: m>x$} is a non empty set of positive integers so it has a minimum element $k$. Then $[x] :=k-1$ (if $k -1 > x$ then $k-1$ is an element of $X$ but it can't be so $k-1 \le x$)

  • $\begingroup$ If $x$ is not an integer, would't it be $[x] = k-1$? We're supposed to have $[x]\leq x$, after all. $\endgroup$ – Arthur Aug 22 '16 at 14:29
  • $\begingroup$ @Arthur right! see edit $\endgroup$ – user290300 Aug 22 '16 at 14:30

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