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What is the smallest real number $M$ so that $x>M$ implies that $[1/x]<10^{-6}$

My Intuition:

$$\begin{align}\frac{1}{x}<\frac{1}{10^6}\end{align}$$

$$\begin{align}10^6<x\end{align}$$

and we kno that $x>M$ , so it seems obvious to me the smallest real value of $M$ is :

$10^6+1=1000001$

Am I wrong? I don't know the correct answer to this. Since this was a question that was given to a friend to do as homework after she studied Limits in her calculus class. Because I used only simple algebra, this is strange and makes me suspicious about my reasoning because I didnt use calculus.

Manahil
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    Do square brackets mean just brackets or maybe a floor function? – Adayah Sep 29 '14 at 17:27
  • It seems that $M = 10^6$ works just as well. This question should be easy (and certainly does not require calculus), but is a useful stepping stone in understanding the rigorous definition of a limit. – Ben Grossmann Sep 29 '14 at 18:18
  • @Omnomnomnom , Im curious. How would $M=10^6$ work? Won't that violate all the inequalities written??? – M.S.E Sep 29 '14 at 18:58
  • @Tharindu it is indeed the case that $x > M = 10^6 \implies \frac 1x < 10^{-6}$, and there is no smaller real number satisfying this condition. – Ben Grossmann Sep 29 '14 at 19:00
  • I dont understand :/ If $x>10^6$ , and $x>M$ then it means $M>10^6$ , So you mean to say : $10^6>10^6$ which isnt correct. – M.S.E Sep 29 '14 at 19:05

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