If $\ x$ is an arbitrary real number, prove that there is exactly one integer $n$ which satisfies the inequalities $n\leq x < n + 1$.
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It is obvious to see this, but I can't prove it. – Daniel Fugisawa Oct 01 '15 at 02:06
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Step 1: prove there is no integer $m$ such that $n<m<m+1$. – vadim123 Oct 01 '15 at 02:09
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1what did you try? – Ittay Weiss Oct 01 '15 at 02:09
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1@vadim123, I think you meant $n<m<n+1$. – JRN Oct 02 '15 at 03:30
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Wow! You're right. Very good answers there. Thank you. – Daniel Fugisawa Oct 03 '15 at 23:07
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Assume $x>0$, the general case then follows easily.
Existence: By the Archimedean property, the set $\{k \in \mathbb{N}: k>x \}$ is non-empty, and hence contains a least element, call it $n$. Since $n-1<n$, it follows that $n-1\le x$.
Uniqueness: Use that, if $a<b$ are integers, then $b-a<1 \iff a=b$.
Reveillark
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Thank you so much. I developed your ideas and made the demonstration step-by-step. I'm self-studying for the first time Calculus (Apostol), and LA (Axler), in order to prepare for a Real Analysis course. That's why such a basic question. – Daniel Fugisawa Oct 01 '15 at 02:59
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Hey! The set you described with a least element, has an infimum. It follows from the uniqueness of infS that n must be unique. Right? – Daniel Fugisawa Oct 02 '15 at 00:22
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The ser has a least element, which in particular is the infimum of the set, which is unique. However it is not immediate from this uniqueness that the integer with the you want is also unique. – Reveillark Oct 02 '15 at 00:46