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This is exercise 4 from chapter I 3.12 from Tom Apostol's "Calculus" (2nd edition):

If $x$ is an arbitrary real number, prove that there is exactly one integer $n$ which satisfies the inequalities $n \le x \lt n+1$. This $n$ is called the greatest integer in $x$ and is denoted by $\lfloor x \rfloor$.

Up to this point in the book induction, nor well-ordering principle isn't introduced. All that's available is some basic theorems deduced from 10 axioms of real numbers. For example I can't claim that for every real number $x$ there is $n$ satisfying $x \le n \lt x + 1$ (which would be enough for me to make the proof).

My aim at the proof was:

Let S be the set of all positive integers $n$ which satisfy $n \lt x$.

From Theorem I.29 (For every real $x$ there exists a positive integer $n$ such that $n \gt x$) S is not empty.

Notice that $y < x$ for every $y \in S$, so $x$ is an upper bound of $S$.

From Axiom 10 (Every upper-bounded set of real numbers has a supremum) $S$ has supremum $L$ and $L \le x$.

I don't know how to prove that this supremum is an integer.

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  • $\begingroup$ $n$ needs not to be positive - take $x=-2.3$, then $n=-3$ $\endgroup$ – vrugtehagel Feb 15 '16 at 21:00
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You should take $S$ to be the set of all integers $n$ (not necessarily non-negative) satisfying $n \le x$ and let $L = \sup S$. By definition of sup, as you point out, $L \le x$.

Why must $L$ be an integer? We can avoid that question as follows: since $L = \sup S$, there exists (by definition of the supremum) an element $n \in S$ satisfying $L-1 < n \le L$.

By the transitive property you get $n \le x$.

On the other hand, $L < n+1$ implies that $n+1 \notin S$ because $L = \sup S$. The way $S$ is defined leads to $n+1 > x$.

Thus $n \le x < n+1$.

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  • $\begingroup$ How in the second step you prove that there exists $n$ which satisfies $L - 1 \lt n \le L$? $\endgroup$ – Robert Kusznier Feb 15 '16 at 21:37
  • $\begingroup$ If there was no such $n$, then $L-1$ would be an upper bound of $S$! $\endgroup$ – user312938 Feb 15 '16 at 21:49
  • $\begingroup$ True. Thanks for the help! $\endgroup$ – Robert Kusznier Feb 15 '16 at 22:21
  • $\begingroup$ (Do you mean the transitive property, instead of the commutative property?) $\endgroup$ – user84413 Feb 16 '16 at 0:26
  • $\begingroup$ Oops, I will fix that. $\endgroup$ – user312938 Feb 16 '16 at 15:23

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