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.

  • $\begingroup$ $n$ needs not to be positive - take $x=-2.3$, then $n=-3$ $\endgroup$ – vrugtehagel Feb 15 '16 at 21:00

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 $n \le x < n+1$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.