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Prove that there is a largest integer $n$ such that $n \le x$ for any fixed rational $x$.

What about any fixed real number $x$?

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Where did you get this problem? This looks like it could be a homework problem. Did you try it and are stuck? If you show your working, people can help you better. –  Aryabhata Dec 24 '10 at 7:52
    
Oh, it's not a homework problem. I teach myself, for me there is no such thing as "homework". :) It's just a sentence in an analysis textbook with no proof given. –  qed Dec 24 '10 at 12:54
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As Christian Blattner says, in questions like this it is difficult to know what can be used and what can not in the proof. It is "obviously true" (not denigrating the question), because so much of what we do depends on it. But if you are constructing the numbers, there are different routes, which lead to different answers. –  Ross Millikan Dec 24 '10 at 15:24
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4 Answers

up vote 4 down vote accepted

The answer to your question depends on the set of facts you take for granted. In terms of elementary number theory I would argue as follows: Let $x=p/q$ with $q>0$. Then the $q$ consecutive numbers $p-q+1$, $p-q+2$, $\ldots$, $p$ form a complete residue system mod$\thinspace q$, so there is exactly one among them, say $r$, that is divisible by $q$. It follows that $n:=r/q$ is the largest integer $≤x$.

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I like this answer because it doesn't make use of the extra properties which $\mathbb{R}$ has. It seems preferable to me to avoid special properties of $\mathbb{R}$ to prove this fact which certainly remains true if we totally ignore the existence of $\mathbb{R}$. –  Zach Conn Dec 24 '10 at 17:44
    
Thank you, this proves it in a beautiful way. –  qed Dec 26 '10 at 6:20
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(Sorry to post this as an answer, but I can't comment yet). Do you mean that given a fixed rational number $x$, the set $\lbrace n: n\leq x, n \in \mathbb Z\rbrace$ has a maximum? (Hint: Use the well-ordering principle)

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Yeah. That is what I meant. –  qed Dec 24 '10 at 12:49
    
I got your point. Consider $A=\{m:m>x,\; m\in Z\}$ , then A is bounded below and has a least element. Let $m_{0}$ be the least element of A, then $m_{0}-1$ is not in A, hence $m_{0}-1 \le x$, and it is the greatest integer satisfying that condition. –  qed Dec 24 '10 at 13:29
    
But I'll wait for a different proof. Especially a proof using Archimedean property, since the textbook says this result is based on it. –  qed Dec 24 '10 at 13:31
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First, as a clarification pointed out by Dactyl, you probably want a proof of:

For any rational (or real) $x$, there is a largest integer $n$ such that $n \leq x$.

The reversal of words makes the claim false because it says that there is a fixed largest integer $n$ less than or equal to arbitrarily large $x$.

Also for Dactyl's hint, you have to be careful when $x$ is negative because the well-ordering principle is for Natural numbers. Specifically, not every nonempty subset of the integers has a smallest element (e.g., the integers themselves have no least element). However, every nonempty subset of the integers that has a lower bound has a least element, and you can prove this by induction on the absolute value of an integral lower bound.

Of course, $n$ is typically used to denote a Natural number so maybe you're restricting yourself to nonnegative $x$ anyway.


Edit (in response to request for different proof):

Consider $\{n \in \mathbb{N}| |x| < n\}$. The Archimedean property tells us that for any Real number, we can find a Natural number that's larger so this set is non-empty. Then by the well-ordering of the Natural numbers, there exists a least element in this set, call it $n_0$. If $x \geq 0$, verify that $n_0 - 1$ is the desired integer, and if $x < 0$, verify that that $1 - n_0$ is the one you want if $x$ is an integer and $-n_0$ otherwise.

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Since the problem is tagged "real analysis", I'd just use the supremum of the set $\{ n \in \mathbf{Z} : n \le x \}$. Of course, you need to prove that this set is not empty and that the supremum is actually a maximum.

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