Prove the existence of a largest integer less than or equal to a rational number Prove that there is a largest integer $n$ such that $n \le x$ for any fixed rational $x$.
What about any fixed real number $x$?
 A: 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 $≤x$.
A: (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)
A: 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.
A: 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.
