How to show that $\forall x\in \mathbb{R},\space \exists!n\in \mathbb{Z}$ such that $n\leq x < n+1$?

Question

How can I show that, $$\forall x\in \mathbb{R},\space \exists!n\in \mathbb{Z} \text{ such that, } \space n\leq x < n+1$$ which will be denoted later as $E(x)$ or $[x]$?

In my textbook the the prof prove it like that: Let $A = \{k\in \mathbb{Z},k\leq x \}$ $A$ is a non empty part of $\mathbb{Z}$ upper bounded (by x) in $\mathbb{R}$ so $A$ admits a largest element therefore $\exists!n\in \mathbb{Z}$ such that, $n\leq x<n+1$ but I can't get this part

therefore $\exists!n\in \mathbb{Z}$ such that, $n\leq x<n+1$

Explaining the points at which you have trouble understanding in the proof is what is needed in the question. It will greatly increase the chance to get an appropriate answer. — Sangchul Lee, Oct 17 '12 at 14:29
@sos440 are you suggesting that i should put the proof whit question? — mohamez, Oct 17 '12 at 14:48
Add the proof to the question (Edit-Button) and mark the steps you don't understand. — ulead86, Oct 17 '12 at 15:04

score 3 · Accepted Answer · answered Oct 17 '12 at 15:10

Assuming that you agree with the existence of the largest element in $A$, the part of the proof in question follows as an easy consequence.

Let $n_0$ be the largest element in the set $A$. Then it satisfies the following properties:

$n_0 \leq x$, since $n_0$ is an element of $A$.
If $n < n_0$, then $n+1 \leq x$. This is clear from the inequality $n+1 \leq n_0 \leq x$.
If $n > n_0$, then $n > x$. Indeed, assume $n > n_0$. Then $n$ is not a member of $A$, otherwise the maximality of $n_0$ is contradicted. Since $n \notin A$, we have $n > x$.

Since $n_0 + 1 > n_0$, we have $n_0 \leq x < n_0 + 1$, which proves the existence. Also, both 2 and 3 excludes the possibility that $n \leq x < n+1$ is satisfied for some $n$ other than $n_0$. This proves the uniqueness.

If you are questioning the existence of the largest element of $A$, you may exploit the well-ordering principle of $\Bbb{N}$ with a slight modification.

Why he choose $A = {k\in \mathbb{Z},k\leq x }$ ? – mohamez Oct 17 '12 at 15:46 — mohamez, Oct 17 '12 at 15:46

score 1 · Answer 2 · edited Apr 13 '17 at 12:20

1

Hint $\ $ Apply continuous mathematical induction.

Remark $\ $ See the linked thread for various approaches to this sort of induction.

edited Apr 13 '17 at 12:20

Community

1

answered Oct 17 '12 at 15:23

Bill Dubuque

272,048

Jean-Sébastien · Answer 3 · 2012-10-17T15:05:42.640

0

What the proof says is that there is some set $A$, made of integer that are all less or equal to $x$. The set being bounded, it has a largest element, say $N$. In particular, $N\leq x$ by definition. Now why can we conclude that $$ N\leq x<N+1? $$

Suppose $x\geq N+1$, then certainly $N+1$ would be in our set $A$, which contradict the fact that $N$ was the maximal element. Therefore we conclude that $N\leq x<N+1$

edited Oct 17 '12 at 15:05

answered Oct 17 '12 at 14:20

Jean-Sébastien

4,623
1
17
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Backwards. This would be proved as a consequence of the result in the question. – GEdgar Oct 17 '12 at 14:56
it depends on how you see the problem, I agree that it does not go with the added information though – Jean-Sébastien Oct 17 '12 at 14:59
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Downvote rescinded. Another point (for the uniqueness): you need that there are no integers with distance less than 1 from each other. – GEdgar Oct 17 '12 at 15:15

How to show that $\forall x\in \mathbb{R},\space \exists!n\in \mathbb{Z}$ such that $n\leq x < n+1$?

3 Answers3