Proof that intervals of the form $[x, x+1)$ or $(x, x+1]$ must contain a integer. Show that any real interval of the form $[x, x+1)$ or $(x, x+1]$ must contain a integer.
Here is my proof (by contradiction)
We start by saying, assume the interval of the form $[x, x+1)$ or $(x, x+1]$ does not contain a integer. Now suppose $x$ is an element of $\mathbb{Z}$, then x is in the interval $[x, x+1)$ and since $x$ is an integer, then $x+1$ must also be an integer.  Thus $(x, x+1]$ must also contain a integer. This is a contradiction, it contradicts our assumption.  Therefore the statement is true.
Can someone help me on this.  Thanks in advance.
 A: These two statements are equivalent:


*

*$[x, x+1)$ or $(x, x+1]$ always contains an integer

*$[x, x+1]$ always contains an integer.


For the proving, we can suppose, that $x \notin \mathbb{Z}$(otherwise $x$ is a trivial solution). Then, we can find an $y$, such that, $y > x, y \in \mathbb{Z}$. Take the following set: {$y \in \mathbb{Z}, y > x$}. Since this set is bounded from below(and obviously not empty), it must have a minimal element, let this element be $m$.
We prove, that $m \in [x,x+1]$.
Since $x \notin \mathbb{Z},  |m-x| < 1$. Therefore, $m \in [x,x+1], m \in \mathbb{Z}$. QED.
A: As you've observed, if $x$ is an integer, then so is $x+1$, so in that case we are done. Otherwise let $n$ be the smallest integer such that $n>x$, which must exist because there are integers greater than $x$ and the interval $(x,\infty)\cap\mathbb Z$ is well ordered. I claim that $n<x+1$. If this were not true then we'd have $n-1>x$, contradicting the choice of $n$. Thus $n\in (x,x+1)$ and we are done.
A: Note that
$$ 0 \le \{ x \} < 1,$$
where $\{ x \}$ is the fraction of $x$, thus $\{ x \} = x - \lfloor x \rfloor$, and $\lfloor x \rfloor$ is the floor of $x$.

Thus
$$0 \le x - \lfloor x \rfloor < 1,$$
or
$$x \ge \lfloor x \rfloor > x - 1.$$
Therefore
$$\lfloor x \rfloor \in ( x - 1, x ],$$
but also
$$\lfloor x+1 \rfloor \in ( x, x + 1 ],$$
and $\lfloor x+1 \rfloor$ is an integer.

But we also have
$$0 \le - x - \lfloor - x \rfloor < 1,$$
or
$$x \le - \lfloor - x \rfloor < x + 1.$$
Therefore
$$- \lfloor - x \rfloor \in [ x , x + 1 ),$$
and $- \lfloor - x \rfloor$ is an integer.

So there is always an integer element of either $ [ x, x + 1 ) $ or $ ( x, x + 1 ]$.
A: Assume $[x,x+1)$ does not contain an integer. Then $x$ is not an integer and neither is $x+1$. Then $[x,x+1]$ does not contain an integer. Assume WOLG, that $x=0.a_1a_2...$ , i.e., $x>0$. Then we must have $x+1 <1 $ (otherwise there would be an integer in the interval). But $x>0$ , so $x+1>1$, so there must be an integer in the interval. Similar argument for $(x,x+1]$.
A: You have proved the statement only for $x\in\Bbb Z$.
If $x\notin\Bbb Z$, consider the set $A=\{t\in\Bbb Z: t>x\}$. This set is bounded below and not empty, and it is well-ordered. Let $n=\min A$. Try to prove that $n$ is in the interval.
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