If $b-a>1$ then there is a $k\in \mathbb{Z}$ such that $aGiven $a, b \in \mathbb{R}$, such that $b-a>1$, there is at least one $k\in \mathbb{Z}$ such that $a<k<b$.
My attempt:
Consider $E:=(a,b)\cap \mathbb{N}$. We need to show that $E$ is not empty. Set $b:=a+1+\varepsilon$. Then $(a,b)=(a, a+1+\varepsilon)$ for some $\varepsilon > 0$. But $(a, a+1+\varepsilon)$ contains the elements $a, a+1, a+1+\varepsilon$, and $a<a+1<a+1+\varepsilon$. Set $k:=\min\{(a+1)\cap \mathbb{N}\ne \emptyset\}$. Thus $k$ exists.
I know this proof is lame, but I have no idea how else to develop this. Would appreciate some hints.
 A: HINT: $E$ can certainly be empty: consider the case $a=-3,b=-1$. How your proof goes will depend on what you’re allowed to assume or have already proved. The hypothesis implies that $a<b$. Let $A=\{n\in\Bbb Z:b\le n\}$. If you are allowed to use the fact that every non-empty set of integers that is bounded below has a least element, let $m=\min A$, and use $m$ to find an integer between $a$ and $b$. 
If not, you’ll have to work a bit harder. Let $A=\{n\in\Bbb N:|b|\le n\}$, and let $m=\min A$. You can use this $m$ to find an integer between $a$ and $b$, but you’ll need to consider two cases, one for $b\ge 0$ and one for $b<0$.
A: Assume $a<b$
now $A=ceil(a)$, i.e. $A$ smallest integer s.t. $A>=a$.
Then $A>=a$ by definition of $A$. We have to prove that $A<b$.
Suppose not, then, $A>b$, then $A-a>A-b+1>1$.
That means $A>a+1$. So $A-1>a$, but $A-1$ is also integer contradicting the $A=ceil(a)$ assumption above. 
A: It depends on what your basic tools are.
The archimedian property states: For any real, $a$ there exist an integer $n$ such that $n \le a < n + 1$.  So if $b - a > 1$ then $b > a+1$ and we have:
$n \le a < n + 1 \le a + 1 < b$.  So $k = n+1$ is a integer that satisfies.
Of course we have to prove the archimedian property...
The reals have the least upper bound property.  Let $S = \{n \in \mathbb Z| n \le a\}$.  S is bounded above by $a$ so let $n = \sup S \le a$.  Let $1 > \epsilon > 0$ then $v - \epsilon$ is not an upper bound so there is an integer $n \le v \le a$.  $n + 1> v$ is an integer but $n +1 > v = \sup S$ so $n + 1 \not \in S$ so $n + 1 > a$.  So $n \le a < n+1 $. 
Hmmm... I guess we need to prove $S$ isn't empty... sigh...
A: As whole numbers are not upper bounded, let n be the smaller whole number greater than a. If n<b it is done.
So suppose n>b , then:
(1) n-1 > b-1
(2) and by hypothesis b-a > 1 , so b-1 > a
Therefore ( using (1) and (2)), n-1 > b-1 > a
So i found another whole number n-1 greater than a , which is absurd cause n was the minimum whole number greater than a , and n-1 < n
