Showing for any real number $\lfloor a\rfloor+1>a$ This seem a simple proposition
For any real number a $\lfloor a\rfloor+1>a$
For any example
$\lfloor 2.9\rfloor=2$
$\lfloor 3.1\rfloor=3$
$\lfloor 4\rfloor=4$
I think this is obvious. Because a number with decimals and take out the decimals and add 1 it would be bigger than the number with decimals.
 A: Assume the contrary, $\lfloor a\rfloor +1 \le a$.
Then consider $\lfloor a\rfloor + 1$, which is a larger integer less than or equal to $a$. This contradicts to the definition of floor function: 

$\lfloor a\rfloor$ is the maximum of those integers that are smaller or equal to $a$.

A: The proposition is correct. For each $x \in \mathbb{R}$ the number $\lfloor x \rfloor$ is defined as the unique integer $a$ that satisfies the inequality $a \le x < a + 1$. Therefore $x < \lfloor x \rfloor + 1$.
A: In addition to the previous answers, it may also help to first think of what $a-\lfloor{a}\rfloor$ equals. For instance, if $a=2.9$, as you have given, $a-\lfloor{a}\rfloor=0.9.$
If $a=-3.1$, $\lfloor{a}\rfloor=-4$, so that $a-\lfloor{a}\rfloor=0.1$ (why?). 
If $a=2$, $\lfloor{a}\rfloor=2$, so that $a-\lfloor{a}\rfloor=0$. 
We see then that $a-\lfloor{a}\rfloor$ is the "decimal part" of your real number $a$, so to speak. As such, $0 \leq a-\lfloor{a}\rfloor < 1$, meaning that $$\lfloor{a}\rfloor \leq a < \lfloor{a}\rfloor+1$$
which is what we wanted to show (actually, a little bit more!). 
