help me verify my proof ($\lfloor x-1 \rfloor = \lfloor x \rfloor - 1$) prove the following statement:
$\forall x \in \mathbb{R}, \lfloor x-1 \rfloor = \lfloor x \rfloor - 1$
suppose $x \in \mathbb{Z}$, then $\lfloor x-1 \rfloor = x-1 $ and $ \lfloor x \rfloor -1 = x-1 $ since the floor of any integer is itself.
suppose $x \in \mathbb{R} $, then $\lfloor x-1 \rfloor$ will give an integer that is also given when taking $\lfloor x \rfloor -1$. 
eg. $\lfloor 1.5-1 \rfloor = \lfloor .5\rfloor = 0 = \lfloor 1.5 \rfloor - 1$
i think ive almost got this proof correct but something about it just doesnt seem quite right. Can someone please help me verify? 
-thanks

Good start.
You have shown that for $x \in \mathbb{Z}$, $\lfloor x-1 \rfloor = x-1 $ and $ \lfloor x \rfloor -1 = x-1 $.
For $x \in \mathbb{R} $, let $x=y+\delta$, where $y \in \mathbb{Z}$ and $0 \le \delta <1$.
Then $\lfloor x-1 \rfloor = \lfloor y+ \delta -1 \rfloor = \lfloor y-1 + \delta \rfloor = y-1$.
and $\lfloor x \rfloor-1 = \lfloor y+ \delta  \rfloor -1= y-1$.
