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

You have simply restated what you want to prove.
By definition, $n=\lfloor y\rfloor$ if and only if $n$ is an integer and $n\le y<n+1$.
Let $m=\lfloor x\rfloor$. Then $m$ is an integer and $m\le x<m+1$, so $m-1$ is an integer and
$$
m-1\le x-1<m
$$
Therefore $m-1=\lfloor x-1\rfloor$.

More generally, the same argument shows that if $k$ is an integer, then for every real $x$ we have
$$
\lfloor x+k\rfloor = \lfloor x\rfloor + k
$$
