How to prove this property of floor function? $
\left\lfloor { - x} \right\rfloor  =  - \left\lfloor x \right\rfloor
$
if $x \in \mathbb{Z}$ and
$ \left\lfloor { - x}\right\rfloor = - \left\lfloor x \right\rfloor -1 $ otherwise
This is an exercise from Tom Apostol's book "Calculus Volume I" section 1.11 number 4. He defined $\left\lfloor x \right\rfloor$ as the greatest integer $\leqslant x$.
I have tried it but I don't get it. Could you help me?.
 A: Use that you define $[x]$ such that $[x]\leq x<[x]+1$. Now replace this into all the statements you want to prove.
A: Let, $x  =  y+z$  where y is an integer number and $1>z>0$. So $y = \left\lfloor x\right\rfloor$.
multiplying both sides of $x  =  y+z$ with -1, we get,
$-x=-y-z$
So $-x<-y$ and the difference between $-y$ and $-x$ is $-z$ where, $-1<z<0$ So, $\left\lfloor -z\right\rfloor=-1$.
$
\left\lfloor { - x} \right\rfloor  = -y-1= - \left\lfloor x \right\rfloor-1
$ as $\left\lfloor {x} \right\rfloor = y$
A: Lets make 2 cases:CASE 1 :**$
\left\lfloor { - x} \right\rfloor  =  - \left\lfloor x \right\rfloor
$
if $x \in \mathbb{Z}$  Whenever we have an Integer inside this function it remains unchanged so bringing the minus sign out is completely legal.
 **CASE 2:
$ \left\lfloor { - x}\right\rfloor = - \left\lfloor x \right\rfloor -1 $ otherwise.
 We know [-x] = -x - {-x} where{.} denotes the fractional part of x
Take minus common in RHS to get [-x] = -(x+{-x})
Now we know that {-x} = 1-{x} when x is not an integer (By the very nature of{.}) 
 Now substitute this to the above equation to get the desired result. 
