Proving a simple property of Floor function I have to prove the following property of Floor function:

For any real number $x$, $x$ not being an integer, $\lfloor x \rfloor + \lfloor -x \rfloor = -1$.

Now, we know from the definition of floor that $\lfloor x \rfloor$ is the unique integer $n$ such that $n \leq x < n+1$. The trouble is writing $\lfloor -x \rfloor$. If I imagine a negative real number on the number line, it is obvious that $-n-1 \leq -x < -n$. Then, simply adding the two yields -1.
My problem is that I can't seem to arrive at $\lfloor -x \rfloor$ from the definition. For instance, if $x$ is a postive real number, then the floor is given by
$$n \leq x < n+1$$
Multiplying by -1 throughout, 
$$-n \geq -x > -n-1$$
$$\Rightarrow -n-1 < x \leq -n$$
Feels like I'm almost there, but this does not match the definition (the $\leq$ has appeared on the upper bound). What am I doing wrong?
 A: Put
$$\lfloor x\rfloor=:n,\qquad \lfloor-x\rfloor=: n'\ .$$
Then the definition of $\lfloor\cdot\rfloor$ implies together with $x\notin{\mathbb Z}$ that we have $$n<x<n+1\qquad\wedge\qquad n'<-x<n'+1\ .$$  Adding the left parts of these two squeezes gives $n+n'<0$, and adding the right parts leads to $n+n'>-2$. All in all only $n+n'=-1$ remains possible.
A: Let us show that $-1-[-x]$ satisfies the definition of $[x]$, for $x$ not an integer. The definition of $[x]$ would be $$[x]\leq x<[x]+1,$$
i.e. we need to check that $$-1-[-x]\leq x<-[-x].$$ 
The inequality on the right, $[-x]<-x$, is the first inequality of the definition of $[-x]$, i.e. $[-x]\leq-x<[-x]+1$, and therefore it is true. Notice the equality doesn't hold on the left, because it is not an integer.
The inequality on the left is $-x\leq [-x]+1$, which is equivalent to $-x<[-x]+1$, again because it is not an integer. And this is the other part of the definition of $[-x]$, and therefore it is true.
This proves that $$-1-[-x]=[x].$$
