Floor function simplification identities I can't seem to find any identity(if any)for division/multiplication involving floor functions: for example
$$\lfloor{\frac{n-1}{2}}\rfloor\cdot 2$$
I know does not simplify down to $$n-1$$. 
 A: You could consider separately what happens when $n$ is an even number and when $n$ is an odd number.
If $n=2k$, then
\begin{align}
   2\left\lfloor \dfrac{n-1}{2} \right\rfloor
      &=  2\left\lfloor \dfrac{2k-1}{2} \right\rfloor\\
      &=  2\left\lfloor k - \dfrac 12 \right\rfloor\\
      &=  2(k-1) \\
      &= 2\left( \dfrac n2 - 1 \right) \\
      &= n - 2
\end{align}
If $n=2k+1$, then
\begin{align}
   2\left\lfloor \dfrac{n-1}{2} \right\rfloor
      &=  2\left\lfloor \dfrac{2k+1-1}{2} \right\rfloor\\
      &=  2\left\lfloor k \right\rfloor\\
      &=  2k \\
      &=  n - 1
\end{align}
A: The floor function involves rounding down to the greatest integer less than or equal to the number. Thus, it involves a reduction of $0$ up to just less than $1$. In particular, this means for all real $x$ that
$$x - 1 \lt \lfloor x \rfloor \le x \tag{1}\label{eq1}$$
Thus, with your example, you have
$$\frac{n - 3}{2} \lt \left\lfloor \frac{n-1}{2} \right\rfloor \le \frac{n-1}{2} \tag{2}\label{eq2}$$
Multiplying by $2$ gives
$$n - 3 \lt \left\lfloor \frac{n-1}{2} \right\rfloor \cdot 2 \le n-1 \tag{3}\label{eq3}$$
Among integers, the result can potentially be either $n - 2$ or $n - 1$, depending on the value of $n$. For example, if $n = 2$, then the result is $0$, i.e., $n - 2$.
A: We have that 
$$
x = \frac{x}
{2} + \frac{x}
{2} = 2\left\lfloor {\frac{x}
{2}} \right\rfloor  + 2\left\{ {\frac{x}
{2}} \right\} = 2\left\lfloor {\frac{x}
{2}} \right\rfloor  + x\bmod 2
$$
so
$$
2\left\lfloor {\frac{x}
{2}} \right\rfloor  = x - 2\left\{ {\frac{x}
{2}} \right\} = x - x\bmod 2
$$
which means that for any $x$
$$
x - 2 < 2\left\lfloor {\frac{x}
{2}} \right\rfloor  \leqslant x
$$
