<s>[INST] Logaritmic equation with floor brackets $ \left \lfloor \frac{x-1}{2} -\left \lfloor \frac{x}{2} \right \rfloor \right \rfloor=\log_{10}x$ Solve equation
$ \left \lfloor \frac{x-1}{2} -\left \lfloor \frac{x}{2} \right \rfloor \right \rfloor= \log_{10}x$
I dont understand solution explained in the book.
$ \left \lfloor \frac{x-1}{2} -\left \lfloor \frac{x}{2} \right \rfloor \right \rfloor= \left \lfloor \frac{x}{2}- \left \lfloor \frac{x}{2} \right \rfloor -\frac{1}{2} \right \rfloor=\left \lfloor \left \{ \frac{x}{2} \right \}-\frac{1}{2} \right \rfloor$
I dont understand why is
$\left \lfloor \frac{x}{2}- \left \lfloor \frac{x}{2} \right \rfloor -\frac{1}{2} \right \rfloor=\left \lfloor \left \{ \frac{x}{2} \right \}-\frac{1}{2} \right \rfloor$
 [/INST] $\{x\}$, the fractional part of x, is defined as $\{x\} = x - \lfloor x\rfloor$. (For example, $\{3.14\} = 0.14$, $\{-0.6\} = 0.4$, $\{1\} = 0$, etc. Thus, $\boxed{\frac{x}{2}-\left\lfloor\frac{x}{2}\right\rfloor = \bigg\{\frac{x}{2}\bigg\}.}$
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