I have 2 questions about the floor functions:
1) $\left\lfloor \frac{x-1}{3}\right\rfloor =\left\lfloor \frac{x}{3}+\frac{2}{3}\right\rfloor -1$
2) $\left\lfloor \frac{x+1}{3}\right\rfloor =\left\lfloor \frac{x}{3}+\frac{1}{3}\right\rfloor$
As we know that the definitions and properties of floor functions are:
1) $\lfloor x\rfloor =m$ if $m\leq x<m+1$ and
2) $\lfloor m+x\rfloor =\lfloor x\rfloor +m$ if $m$ is an integer.
Questions:
1) Why the first floor function above has to +1 inside the floor brackets and -1 outside the floor brackets: $\left\lfloor \frac{x-1}{3}\right\rfloor$ = $\left\lfloor \frac{x-1}{3}+1\right\rfloor -1$ = $\left\lfloor \frac{x}{3}+\frac{2}{3}\right\rfloor -1$
2)Why the second floor function above doesn't need to add or minus 1 inside or outside the floor brackets: $\left\lfloor \frac{x+1}{3}\right\rfloor$ = $\left\lfloor \frac{x}{3}+\frac{1}{3}\right\rfloor$
Does anyone here know the reason? Thank you.