Showing that $\lfloor\frac{x-1}3\rfloor=\lfloor\frac{x}3+\frac23\rfloor-1$ and $\lfloor\frac{x+1}3\rfloor=\lfloor\frac{x}3+\frac13\rfloor$.

Question

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.

Answer 1

An integer number can freely cross the floor delimiters. For all real $a$ and integer $n$,

$$\lfloor a+n\rfloor=\lfloor a\rfloor+n.$$

This is enough to justify the claims.

Answer 2

I did it finally.

Definition: $\lfloor x\rfloor =m$ if $m\leq x<m+1$.

So, $\left\lfloor \frac{x-1}{3}\right\rfloor =\frac{m-1}{3}$ if $\frac{m-1}{3}\leq \frac{x-1}{3}<\frac{m-1}{3}+1$.

Then, $\left\lfloor \frac{x-1}{3}\right\rfloor =\frac{m-1}{3}=\left\lfloor \frac{x-1}{3}\right\rfloor$, where $\lfloor x\rfloor =m$.

Next, $\left\lfloor \frac{x+2}{3}\right\rfloor -1=\frac{m+2}{3}-1$ if $\frac{m+2}{3}-1\leq \frac{x+2}{3}-1<\frac{m+2}{3}$.

$\frac{m+2}{3}-1=\frac{1}{3} (m+2-3)=\frac{m-1}{3}=\left\lfloor \frac{x-1}{3}\right\rfloor$.

Thus, $\left\lfloor \frac{x-1}{3}\right\rfloor =\left\lfloor \frac{x+2}{3}\right\rfloor -1$.

Answer 3

If n is an integer, then $\lfloor x + n \rfloor = \lfloor x \rfloor + n$.

\begin{align} \left\lfloor\frac{x-1}3 \right\rfloor &= \left\lfloor\frac{x-1}3 \right\rfloor + 1 - 1 \\ &= \left\lfloor\frac{x-1}3 + 1\right\rfloor - 1 \\ &= \left\lfloor\frac x3 + \frac 23 \right\rfloor - 1 \end{align}

The second is true because $\dfrac{x+1}3 = \dfrac x3 + \dfrac 13$.