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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.

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  • $\begingroup$ This is hard to read. here is a good tutorial on formatting for this site. $\endgroup$
    – lulu
    Sep 9, 2019 at 12:23
  • $\begingroup$ Thank you so much. $\endgroup$
    – Kuan
    Sep 9, 2019 at 13:35

3 Answers 3

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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.

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  • $\begingroup$ The problems are $\frac{-1}{3}$ of $\left\lfloor \frac{x-1}{3}\right\rfloor$ is not an integer, where $\frac{x}{3}$ is real number. Also, $\frac{1}{3}$ of $\left\lfloor \frac{x+1}{3}\right\rfloor$ is not an integer, where $\frac{x}{3}$ is a real number.. $\endgroup$
    – Kuan
    Sep 9, 2019 at 13:32
  • $\begingroup$ @Kuan: think twice. Also note that the question 2) has absolutely nothing to do with the floor. $\endgroup$
    – user65203
    Sep 9, 2019 at 13:35
  • $\begingroup$ Is this because $\left\lfloor -\frac{1}{3}\right\rfloor =-1$ of $\left\lfloor \frac{x-1}{3}\right\rfloor$? But, can we do it likes $\left\lfloor \frac{x-1}{3}\right\rfloor =\left\lfloor \frac{x}{3}\right\rfloor +\left\lfloor -\frac{1}{3}\right\rfloor$? This is because $\left\lfloor \frac{x-1}{3}\right\rfloor \neq \left\lfloor \frac{x}{3}\right\rfloor +\left\lfloor -\frac{1}{3}\right\rfloor$. For example: let x=4, $\left\lfloor \frac{x-1}{3}\right\rfloor =1$ and $\left\lfloor \frac{x}{3}\right\rfloor +\left\lfloor -\frac{1}{3}\right\rfloor =1-1=0$. $\endgroup$
    – Kuan
    Sep 9, 2019 at 13:48
  • $\begingroup$ @Kuan: use the given property !!! $\endgroup$
    – user65203
    Sep 9, 2019 at 14:12
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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 $\dfrac{x+1}3 = \dfrac x3 + \dfrac 13$.

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

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