Tag: Floor

Consider the continuity of x(-1)^[1/x]

by RoRi

August 31, 2015

Define:

$f(x) = x(-1)^{\left \lfloor \frac{1}{x} \right \rfloor} \qquad \text{for } x \neq 0,$

where $\lfloor \ \rfloor$ denotes the greatest integer function, or floor function. Sketch the graph of $f(x)$ for $-2 \leq x \leq -\frac{1}{5}$ and $\frac{1}{5} \leq x \leq 2$ . Evaluate

$\lim_{x \to 0^+} f(x) \qquad \text{and} \qquad \lim_{x \to 0^-} f(x).$

Is it possible to define $f(0)$ in a way that makes $f(x)$ continuous at 0.

First, we sketch the graph of $f$ on the requested intervals.

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As $x \to 0^+$ , $f(x) \to 0$ .
As $x \to 0^-$ , $f(x) \to 0$ .

If we define $f(0) = 0$ then $f$ is continuous at 0.

Consider the continuity of (-1)^[1/x]

by RoRi

August 31, 2015

Define:

$f(x) = (-1)^{\left \lfloor \frac{1}{x} \right \rfloor} \qquad \text{for } x \neq 0,$

where $\lfloor \ \rfloor$ denotes the greatest integer function, or floor function. Sketch the graph of $f(x)$ for $-2 \leq x \leq -\frac{1}{5}$ and $\frac{1}{5} \leq x \leq 2$ . Evaluate

$\lim_{x \to 0^+} f(x) \qquad \text{and} \qquad \lim_{x \to 0^-} f(x).$

Is it possible to define $f(0)$ in a way that makes $f(x)$ continuous at 0.

First, we sketch the graph of $f$ on the requested intervals.

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As $x \to 0^+$ , $f(x)$ alternates between $+1$ and $-1$ .
As $x \to 0^-$ , $f(x)$ alternates between $+1$ and $-1$ .

There is no way to define $f(0)$ to make $f$ continuous at 0 since $f(x)$ will take both values $+1$ and $-1$ no matter how small we choose our $\delta > 0$ . (So, if we were to try to define $f(0) = 1$ , then for $\varepsilon = \frac{1}{2} > 0$ there is no $\delta> 0$ such that $f(x) < \varepsilon$ whenever $|x| < \delta$ , and similarly if we try to define $f(0) = -1$ .)

Consider the limit and continuity of the floor function of (1/x)

by RoRi

August 30, 2015

Define:

$f(x) = \left\lfloor \frac{1}{x} \right\rfloor \qquad \text{for } x \neq 0,$

where $\lfloor \ \rfloor$ denotes the greatest integer function, or floor function. Sketch the graph of $f(x)$ for $-2 \leq x \leq -\frac{1}{5}$ and $\frac{1}{5} \leq x \leq 2$ . Evaluate

$\lim_{x \to 0^+} f(x) \qquad \text{and} \qquad \lim_{x \to 0^-} f(x).$

Is it possible to define $f(0)$ in a way that makes $f(x)$ continuous at 0.

First, we sketch the graph of $f$ on the requested intervals.

Rendered by QuickLaTeX.com

As $x \to 0^+$ , $f(x)$ takes on arbitrarily large positive values.
As $x \to 0^-$ , $f(x)$ takes on arbitrarily large negative values.

There is no way to define $f(0)$ to make $f$ continuous at 0.

Establish some properties of the integral of the floor function

by RoRi

August 24, 2015

Define a function

$f(x) = \begin{dcases} x - [x] - \frac{1}{2} & \text{if } x \notin \mathbb{Z} \\ 0 & \text{if } x \in \mathbb{Z}. \end{dcases}$

where $[x]$ denotes the greatest integer less than or equal to $x$ , also called the floor function.

Then, define a function

$P(x) = \int_0^x f(t) \, dt \qquad \text{for all } x \in \mathbb{R}.$

Draw the graph of $f(x)$ for $x \in [-3,3]$ and prove that $f$ is periodic with period 1.
For $0 \leq x \leq 1$ , prove
$P(x) = \frac{1}{2} (x^2 - x),$

and prove that $P$ is also periodic with period 1.
Give a formula for $P(x)$ in terms of the floor of $x$ .
Find a $c \in \mathbb{R}$ such that
$\int_0^1 (P(t) + c) \,dt = 0.$
For the constant $c \in \mathbb{R}$ from part (d), define
$Q(x) = \int_0^x (P(t) + c) \, dt.$

Prove that $Q$ is periodic with period 1 and that if $0 \leq x \leq 1$ , we have,

$Q(x) = \frac{x^3}{6} - \frac{x^2}{4} + \frac{x}{12}.$

The graph of $f$ is as follows:

Then, we prove that $f$ is periodic with period 1.
Proof. We compute to show $f(x+1) = f(x)$ (where I’ve replaced Apostol’s notation $[x]$ with $\lfloor x \rfloor$ since this less likely to cause confusion, also we use the solution to this exercise in the second line)

$\begin{align*} f(x+1) &= (x+1) - \lfloor x+1 \rfloor - \frac{1}{2} & (\lfloor x \rfloor \text{ denotes floor})\\ &= x - (\lfloor x \rfloor + 1) + \frac{1}{2} & (\lfloor x+1 \rfloor = \lfloor x \rfloor + 1)\\ &= x - \lfloor x \rfloor - \frac{1}{2} \\ &= f(x). \qquad \blacksquare \end{align*}$
Proof. First, we establish the requested formula,
$\begin{align*} P(x) = \int_0^x f(t) \, dt &= \int_0^x \left( t - \lfloor t \rfloor - \frac{1}{2} \right) \, dt \\ &= \int_0^x t \, dt - \int_0^x \lfloor t \rfloor \, dt - \int_0^x \frac{1}{2} \, dt \\ &= \left. \frac{t^2}{2} \right|_0^x - \left. \frac{1}{2} \int_0^x & (\lfloor t \rfloor = 0 \text{ for all } x \in [0,1))\\ &= \frac{1}{2} (x^2 - x). \end{align*}$

This was the requested formula. Next, to show $P$ is periodic with period 1, we show $P(x+1) = P(x)$ for any $x$ ,

$\begin{align*} P(x+1) = \int_0^{x+1} f(t) \, dt &= \int_0^1 f(t) \, dt + \int_1^{x+1} f(t) \, dt \\ &= P(1) + \int_0^x f(t+1) \, dt. \end{align*}$

Then, using the formula in the first part of (b) we have

$P(x) = \frac{1}{2} (x^2 - x) \quad \implies \quad P(1) = \frac{1}{2} (1^2 - 1) = 0.$

Further, from part (a) we know $f$ is periodic with period 1, so $f(t+1) = f(t)$ , and we have,

$\begin{align*} P(x+1) &= 0 + \int_0^x f(t) \, dt \\ &= P(x), \end{align*}$

by definition of $P(x)$ . Thus, $P$ is periodic with period $1. \qquad \blacksquare$
To express $P(x)$ in terms of of $\lfloor x \rfloor$ we compute as follows:
$\begin{align*} P(x) = \int_0^x f(t) \, dt &= \int_0^1 f(t) \, dt + \int_1^2 f(t) \, dt + \cdots + \int_{\lfloor x \rfloor}^x f(t) \, dt \\ &= 0 + 0 + \cdots + \int_{\lfloor x \rfloor}^x f(t) \, dt \\ &= \int_0^{x - \lfloor x \rfloor} f(t) \, dt, \end{align*}$

where $f(t + \lfloor x \rfloor) = f(t)$ since $\lfloor x \rfloor$ is an integer by definition, and $f$ has period 1, so $f(t+n) = f(t)$ for any integer $n$ . Continuing,

$\begin{align*} P(x) &= \int_0^{x - \lfloor x \rfloor} \left( t - \lfloor t \rfloor - \frac{1}{2} \right) \, dt \\ &= \left. \frac{t^2}{2} \right|_0^{x - \lfloor x \rfloor} - \left. \frac{t}{2} \right|_0^{x - \lfloor x \rfloor}. \end{align*}$

Here, we know the $\lfloor t \rfloor$ term in the integral is zero since $\lfloor t \rfloor = 0$ for all $t \in [0, \lfloor x \rfloor ]$ since $x - \lfloor x \rfloor < 1$ . Then,

$\begin{align*} P(x) &= \frac{1}{2} \left( \left( x - \lfloor x \rfloor\right)^2 - x + \lfloor x \rfloor \right) \\ &= \frac{1}{2} \left( x - \lfloor x \rfloor\right)^2 - \frac{1}{2} \left( x - \lfloor x \rfloor \right). \end{align*}$

This is the requested expression of $P(x)$ in terms of $\lfloor x \rfloor$ .
Here we compute the integral, using the formula for $P(x)$ we established in part (c), and solve for the requested constant $c$ ,
$\begin{align*} \int_0^1 \left( P(t) + c \right) \, dt = 0 && \implies && \int_0^1 \left( \frac{1}{2} \left(t - \lfloor t \rfloor \right)^2 - \frac{1}{2}\left(t - \lfloor t \rfloor \right) + c \right) \, dt &= 0 \\ && \implies && \frac{1}{2} \int_0^1 t^2 \, dt - \frac{1}{2} \int_0^1 t \, dt + \int_0^1 c \, dt &= 0 \\ && \implies && \frac{1}{6} - \frac{1}{4} + c &= 0 \\ && \implies && c = \frac{1}{12}. \end{align*}$
First we give the proof that $Q$ is periodic with period 1.
Proof. We compute,

$\begin{align*} Q(x+1) &= \int_0^{x+1} \left( P(t) + \frac{1}{12} \right) \, dt \\ &= \int_0^1 \left( P(t) + \frac{1}{12} \right) \,dt + \int_1^{x+1} \left( P(t) + \frac{1}{12} \right) \, dt \\ &= 0 + \int_0^x \left( P(t+1) \frac{1}{12} \right) \, dt & (\text{Exp/Cont of interval of int})\\ &= \int_0^x \left( P(t) + \frac{1}{12} \right) \, dt &(P \text{ is periodic with period } 1) \\ &= Q(x). \qquad \blacksquare \end{align*}$

Next, we establish the requested formula. If $0 \leq x \leq 1$ , we have

$\begin{align*} Q(x) &= \int_0^x \left( P(t) + \frac{1}{12} \right) \, dt \\ &= \int_0^x \left( \frac{1}{2} \left( t - \lfloor t \rfloor\right)^2 - \frac{1}{2} \left( t - \lfloor t \rfloor \right) + \frac{1}{12} \right) \, dt \\ &= \frac{1}{2} \int_0^x t^2 \,dt - \frac{1}{2} \int_0^x t \, dt + \frac{1}{12} \int_0^x dt \\ \intertext{where all of the $\lfloor t \rfloor$ terms are 0 for $0 \leq t \leq 1$} \\ &= \left. \frac{t^3}{6} \right|_0^x - \left. \frac{t^2}{4} \right|_0^x + \left. \frac{t}{12} \right|_0^x \\ &= \frac{x^3}{6} - \frac{x^2}{4} + \frac{x}{12}. \end{align*}$

as requested.

Find some formulas for the integral of the step function [t]^2

by RoRi

July 18, 2015

For $n \in \mathbb{Z}_{>0}$ , prove
$\int_0^n [t]^2 \, dt = \frac{n(n-1)(2n-1)}{6}.$
For $x \in \mathbb{R}$ , with $x \geq 0$ , define
$f(x) = \int_0^x [t]^2 \, dt.$

Draw the graph of $f$ on the interval $[0,3]$ .
Find all real $x > 0$ such that
$\int_0^x [t]^2 \, dt = 2(x-1).$

Proof. Let $P = \{ 0,1, \ldots, n \}$ . Then $P$ is a partition of $[0,n]$ and $[t]^2$ is constant on the open subintervals of $P$ . Further, $[t]^2 = (k-1)^2$ for $k-1 < t < k$ . So,
$\begin{align*} \int_0^n [t]^2 \, dt &= \sum_{k=1}^n (k-1)^2 \cdot (k-(k-1)) \\ &= \sum_{k=1}^n (k-1)^2 \\ &= \sum_{k=0}^{n-1} k^2 \\ &= \frac{n^3}{3} - \frac{n^2}{2} + \frac{n}{6} \\ &= \frac{n(n-1)(2n-1)}{6}. \end{align*}$

The second to last line follows from this exercise (I.4.7, #6) $. \qquad \blacksquare$
The graph is:
By inspection, we have, $x = 1, \frac{5}{2}$ .

Compute some integrals of step functions

by RoRi

July 18, 2015

Let $n \in \mathbb{Z}_{>0}$ , prove
$\int_0^n [t] \, dt = \frac{n(n-1)}{2}.$
Let $x \in \mathbb{R}_{>0}$ , and define
$f(x) = \int_0^x [t] \, dt.$

Draw the graph of $f$ for $0 \leq x \leq 4$ .

Proof. Let $P = \{ 0,1,2, \ldots, n \}$ be a partition of $[0,n]$ . Then, by the definition of the greatest integer function, $[t]$ is constant on the open subintervals of $P$ , so
$\begin{align*} \int_0^n [t] \, dt &= \sum_{k=1}^n (k-1)(k-(k-1)) & (s_k = k-1 \text{ since } [t]=k-1 \text{ if } k-1 < t < k)\\ &= \sum_{k=1}^n (k-1) \\ &= \sum_{k=0}^{n-1} k \\ &= \frac{n(n-1)}{2}. \end{align*}$

The final equality follows from here (I.4.7, #5) $. \qquad \blacksquare$
The graph is:

Prove a formula for the sum of [(na)/b] where a,b are relatively prime integers

by RoRi

July 17, 2015

Let $a,b$ be relatively prime positive integers (i.e., they have no common factors other than 1). Then we have the formula

$\sum_{n=1}^{b-a} \left[ \frac{na}{b} \right] = \frac{(a-1)(b-1)}{2}.$

The sum is 0 when $b=1$ .

Prove this result by a geometric argument.
Prove this result by an analytic argument.

Proof. We know by the previous exercise (1.11, #6) that
$\sum_{n=1}^{b-1} \left[ \frac{na}{b} \right] = \begin{array}{l} \text{\# of lattice points inside a right triangle}\\ \text{of base } b \text{ and height } a \ (\text{letting } f(x) = \frac{ax}{b}).\end{array}$

Further, from this exercise (1.7, #4), we know

$a(T) = I_T + \frac{1}{2} B_T - 1$

where $I_T =$ number of interior lattice points, and $B_T =$ number of boundary lattice points. We also know by the formula for the area of a right triangle that

$a(T) = \frac{1}{2} (ab).$

Thus, we have,

$I = \sum_{n=1}^{b-1} \left[ \frac{na}{b} \right] = \frac{1}{2} (ab) - \frac{1}{2}B_T + 1.$

Then, to calculate $B_T$ we note there are no boundary points on the hypotenuse of our right triangle (since $a$ and $b$ have no common factor). (This follows since if there were such a point then $\frac{na}{b} \in \mathbb{Z}$ for some $n < b$ , but then we would have $a$ divides $b$ , contradicting that they have no common factor.) Thus, $B_T = a + b + 1$ . So,

$\begin{align*} \sum_{n=1}^{b-1} \left[ \frac{na}{b} \right] &= \frac{1}{2} (ab) - \frac{1}{2} (a+b+1) +1 \\ &= \frac{1}{2} (ab - a - b + 1) \\ &= \frac{1}{2} (a-1)(b-1). \qquad \blacksquare \end{align*}$
Proof. To derive the result analytically, first, by counting in the other direction we have,
$\sum_{n=1}^{b-1} \left[ \frac{na}{b} \right] = \sum_{n=1}^{b-1} \left[ \frac{a(b-n)}{b} \right].$

Then,

$\begin{align*} && \sum_{n=1}^{b-1} \left[ \frac{a(b-n)}{b} \right] &= \sum_{n=1}^{b-1} [a] - \sum_{n=1}^{b-1} \left[ \frac{na}{b} \right] - \sum_{n=1}^{b-1} 1 \\ \implies && 2 \sum_{n=1}^{b-1} \left[ \frac{a(b-n)}{b} \right] &= \sum_{n=1}^{b-1} [a] - \sum_{n=1}^{b-1} 1 = (b-1)a - (b-1)\\ \implies && \sum_{n=1}^{b-1} \left[ \frac{na}{b} \right] &= \frac{1}{2} (a-1)(b-1). \qquad \blacksquare \end{align*}$

Formula for counting lattice points in the ordinate set of a function

by RoRi

July 17, 2015

For a nonnegative function $f$ defined on an interval $[a,b]$ define the set

$S = \{ (x,y) \mid a \leq x \leq b, \ 0 < y \leq f(x) \}.$

(i.e., the region enclosed by the graph of the function and the $x$ -axis between the vertical lines at $x=a$ and $x=b$ ). Then prove

$\text{\# of lattice points in } S = \sum_{n=a}^b [f(n)],$

where $[f(n)]$ is the greatest integer less than or equal to $f(n)$ .

We can help ourselves by drawing a picture to get a good idea of what is going on, then turn that intuition into something more rigorous. The picture is as follows:

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In the picture, we can see the number of lattice points in the ordinate set of $f(x)$ (not including the $x$ -axis since the question stipulates $S$ contains the points $0 < y \leq f(x)$ ). At each integer between $a$ and $b$ , we count the number of lattice points beneath $[f(x)]$ , the greatest integer less than or equal to $f(x)$ . Then we need to turn this intuition from the picture into a proof:

Proof. Let $n \in \mathbb{Z}$ with $a \leq n < b$ . We know such an $n$ exists since $a,b \in \mathbb{Z}$ with $a < b$ . Then, the number of lattice points in $S$ with first element $n$ is the number of integers $y$ such that $0 < y \leq f(n)$ . But, by definition, this is $[f(n)]$ (the greatest integer less than or equal to $f(n)$ ). Summing over all integers $n, \ a \leq n \leq b$ we have,

$\text{\# of lattice points in } S = \sum_{n=a}^b [ f(n) ]. \qquad \blacksquare$

Deduce and prove a formula for [nx]

by RoRi

July 16, 2015

Use the previous exercise (1.11, #4 parts (d) and (e)) to deduce a formula for $[nx]$ and prove this formula is correct.

Claim:

$[nx] = \sum_{k=0}^{n-1} \left[ x + \frac{k}{n} \right].$

Proof. Let $[x] = m$ , then we have

$m \leq x < m+1 \quad \implies \quad nm \leq nx < nm+n.$

Thus, there exists some $j \in \mathbb{Z}$ with $0 \leq j < n$ such that

$nm +j \leq nx < nm+j+1.$

Hence, $[nx] = nm+j$ . Thus,

$m+\frac{j}{n} \leq x < m + \frac{j+1}{n}.$

Then, for each $k \in \mathbb{Z}$ with $0 \leq k < n-j$ we have

$\begin{alignat*}{2} &m+\frac{j+k}{n} \leq x + \frac{k}{n} < m+\frac{j+k+1}{n} & \qquad \qquad & (\text{adding } \frac{k}{n}) \\ \implies \quad & m \leq x + \frac{k}{n} < m+1 & & (\frac{j+k}{n} < 1 \text{ since } k < n-j) \\ \implies \quad &\left[ x + \frac{k}{n} \right] = m = [x] && \text{for } 0 \leq k < n-j. \end{alignat*} \end{align*}$

Then, for $n-j \leq k < n$ , we have

$\begin{alignat*}{3} &m+\frac{j+k}{n} \leq x + \frac{k}{n} < m + \frac{j+k+1}{n} \\ \implies \quad & m+1 \leq x + \frac{k}{n} < m+2 & \qquad \qquad & (\frac{j+k}{n} \geq 1 \text{ since } n-j \leq k) \\ \implies \quad & \left[ x + \frac{k}{n} \right] = m+1 = [x] + 1 && \text{for } n-j \leq k < n. \end{alignat*}\end{align*}$

So,

$\begin{align*} \sum_{k=0}^{n-1} \left[ x+ \frac{k}{n} \right] &= \sum_{k=0}^{n-j-1} \left[ x + \frac{k}{n} \right] + \sum_{k=n-j}^{n-1} \left[ x + \frac{k}{n} \right] \\ &= (n-j)[x] + j([x] + 1) \\ &= n[x] + j \\ &= nm+j \\ &= [nx]. \qquad \blacksquare \end{align*}$

Prove some properties of the greatest integer function

by RoRi

July 16, 2015

For any $x \in \mathbb{R}$ we denote the greatest integer less than or equal to $x$ by $[x]$ . Prove the following properties of the function $[x]$ :

$[x+n] = [x] + n$ for any integer $n$ .
$[-x] = \displaystyle{\begin{cases} -[x] & \text{if } x \text{ is an integer,} \\ -[x]-1 & \text{otherwise.} \end{cases}}$
$[x+y] = [x]+[y]$ or $[x]+[y]+1$ .
$[2x] = [x] + \left[ x + \frac{1}{2} \right]$ .
$[3x] = [x] + \left[ x + \frac{1}{3} \right] + \left[ x + \frac{2}{3} \right]$ .

Proof. Let $[x+n] = m$ for some integer $m$ . Then,
$\begin{align*} m \leq x+n < m+1 &&\implies && m-n &\leq x < m-n+1 \\ &&\implies && [x] &= m-n \\ &&\implies && [x] + n &= m. \end{align*}$

But, we defined $m=[x+n]$ . Thus, $[x+n] = [x] + n$ for any $n \in \mathbb{Z}. \qquad \blacksquare$
Proof. If $x \in \mathbb{Z}$ , then $x = n$ for some $n \in \mathbb{Z}$ . Hence, $[x]=n$ and
$-x=-n \quad \implies \quad [-x]=-n \quad \implies \quad [-x]=-[x].$

On the other hand, if $x \notin \mathbb{Z}$ , then let $[x] = n$ . This gives us,

$\begin{align*} n \leq x < n+1 &&\implies && -n-1 &< -x < n &(n \neq x \text{ since } x \notin \mathbb{Z}) \\ &&\implies && [-x] &= -n-1 = -[x]-1. \qquad \blacksquare \end{align*}$
Proof. Let $[x] = m$ and $[y] = n$ , then we have
$m \leq x < m+1 \qquad \text{and} \qquad n \leq y < n+1.$

So, adding, we obtain,

$m+n \leq x+y < m+n+2.$

Thus,

$[x+y] = m+n = [x] + [y] \qquad \text{or} \qquad [x+y] = m+n+1 = [x] + [y] + 1. \qquad \blacksquare$
Proof. By part (c) we have,
$[2x] = [x+x] = [x] + [x] \qquad \text{or} \qquad [x] + [x] + 1.$

If $[2x] = [x] + [x]$ , then let $[x] = n$ ,

$\begin{align*} && [2x] &= 2n \\ \implies && 2n &\leq 2x < 2n+1 \\ \implies && n &\leq x < n + \frac{1}{2} \\ \implies && n &\leq x + \frac{1}{2} < n+1 \\ \implies && \left[ x + \frac{1}{2} \right] &=n . \end{align*}$

Thus, $[2x] = 2n = n+n = [x] + \left[ x + \frac{1}{2} \right]$ .
On the other hand, if $[2x] = [x] + [x] + 1$ , then let $[x] = n$ , and

$\begin{align*} && [2x] &= 2n+1 \\ \implies && 2n+1 &\leq 2x < 2n+2 \\ \implies && n+\frac{1}{2} &\leq x < n+1 \\ \implies && n +1 &\leq x + \frac{1}{2} < n+2 \\ \implies && \left[ x + \frac{1}{2} \right] &= n+1. \end{align*}$

Thus, $[2x] = n+n+1 = [x] + \left[ x + \frac{1}{2} \right]. \qquad \blacksquare$
Proof. By part (c) we have
$[3x] = [x+x+x] = [x+x] + [x] \qquad \text{or} \qquad [x+x]+[x] + 1.$

And,

$[x+x] = [x] + [x] \qquad \text{or} \qquad [x]+[x]+1.$

So, putting these together we have,

$[3x] = [x]+[x]+[x], \qquad \text{or} \qquad [x]+[x]+[x]+1, \qquad \text{or} \qquad [x]+[x]+[x]+2.$

If $[3x] = [x] + [x] + [x]$ , then, let $[x] = n$ , so

$\begin{align*} && 3n &\leq 3x < 3n+1 \\ \implies && n &\leq x < n + \frac{1}{3} \\ \implies && n &\leq x + \frac{1}{3} < n+1 & \text{and}&& n &\leq x+\frac{2}{3} < n+1 \\ \implies && [x] &= \left[ x+ \frac{1}{3} \right] = \left[ x+ \frac{2}{3} \right] = n. \end{align*}$

Thus, $[3x] = 3n = [x] + \left[ x + \frac{1}{3} \right] + \left[ x + \frac{2}{3} \right]$ .
Next, if $[3x] = [x] + [x] + [x] + 1$ , then let $[x] = n$ , giving us,

$\begin{align*} && 3n+1 &\leq 3x < 3n+1 \\ \implies && n + \frac{1}{3} &\leq x < n + \frac{2}{3} \\ \implies && n + \frac{1}{3} & \leq x + \frac{1}{3} < n+1 &\implies \quad \left[ x + \frac{1}{3} \right] &=n \\ \text{and}, \ \implies && n + 1 &\leq x + \frac{2}{3} < n+2 &\implies \quad \left[ x + \frac{2}{3} \right] &= n+1. \end{align*}$

Thus,

$[3x] = 3n+1 \quad \implies \quad [3x] = [x] + \left[ x + \frac{1}{3} \right] + \left[ x + \frac{2}{3} \right].$

Finally, if $[3x] = [x] + [x] + [x] + 2$ , then let $[x] = n$ , and we have

$\begin{align*} && 3n+2 &\leq 3x < 3n+3 \\ \implies && n+\frac{2}{3} &\leq x < n+1 \\ \implies && n+1 &\leq x+\frac{1}{3} < n+2 & \text{and} && n+1 \leq x+\frac{2}{3} < n+2. \end{align*}$

So,

$[3x] = 3n+2 = [x] + \left[ x + \frac{1}{3} \right] + \left[ x + \frac{2}{3} \right]. \qquad \blacksquare$