Exercise 4.1.3 (Some equations)

Proof.

(a)

Let $$S=\{x\in ℝ\mid \lfloor x\rfloor +\lfloor x\rfloor =\lfloor 2x\rfloor \}.$$

If $x\in ℝ$, we write $x=n+\nu$, where $n\in \mathbb{Z}$ and $0\le \nu <1$. Then $n=\lfloor x\rfloor$. Moreover $\lfloor x\rfloor +\lfloor x\rfloor =2n$, and $2x=2n+2\nu$, so $\lfloor 2x\rfloor =2n+\lfloor 2\nu \rfloor$. Since $0\le 2\nu <2$,

$$\begin{array}{llll}\hfill x\in S& ⟺\lfloor 2\nu \rfloor =0& \hfill & \\ \hfill & ⟺0\le \nu <1∕2.& \hfill & \end{array}$$

In conclusion,

$$\begin{array}{lll}\hfill S=\{x\in ℝ\mid \lfloor x\rfloor +\lfloor x\rfloor =\lfloor 2x\rfloor \}& & \hfill \\ \hfill & =\underset{n\in \mathbb{Z}}{\bigcup }\left[n,n+\frac{1}{2}\right[.& \hfill & \end{array}$$

(b)

By theorem 4.1.3, for every real $x$, $\lfloor x+3\rfloor =\lfloor x\rfloor +3$. Therefore $$\{x\in ℝ\mid \lfloor x+3\rfloor =3+\lfloor x\rfloor \}=ℝ.$$

(c)

By definition of the greatest integer function, the condition $\lfloor x+3\rfloor =x+3$ is equivalent to $x+3\in \mathbb{Z}$, so is equivalent to $x\in \mathbb{Z}$. $$\{x\in ℝ\mid \lfloor x+3\rfloor =3+x\}=\mathbb{Z}.$$

(d)

Let $$T=\{x\in ℝ\mid \lfloor x+\frac{1}{2}\rfloor +\lfloor x-\frac{1}{2}\rfloor =\lfloor 2x\rfloor \}.$$

If $x\in ℝ$, we write $x=n+\nu$, where $n\in \mathbb{Z}$ and $0\le \nu <1$.

• If $0\le \nu <1∕2$, then

  $$\begin{array}{llll}\hfill & n\le x+\frac{1}{2}=n+\nu +\frac{1}{2}<n+1,& \hfill & \\ \hfill & n-1\le x-\frac{1}{2}=n+\nu -\frac{1}{2}<n,& \hfill & \\ \hfill & 2n\le 2x=2n+2\nu <2n+1.& \hfill & \end{array}$$

  Therefore $\lfloor x+\frac{1}{2}\rfloor =n,\lfloor x-\frac{1}{2}\rfloor =n-1,\lfloor 2x\rfloor =2n$, thus $\lfloor x+\frac{1}{2}\rfloor +\lfloor x-\frac{1}{2}\rfloor =2n-1\ne 2n=\lfloor 2x\rfloor$, so $x\notin T$.

• If $1∕2\le \nu <1$, then

  $$\begin{array}{llll}\hfill & n+1\le x+\frac{1}{2}=n+\nu +\frac{1}{2}<n+2,& \hfill & \\ \hfill & n\le x-\frac{1}{2}=n+\nu -\frac{1}{2}<n+1,& \hfill & \\ \hfill & 2n+1\le 2x=2n+2\nu <2n+2.& \hfill & \end{array}$$

  Therefore $\lfloor x+\frac{1}{2}\rfloor =n+1,\lfloor x-\frac{1}{2}\rfloor =n,\lfloor 2x\rfloor =2n+1$, thus $\lfloor x+\frac{1}{2}\rfloor +\lfloor x-\frac{1}{2}\rfloor =2n+1=\lfloor 2x\rfloor$, so $x\in T$.

  In conclusion,

  $$\begin{array}{llll}\hfill T& =\{x\in ℝ\mid \lfloor x+\frac{1}{2}\rfloor +\lfloor x-\frac{1}{2}\rfloor =\lfloor 2x\rfloor \}& \hfill & \\ \hfill & =\underset{n\in \mathbb{Z}}{\bigcup }\left[n+\frac{1}{2},n\right[.& \hfill & \end{array}$$

• By definition of the greatest integer function, for every $x\in ℝ$,

  $$\begin{array}{llll}\hfill \lfloor 9x\rfloor =9& ⟺9\le 9x<10& \hfill & \\ \hfill & ⟺1\le x<\frac{10}{9}.& \hfill & \end{array}$$

Therefore

$$\{x\in ℝ\mid \lfloor 9x\rfloor =9\}=\left[1,\frac{10}{9}\right[.$$