Floor && (Ceiling)
Suppose
\(
x\in R
\)
floor
\(
\exists!n\in Z
\rightarrow
\)
\(
n\leq x < n +1
\), sign as
\(
n =
\lfloor x \rfloor
\)
ceiling
\(
\exists!n\in Z
\rightarrow
\)
\(
x\leq n < x +1
\), sign as
\(
n =
\lceil x \rceil
\)
\(
-\lfloor x \rfloor
=
\lceil -x \rceil
\)
\(
-\lceil x \rceil
=
\lfloor -x \rfloor
\)
Exercise
Determine true or false.
\(
\lfloor 2x \rfloor
\)
\(
=\lfloor x \rfloor+
\lfloor x+\frac{1}{2} \rfloor
\)
solution
Suppose
\(
x=z+r, z\in Z, 0\leq r <1
\)
\(
(\rightarrow )
\)
\(
\lfloor 2x \rfloor
=\lfloor 2z+2r \rfloor
=2z+\lfloor 2r \rfloor
=
\begin{cases}
2z \text{, if } r< \frac{1}{2}\\
2z+1 \text{, if } r\geq \frac{1}{2}\\
\end{cases}
\)
\(
(\leftarrow)
\)
\(
\lfloor x \rfloor + \lfloor x+\frac{1}{2} \rfloor
=
\lfloor z+r \rfloor + \lfloor z+r+\frac{1}{2} \rfloor
=
z+z+\lfloor r+\frac{1}{2} \rfloor
=
\begin{cases}
2z \text{, if } r< \frac{1}{2}\\
2z+1 \text{, if } r\geq \frac{1}{2}\\
\end{cases}
\)
\(
\therefore
\lfloor 2x \rfloor
\)
\(
=\lfloor x \rfloor+
\lfloor x+\frac{1}{2} \rfloor
\)
Theorem
Suppose \( x,y\in R \)
\(
x-1<\lfloor x \rfloor\leq x
\)
-
proof
\(
m =\lfloor x \rfloor
\)
\(
\rightarrow
(x-1< m) \wedge (m\leq x)
\)
\(
\rightarrow
(x-1<\lfloor x \rfloor) \wedge (\lfloor x \rfloor\leq x)
\)
\(
\therefore
x-1<\lfloor x \rfloor\leq x
\)
\(
\lfloor x+n \rfloor=\lfloor x \rfloor +n
,\forall n\in Z
\)
-
proof
Suppose \( m=\lfloor x \rfloor \)
\(
m\leq x < m+1
\)
\(
\rightarrow
m+n\leq x+n < m+1+n
\)
\(
\rightarrow
\lfloor x+n \rfloor
=m+n,
\forall n\in Z(
\lfloor x+n \rfloor=
\lfloor x \rfloor+n
)
\)
\(
\therefore
\lfloor x+n \rfloor=\lfloor x \rfloor +n
,\forall n\in Z
\)
\(
\lfloor x \rfloor+\lfloor y \rfloor
\leq
\lfloor x+y \rfloor
\leq
\lfloor x \rfloor+\lfloor y \rfloor +1
\)
-
proof
Suppose \( m=\lfloor x \rfloor , n=\lfloor y \rfloor\)
\(
\rightarrow m\leq x < m+1,
n\leq y < n+1
\)
\(
\rightarrow
m+n\leq x+y < m+n+2
\)
\(
\rightarrow
m+n\leq \lfloor x+y\rfloor {\color{red}\leq} m+n+1
\)
\(
\rightarrow
\lfloor x \rfloor+\lfloor y \rfloor\leq \lfloor x+y\rfloor \leq \lfloor x \rfloor+\lfloor y \rfloor+1
\)
\(
\lfloor x \rfloor + \lfloor -x \rfloor
=
\begin{cases}
0\text{ ,if } x\in Z\\
-1\text{ , otherwise}
\end{cases}
\)
-
proof
hint:
-
\(
\lfloor -r \rfloor
=
\lfloor -1+r \rfloor
, r\in R^+ \text{ and } 0< r<1
\)
-
\(
\lfloor -r \rfloor =
\lfloor r \rfloor,
r=0
\)
-
\(
\lfloor r \rfloor
=
\lfloor 1-r \rfloor
=0
, r\in R^+ \text{ and } 0< r<1
\)
Suppose
\(
x = z+r , z\in Z, 0\leq r < 1
\)
\(
\rightarrow
\lfloor x \rfloor+\lfloor -x \rfloor
=
z+\lfloor -z-r \rfloor
\)
\(
=
z+\lfloor -z{\color{blue}-1+1}-r \rfloor
\)
\(
=
z-z-1+\lfloor 1-r \rfloor
\)
\(
=
-1+\lfloor 1-r \rfloor
\)
\(
\Rightarrow
\begin{cases}
0 \;\;,x\in Z \rightarrow r=0\rightarrow \lfloor 1-r \rfloor=1\\
-1,x\not\in Z \rightarrow \lfloor 1-r \rfloor r =0\\
\end{cases}
\)
Exercise
Suppose \( x\in R, x\geq 0 \),
Show that
\(
\lfloor \sqrt{\lfloor x \rfloor}\rfloor
=
\lfloor \sqrt{x} \rfloor
\)
solution
Suppose
\( m=\lfloor \sqrt{\lfloor x \rfloor} \rfloor \)
\(
\rightarrow
m\leq \sqrt{\lfloor x \rfloor} < m+1
\)
\(
\rightarrow
m^2\leq \lfloor x \rfloor < (m+1)^2
\)
\(
\rightarrow
m^2\leq x < (m+1)^2
\)
\(
\rightarrow
m\leq \sqrt{x} < m+1
\)
\(
\therefore m=\lfloor \sqrt{x} \rfloor
=
\lfloor \sqrt{\lfloor x \rfloor}\rfloor
\)