$\lfloor x \rceil = m$ if and only if $m \in {\mathbb Z}$ and $m- \frac{1}{2} \leq x \lt m+\frac{1}{2}$.
by definition: | $\bigl\lfloor \sqrt{n} \bigr\rceil = q$ | $\Leftrightarrow$ | $q - \frac{1}{2} \leq \sqrt{n} \lt q + \frac{1}{2}$ |
by squaring positive numbers |
$\Leftrightarrow$ | $q^2 - q + \frac{1}{4} \leq n \lt q^2 + q + \frac{1}{4}$ | |
since $n, q \in {\mathbb Z}_+$ |
$\Leftrightarrow$ | $q^2 - q \lt n \leq q^2 + q$ | |
adding $q$ |
$\Leftrightarrow$ | $q^2 \lt n + q \leq q^2 + 2q$ | |
since $n, q \in {\mathbb Z}_+$ |
$\Leftrightarrow$ | $q^2 \lt n + q \lt q^2 + 2q + 1$ | |
complete square |
$\Leftrightarrow$ | $q^2 \lt n + q \lt (q + 1)^2$ |
$q$ | $S_q$ | $T_q$ | $f(n)$ |
---|---|---|---|
$1$ | $\{1,2\}$ | $\{2,3\}$ | $n+1$ |
$2$ | $\{3,4,5,6\}$ | $\{5,6,7,8\}$ | $n+2$ |
$3$ | $\{7,8,9,10,11,12\}$ | $\{10,11,12,13,14,15\}$ | $n+3$ |
$4$ | $\{13,14,\ldots,19,20\}$ | $\{17,18,\ldots,23,24\}$ | $n+4$ |
$5$ | $\{21,22,\ldots,29,30\}$ | $\{26,27,\ldots,34,35\}$ | $n+5$ |
$\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ |
$q$ | $\{q^2-q+1,\ldots,q^2+q\}$ | $\{q^2+1,\ldots,q^2+2q\}$ | $n+q$ |
$\bigl\lfloor \sqrt{2n} + \frac{1}{2} \bigr\rfloor = m \ $?
$\bigl\lfloor \sqrt{2n} + \frac{1}{2} \bigr\rfloor = m \ $ if and only if $ \ m \leq \sqrt{2n} + \frac{1}{2} \lt m+1$.
$m \leq \sqrt{2n} + \frac{1}{2} \lt m+1 \ $ if and only if $ \ m - \frac{1}{2} \leq \sqrt{2n} \lt m + \frac{1}{2}$
$m - \frac{1}{2} \leq \sqrt{2n} \lt m + \frac{1}{2} \ $ if and only if $ \ m^2 - m + \frac{1}{4} \leq 2n \lt m^2 + m + \frac{1}{4}$
$m^2 - m + \frac{1}{4} \leq 2n \lt m^2 + m + \frac{1}{4} \ $ if and only if $ \ m^2 - m \lt 2n \leq m^2 + m$
$m^2 - m \lt 2n \leq m^2 + m \ $ if and only if $ \ \dfrac{(m-1)m}{2} \lt n \leq \dfrac{m(m + 1)}{2}$
$\bigl\lfloor \sqrt{2n} + \frac{1}{2} \bigr\rfloor = m \ $ if and only if $ \ \dfrac{m(m-1)}{2} \lt n \leq \dfrac{m(m+1)}{2}$.
$f_1$ | $f_2$ | $f_3$ | $f_4$ | $f_5$ | $f_6$ | $f_7$ | $f_8$ | $f_9$ | |
---|---|---|---|---|---|---|---|---|---|
$\bigcirc$ | $1$ | $1$ | $1$ | $2$ | $2$ | $2$ | $3$ | $3$ | $3$ |
$\square$ | $1$ | $2$ | $3$ | $1$ | $2$ | $3$ | $1$ | $2$ | $3$ |
$f_1$ | $f_2$ | $f_3$ | $f_4$ | $f_5$ | $f_6$ | $f_7$ | $f_8$ | $f_9$ | $f_{10}$ | $f_{11}$ | $f_{12}$ | $f_{13}$ | $f_{14}$ | $f_{15}$ | $f_{16}$ | |
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
$\blacktriangle$ | $0$ | $0$ | $0$ | $0$ | $0$ | $0$ | $0$ | $0$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ |
$\diamond$ | $0$ | $0$ | $0$ | $0$ | $1$ | $1$ | $1$ | $1$ | $0$ | $0$ | $0$ | $0$ | $1$ | $1$ | $1$ | $1$ |
$\bullet$ | $0$ | $0$ | $1$ | $1$ | $0$ | $0$ | $1$ | $1$ | $0$ | $0$ | $1$ | $1$ | $0$ | $0$ | $1$ | $1$ |
$\star$ | $0$ | $1$ | $0$ | $1$ | $0$ | $1$ | $0$ | $1$ | $0$ | $1$ | $0$ | $1$ | $0$ | $1$ | $0$ | $1$ |
s u b s e t |
$\emptyset$ | $\{\star\}$ | $\{\bullet\}$ | $\left\{\begin{array}{c}\!\!\bullet\!\!\\\!\!\star\!\! \end{array}\right\}$ | $\{\diamond\}$ | $\left\{\begin{array}{c}\!\!\diamond\!\!\\\!\!\star\!\!\end{array}\right\}$ | $\left\{\begin{array}{c}\!\!\diamond\!\!\\\!\!\bullet\!\!\end{array}\right\}$ | $\left\{\begin{array}{c}\!\!\diamond\!\!\\\!\!\bullet\!\!\\\!\!\star\!\! \end{array}\right\}$ | $\{\blacktriangle\}$ | $\left\{\begin{array}{c}\!\!\blacktriangle\!\!\\\!\!\star\!\!\end{array}\right\}$ | $\left\{\begin{array}{c}\!\!\blacktriangle\!\!\\\!\!\bullet\!\! \end{array}\right\}$ | $\left\{\begin{array}{c}\!\!\blacktriangle\!\!\\\!\!\bullet\!\!\\\!\!\star\!\! \end{array}\right\}$ | $\left\{\begin{array}{c}\!\!\blacktriangle\!\!\\\!\!\diamond\!\! \end{array}\right\}$ | $\left\{\begin{array}{c}\!\!\blacktriangle\!\!\\\!\!\diamond\!\!\\\!\!\star\!\! \end{array}\right\}$ | $\left\{\begin{array}{c}\!\!\blacktriangle\!\!\\\!\!\diamond\!\!\\\!\!\bullet\!\! \end{array}\right\}$ | $\left\{\begin{array}{c}\!\!\blacktriangle\!\!\\\!\!\diamond\!\!\\\!\!\bullet\!\!\\\!\!\star\!\! \end{array}\right\}$ |
Proposition. For an arbitrary nonempty set $A$ there exist a bijection between the power set of $A$ and the set $\{0,1\}^A$.
Proof. First we define a function $\Phi$ from ${\mathcal P}(A)$ to $\{0,1\}^A$. Let $X$ be an arbitrary subset of $A$. We need to define $\Phi(X)$ as a function from $A$ to $\{0,1\}$. For an arbitrary $a \in A$ we define \[ \bigl(\Phi(X)\bigr)(a) = \begin{cases} 0 & \ \text{if} \quad a \not\in X, \\[7pt] 1 & \ \text{if} \quad a \in X. \end{cases} \] Clearly, $\Phi(X) \in \{0,1\}^A$. (For example, in the above table, if $X = \{\blacktriangle,\bullet,\star\}$, then $\Phi(X) = f_{12}$.)
Next we prove that $\Phi$ is a surjection. Let $f \in \{0,1\}^A$ be arbitrary. We need to define $X \subseteq A$ such that $\Phi(X) = f$. We define $X = \{x \in A \ | \ f(a) = 1 \}$. With this $X$ we need to prove that $\Phi(X) = f$, that is $\bigl(\Phi(X)\bigr)(a) = f(a)$ for all $a \in A$. We distinguish two cases: Case 1: $f(a) = 1$ and Case 2: $f(a) = 0$. (These two cases cover all possibilities since $f(a) \in \{0,1\}$.) Case 1. Assume $f(a) = 1$. Then $a \in X$, and by the definition of $\Phi(X)$ we have $\bigl(\Phi(X)\bigr)(a) = 1$. Thus, in this case $\bigl(\Phi(X)\bigr)(a) = f(a)$. Case 2. Assume $f(a) = 0$. Then $a \not\in X$, and by the definition of $\Phi(X)$ we have $\bigl(\Phi(X)\bigr)(a) = 0$. Thus, in this case $\bigl(\Phi(X)\bigr)(a) = f(a)$. This proves that $\Phi$ is a surjection.
It remains to prove that $\Phi$ is an injection. Let $X, Y$ be arbitrary subsets of $A$ such that $X \neq Y$. We need to prove $\Phi(X) \neq \Phi(Y)$. Since $X \neq Y$ there exists $a \in A$ such that $a \in X$ and $a \not\in Y$ or there exists $b \in A$ such that $b \not\in X$ and $b \in X$. If there exists $a \in A$ such that $a \in X$ and $a \not\in Y$, then $\bigl(\Phi(X)\bigr)(a) = 1$ and $\bigl(\Phi(Y)\bigr)(a) = 0$, hence $\Phi(X) \neq \Phi(Y)$ in this case. If there exists $b \in A$ such that $b \not\in X$ and $b \in Y$, then $\bigl(\Phi(X)\bigr)(b) = 0$ and $\bigl(\Phi(Y)\bigr)(b) = 1$, hence $\Phi(X) \neq \Phi(Y)$ in this case as well. This completes the proof.
A different way to prove this proposition would be to define the function $\Psi: \{0,1\}^A \to {\mathcal P}(A)$ by \[ \Psi(f) = \bigl\{a \in A \,|\, f(a) = 1 \bigr\}, \quad f \in \{0,1\}^A, \] and prove that $\Psi$ is the inverse of $\Phi$. That is, prove that
$\forall \, X \in {\mathcal P}(A) \quad \Psi\bigl(\Phi(X)\bigr) = X$ and $\forall \, f \in \{0,1\}^A \quad \Phi\bigl(\Psi(f)\bigr) = f$.
One could make a claim that the boxed propositions above are obvious. However, it would take some writing to prove them formally.$\forall x\in {\mathbb R} \ \ \forall k \in {\mathbb Z} \quad x \geq k \ \Leftrightarrow \ \lfloor x \rfloor \geq k$
$\forall x\in {\mathbb R} \ \ \forall k \in {\mathbb Z} \quad x \lt k \ \Leftrightarrow \ \lfloor x \rfloor \lt k$
Proof. Let $x \in {\mathbb R}$ and $k \in {\mathbb Z}$ be arbitrary. Assume $x \lt k$. Recall that by the definition of floor we have $\lfloor x \rfloor \leq x$. The last two green boxes and the transitivity property of inequality imply $\lfloor x \rfloor \lt k$. This completes the proof.
Proof. Let $x \in {\mathbb R}$ and $k \in {\mathbb Z}$ be arbitrary. Assume $\lfloor x \rfloor \lt k$. Then $k - \lfloor x \rfloor \gt 0$; that is, $k - \lfloor x \rfloor$ is a positive integer. Since $1$ is the smallest positive integer, we have $k - \lfloor x \rfloor \geq 1$. Therefore $\lfloor x \rfloor + 1 \leq k$. By the definition of floor we have $x \lt \lfloor x \rfloor + 1$. The last two green boxes and the transitivity property of inequality imply that $x \lt k$. This completes the proof.
$\forall x\in {\mathbb R} \ \ \forall k \in {\mathbb Z} \quad x \leq k \ \Leftrightarrow \ \lceil x \rceil \leq k$
$\forall x\in {\mathbb R} \ \ \forall k \in {\mathbb Z} \quad x \gt k \ \Leftrightarrow \ \lceil x \rceil \gt k$
Grid 1 | Grid 2 | Grid 3 | Grid 4 |
---|---|---|---|
In the Gregorian calendar, the current standard calendar in most of the world, most years that are evenly divisible by 4 are leap years. In each leap year, the month of February has 29 days instead of 28. Adding an extra day to the calendar every four years compensates for the fact that a period of 365 days is shorter than a solar year by almost 6 hours. Some exceptions to this rule are required since the duration of a solar year is slightly less than 365.25 days. Years that are evenly divisible by 100 are not leap years, unless they are also evenly divisible by 400, in which case they are leap years.
$P(Y)\ $ stands for: | $Q(Y)\ $ stands for: | $R(Y)\ $ stands for: |
$\dfrac{Y}{4}\ $ is an integer. | $\dfrac{Y}{100}\ $ is an integer. | $\dfrac{Y}{400}\ $ is an integer. |