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@ -5661,3 +5661,937 @@ Omitted
$(m + dk) \mod d = m \mod d$.
Omitted
---
**Exercise Set 4.6**
Page 240
Compute $\lfloor x \rfloor$ and $\lceil x \rceil$ for each of the values of $x$
in 1-4.
1. $37.999$
$\lfloor 37.999 \rfloor = 37$
$\lceil 37.999 \rceil = 38$
2. $\dfrac{17}{4}$
$\dfrac{17}{4} = 4 + \dfrac{1}{4} = 4.25$
$\lfloor \dfrac{17}{4} \rfloor = 4$
$\lceil \dfrac{17}{4} \rceil = 5$
3. $-14.00001$
$\lfloor -14.00001 \rfloor = -15$
$\lceil -14.00001 \rceil = -14$
4. $-\dfrac{32}{5}$
$-\dfrac{32}{5} = -6.4$
$\lfloor -\dfrac{32}{5} \rfloor = -7$
$\lceil -\dfrac{32}{5} \rceil = -6$
5. Use the floor notation to express $259\ div\ 11$ and $259 \mod 11$.
$$ 259\ div\ 11 = \lfloor \frac{259}{11} \rfloor = \lfloor 23.54545454\dots \rfloor = 23 $$
$$ 259 \mod 11 = 259 - 11 \cdot \lfloor \frac{259}{11} \rfloor $$
$$ = 259 - 11 \cdot 23 $$
$$ = 6 $$
6. If $k$ is an integer, what is $\lceil k \rceil$? Why?
$$ \lceil k \rceil = k $$
The ceiling of $k$ is $k$. Since $k$ is already an integer, then given the
definition for ceiling:
$$ \lceil k \rceil = n \Leftrightarrow n - 1 < k \leq n $$
We can substitute in $n = k$:
$$ k - 1 < k \leq k $$
And both parts are true, so:
$$ \lceil k \rceil = k $$
7. If $k$ is an integer, what is $\left\lceil k + \dfrac{1}{2} \right\rceil$?
Why?
$$ \left\lceil k + \frac{1}{2} \right\rceil = k + 1 $$
By the definition of ceiling:
$$ \ceil k + \dfrac{1}{2} \rceil = n \Leftrightarrow n - 1 < k + \frac{1}{2} \leq n $$
Now substitute in $n = k + 1$:
$$ (k + 1) - 1 < k + \frac{1}{2} \leq k + 1 $$
$$ k < k + \frac{1}{2} \leq k + 1 $$
And both parts are true, so:
$$ \lceil k + \frac{1}{2} \rceil = k + 1 $$
8. Seven pounds of raw material are needed to manufacture each unit of a certain
product. Express the number of units that can be produced from $n$ pounds of
raw material using either the floor or the ceiling notation. Which notation
is more appropriate?
$7$ pounds of raw material per product is main ratio.
This means that the number of units $q$ from $n$ pounds can be expressed using
the quotient-remainder theorem as:
$$ n = 7q $$
Where the remainder is removed as no more units can be created from leftover raw
material.
And when converted to floor notation, this is:
$$ q = \left\lfloor \frac{n}{7} \right\rfloor $$
The reason floor is more appropriate is that leftover material would not
realistically be able to be utilized to make another unit of said product.
9. Boxes, each capable of holding 36 units, are used to ship a product from the
manufacturer to a wholesaler. Express the number of boxes that would be
required to ship $n$ units of the product using either the floor or the
ceiling notation. Which notation is more appropriate?
Each box can hold up to $36$ units of product. Shipping $n$ units of product can
be expressed using the quotient remainder theorem as:
$$ n = 36q + r $$
Where $r$ is the remaining products that did not fill another box.
Then we convert to ceiling notation:
$$ q = \left\lceil \frac{n}{36} \right\rceil $$
The ceiling notation is more appropriate as even after packing all boxes full,
you cannot simply not ship the remaining product, so another box filled with the
remaining products is added to the shipment.
10. If 0 = Sunday, 1 = Monday, 2 = Tuesday, ..., 6 = Saturday, then January 1 of
year $n$ occurs on the day of the week given by the following formula:
$$ \left(n + \left\lfloor \frac{n - 1}{4} \right\rfloor - \left\lfloor \frac{n - 1}{100} \right\rfloor + \left\lfloor \frac{n - 1}{400} \right\rfloor\right) \mod 7 $$
a. Use this formula to find January 1 of
i. 2050
6 = Saturday
ii. 2100
5 = Friday
iii. the year of your birth.
Omitted
b. Interpret the different components of this formula.
11. State a necessary and sufficient condition for the floor of a real number to
equal that number.
It is a necessary and sufficient condition for any real number, $x$ to to
satisfy $\lfloor x \rfloor = x$ that $x$ be an integer.
12. Let $S$ be the statement: For any odd integer $n$,
$\left\lfloor \dfrac{n}{2} \right\rfloor = \dfrac{(n - 1)}{2}$. Then $S$ is
true, but the following "proof" is incorrect. Find the mistake.
**"Proof:**
Suppose $n$ is any odd integer. Then $n = 2k + 1$ for some integer $k$.
Consequently,
$$ \left\lfloor \frac{2k + 1}{2} \right\rfloor = \frac{(2k + 1) - 1}{2} = \frac{2k}{2} = k $$
But $n = 2k + 1$. Solving for $k$ gives $k = \dfrac{(n - 1)}{2}$. Hence, by
substitution, $\left\lfloor \dfrac{n}{2} \right\rfloor = \dfrac{(n - 1)}{2}$."
The mistake is in the initial substitution, by setting:
$$ \left\lfloor \frac{2k + 1}{2} \right\rfloor = \frac{(2k + 1) - 1}{2} $$
The author of this proof assumes the what is to be proved.
13. Prove that if $n$ is any even integer, then
$\left\lfloor \dfrac{n}{2} \right\rfloor = \dfrac{n}{2}$.
**Proof:** Suppose $n$ is any even integer.
Since $n$ is an even integer, $n = 2k$ for some integer $k$.
Then, by substitution:
$$ \left\lfloor \frac{n}{2} \right\rfloor = \left\lfloor \frac{2k}{2} \right\rfloor $$
$$ = \left\lfloor k \right\rfloor $$
Because $k$ is an integer, and by the definition of floor, $k \leq k < k + 1$.
We then know that:
$$ \lfloor k \rfloor = k $$
It follows then that:
$$ \left \lfloor \frac{2k}{2} \right\rfloor = \frac{2k}{2} $$
$$ \left \lfloor \frac{n}{2} \right\rfloor = \frac{n}{2} $$
Q.E.D.
14. Show that the following statement is false.
For all real numbers $x$ and $y$,
$\lfloor x - y \rfloor = \lfloor x \rfloor - \lfloor y \rfloor$.
**Proof by Counterexample:**
Let $x = \dfrac{1}{2}$ and $y = \dfrac{3}{4}$
Then:
$$ \lfloor x - y \rfloor = \left\lfloor \frac{1}{2} - \frac{3}{4} \right\rfloor $$
$$ = -1 $$
Then consider:
$$ \lfloor x \rfloor - \lfloor y - \rfloor = \left\lfloor \frac{1}{2} \right\rfloor - \left\lfloor \frac{3}{4} \right\rfloor $$
$$ = 0 - 0 $$
$$ = 0 $$
Thus:
$$ -1 \neq 0 $$
$$ \lfloor x - y \rfloor \neq \lfloor x \rfloor - \lfloor y \rfloor$ $.
Therefore for the given $x$ and $y$, this statement is false.
Q.E.D.
Some of the statements in 15-22 are true and some are false. Prove each true
statement and find a counterexample for each false statement, but do not use
Theorem 4.6.1 in your proofs.
15. For every real number $x$, $\lfloor x - 1 \rfloor = \lfloor x \rfloor - 1$.
**Proof:**
Suppose $x$ is any real number.
By the definition of floor, $x$ can then be expressed as:
$$ \lfloor x \rfloor \Leftrightarrow n \leq x < n + 1 $$
for some integer $n$.
Subtracting $1$ from all sides of the inequality is:
$$ n - 1 \leq x - 1 < n $$
Since $n - 1$ is an integer by the difference of integers, by the definition of
floor:
$$ \lfloor x - 1 \rfloor = n - 1 $$
It follows by substitution then that:
$$ \lfloor x - 1 \rfloor = \lfloor x \rfloor - 1 $$
Q.E.D.
16. For every real number $x$,
$\left\lfloor x^2 \right\rfloor = \lfloor x \rfloor^2$
**Proof by Counterexample:**
Let $x = -\dfrac{1}{2}$
$$ x^2 = \frac{1}{4} $$
$$ \lfloor x^2 \rfloor = \lfloor \frac{1}{4} \rfloor = 0 $$
Now consider:
$$ \lfloor x \rfloor^2 = \lfloor -\frac{1}{2} \rfloor^2 = (-1)^2 = 1 $$
And then:
$$ 0 \neq 1 $$
$$ \lfloor x^2 \rfloor \neq \lfloor x \rfloor^2 $$
Then for the given $x$, $\lfloor x^2 \rfloor \neq \lfloor x \rfloor^2$.
Therefore the statement is false.
Q.E.D.
17. For every integer $n$,
$$
\left\lfloor \dfrac{n}{3} \right\rfloor =
\begin{cases}
\dfrac{n}{3} & \text{if } n \mod 3 = 0 \\
\dfrac{(n - 1)}{3} & \text{if } n \mod 3 = 1 \\
\dfrac{(n - 2)}{3} & \text{if } n \mod 3 = 2 \\
\end{cases}
$$
**Proof:**
Suppose $n$ is any integer.
_Case where $n \mod 3 = 0$:_
Since $n \mod 3 = 0$, then by the definition of mod:
Since $n \mod 3 = 0$, $n$ can be written as:
$$ n = 3q + 0 $$
for some integer $q$ by the definition of mod.
By substitution:
$$ \left\lfloor \frac{n}{3} \right\rfloor = \left\lfloor \frac{3q}{3} \right\rfloor $$
$$ = \lfloor q \rfloor $$
$$ = q \quad \text{ by the definition of floor} $$
Since $q = \dfrac{n}{3}$:
$$ \left\lfloor \frac{n}{3} \right\rfloor = \frac{n}{3} $$
_Case where $n \mod 3 = 1$:_
Since $n \mod 3 = 1$, then by the definition of mod:
Since $n \mod 3 = 1$, $n$ can be written as:
$$ n = 3q + 1 $$
for some integer $q$ by the definition of mod.
By substitution:
$$ \left\lfloor \frac{n}{3} \right\rfloor = \left\lfloor \frac{3q + 1}{3} \right\rfloor $$
$$ = \left\lfloor \frac{3q}{3} + \frac{1}{3} \right\rfloor $$
$$ = \left\lfloor q + \frac{1}{3} \right\rfloor $$
$$ = q \quad \text{ by the definition of floor} $$
Since $q = \dfrac{n - 1}{3}$:
$$ \left\lfloor \frac{n}{3} \right\rfloor = \frac{n - 1}{3} $$
_Case where $n \mod 3 = 2$:_
Since $n \mod 3 = 2$, $n$ can be written as:
$$ n = 3q + 2 $$
for some integer $q$ by the definition of mod.
By substitution:
$$ \left\lfloor \frac{n}{3} \right\rfloor = \left\lfloor \frac{3q + 2}{3} \right\rfloor $$
$$ = \left\lfloor \frac{3q}{3} + \frac{2}{3} \right\rfloor $$
$$ = \left\lfloor q + \frac{2}{3} \right\rfloor $$
$$ = q \quad \text{ by the definition of floor} $$
Since $q = \dfrac{n - 2}{3}$:
$$ \left\lfloor \frac{n}{3} \right\rfloor = \frac{n - 2}{3} $$
Q.E.D.
18. For all real numbers $x$ and $y$,
$\lceil x + y \rceil = \lceil x \rceil + \lceil y \rceil$.
**Proof by Counterexample:**
Let $x = -\dfrac{1}{2}$ and $y = -\dfrac{3}{4}$.
Consider:
$$ \lceil x + y \rceil = \left\lceil -\frac{1}{2} + \left(-\frac{3}{4}\right) \right\rceil $$
$$ = -1 $$
Then:
$$ \lceil x \rceil + \lceil y \rceil = \left\lceil -\frac{1}{2} \right\rceil + \left\lceil -\frac{3}{4} \right\rceil $$
$$ = 0 + 0 $$
$$ = 0 $$
Then:
$$ -1 \neq 0 $$
$$ \lceil x + y \rceil \neq \lceil x \rceil + \lceil y \rceil $$
Therefore, for the given $x$ and $y$, the statement is false.
Q.E.D.
19. For every real number $x$, $\lceil x - 1 \rceil = \lceil x \rceil - 1$.
**Proof:**
Suppose $x$ is any real number.
By the definition of ceiling, $x$ can be expressed as:
$$ \lceil x \rceil = n \Leftrightarrow n - 1 < x \leq n $$
for some integer $n$.
If we then subtract $1$ from the inequality, we get:
$$ n - 2 < x - 1 \leq n - 1 $$
Since $n - 1$ is an integer by the difference of integers, by the definition of
ceiling:
$$ \lceil x - 1 \rceil = n - 1 $$
It follows by substitution then that:
$$ \lceil x - 1 \rceil = \lceil x \rceil - 1 $$
Q.E.D.
20. For all real numbers $x$ and $y$,
$\lceil xy \rceil = \lceil x \rceil \cdot \lceil y \rceil$.
**Proof by Counterexample:**
Let $x = 2$ and $y = \dfrac{1}{2}$.
Then:
$$ \lceil xy \rceil = \left\lceil 2\left(\frac{1}{2}\right) \right\rceil = \lceil 1 \rceil = 1 $$
Then consider:
$$ \lceil x \rceil \cdot \lceil y \rceil = \lceil 2 \rceil \cdot \left\lceil \frac{1}{2} \right\rceil $$
$$ = 2 \cdot 1 $$
$$ = 2 $$
Since:
$$ 1 \neq 2 $$
Then:
$$ \lceil xy \rceil \neq \lceil x \rceil \cdot \lceil y \rceil $$
Thus it has been shown that for at least one given $x$ and one given $y$,
$$ \lceil xy \rceil \neq \lceil x \rceil \cdot \lceil y \rceil $$
Therefore the statement is false.
Q.E.D.
21. For every odd integer $n$,
$\lceil \dfrac{n}{2} \rceil = \dfrac{(n - 1)}{2}$.
**Proof by Counterexample:**
Let $n = 1$.
Consider:
$$ \left\lceil \dfrac{n}{2} \right\rceil = \left\lceil \dfrac{1}{2} \right\rceil $$
$$ = 1 $$
Then:
$$ \frac{n - 1}{2} = \frac{1 - 1}{2} = \frac{0}{2} = 0 $$
Since $1 \neq 0$:
$$ \left\lceil \dfrac{n}{2} \right\rceil \neq \frac{n - 1}{2} $$
Thus it has been shown that there exists some value for $n$ such that:
$$ \left\lceil \dfrac{n}{2} \right\rceil \neq \frac{n - 1}{2} $$
Therefore the statement is false.
Q.E.D.
22. For all real numbers $x$ and $y$,
$\lceil xy \rceil = \lceil x \rceil \cdot \lfloor y \rfloor$.
**Proof by Counterexample:**
Let $x = \dfrac{5}{4}$ and $y = \dfrac{1}{2}$.
Then:
$$ \lceil xy \rceil = \left\lceil \left(\frac{5}{4}\right)\left(\frac{1}{2}\right) \right\rceil $$
$$ = 1 $$
Then:
$$ \lceil x \rceil \cdot \lfloor y \rfloor = \left\lceil \frac{5}{4} \right\rceil \cdot \left\lfloor \frac{1}{2} \right\rfloor $$
$$ = 2 \cdot 0 = 0 $$
So:
$$ 1 \neq 0 $$
$$ \lceil xy \rceil \neq \lceil x \rceil \cdot \lfloor y \rfloor $$
So, it has been shown that there exists a value for $x$ and a value for $y$ such
that:
$$ \lceil xy \rceil \neq \lceil x \rceil \cdot \lfloor y \rfloor $$
Therefore the statement is false.
Q.E.D.
Prove each of the following statements in 23-33.
23. For any real number $x$, if $x$ is not an integer, then
$\lfloor x \rfloor + \lfloor -x \rfloor = -1$.
Suppose $x$ is any real number where $x$ is not an integer.
By the definition of floor, since $x$ is not an integer, $x$ can be expressed
as:
$$ \lfloor x \rfloor = n \Leftrightarrow n < x < n + 1 $$
for $n$ is some integer.
Note that since $x$ is not an integer there is no "or equal to" here.
We can then multiply the inequality by $-1$:
$$ -n > -x > -n - 1 $$
Where $-n - 1$ is an integer by the product and difference of integers. Since
$-n - 1$ is an integer, it then follows:
$$ \lfloor -x \rfloor = -n - 1 $$
$$ \lfloor x \rfloor + \lfloor -x \rfloor = n + (-n - 1) = -1 $$
Therefore:
$$ \lfloor x \rfloor + \lfloor -x \rfloor = -1 $$
Q.E.D.
24. For any integer $m$ and any real number $x$, if $x$ is not an integer, then
$\lfloor x \rfloor + \lfloor m - x \rfloor = m - 1$.
**Proof:**
Suppose $x$ is any real number where $x$ is not an integer, and suppose $m$ is
any an integer.
Since $m$ is an integer and $x$ is not an integer, then $m - x$ is not an
integer by the difference of integers.
It follows then that:
$$ \lfloor m - x \rfloor = n \Leftrightarrow n < m - x < n + 1 $$
for some integer $n$.
Let's then subtract $m$ from the inequality:
$$ n - m < -x < n + 1 - m $$
And multiply the inequality by $-1$:
$$ m - n > x > m - n - 1 $$
Rewritten:
$$ m - n - 1 < x < m - n $$
Since $m - n - 1$ is an integer, this means that:
$$ \lfloor x \rfloor = m - n - 1 $$
By substitution then:
$$ \lfloor x \rfloor + \lfloor m - x \rfloor = (m - n - 1) + (n) $$
$$ \lfloor x \rfloor + \lfloor m - x \rfloor = m - 1 $$
Q.E.D.
25. For every real number $x$,
$\left\lfloor \dfrac{\left\lfloor \dfrac{x}{2}\right\rfloor}{2} \right\rfloor = \left\lfloor \dfrac{x}{4} \right\rfloor$.
**Proof:**
Suppose $x$ is any real number.
Let $n = \left\lfloor \dfrac{x}{2} \right\rfloor$.
Note that $n$ is automatically an integer due to floor always outputting an
integer.
By the definition of floor, since $n$ is an integer:
$$ \left\lfloor \frac{x}{2} \right\rfloor = n \Leftrightarrow n \leq \frac{x}{2} < n + 1 $$
__Case where $n$ is even:_
Since $n$ is even, $n = 2k$ for some integer $k$.
By substitution:
$$ \left\lfloor \frac{x}{2} \right\rfloor = 2k \Leftrightarrow 2k \leq \frac{x}{2} < 2k + 1 $$
We can then multiply out our inequality:
$$ 2(2k) \leq x < 2(2k) + 2 $$
$$ 4k \leq x < 4k + 2 $$
We then divide by $4$:
$$ k \leq \frac{x}{4} < k + \frac{1}{2} $$
By substitution then:
$$ \left\lfloor \frac{x}{4} \right\rfloor = k $$
__Case where $n$ is odd:_
Since $n$ is odd, $n = 2k + 1$ for some integer $k$.
By substitution:
$$ \left\lfloor \frac{x}{2} \right\rfloor = 2k + 1 \Leftrightarrow 2k + 1 \leq \frac{x}{2} < (2k + 1) + 1 $$
We can then multiply out our inequality:
$$ 2(2k + 1) \leq x < 2(2k + 1) + 2 $$
$$ 4k + 2 \leq x < 4k + 4 $$
We then divide by $4$:
$$ k + \frac{1}{2} \leq \frac{x}{4} < k + 1 $$
By substitution then:
$$ \left\lfloor \frac{x}{4} \right\rfloor = k $$
Thus in both cases we have shown the given statement to be true for all real
numbers $x$.
Q.E.D.
26. For every real number $x$, if $x - \lfloor x \rfloor < \dfrac{1}{2}$ then
$\lfloor 2x \rfloor = 2\lfloor x \rfloor$.
Suppose $x$ is any real number where $x - \lfloor x \rfloor < \dfrac{1}{2}$.
$$ x - \lfloor x \rfloor < \frac{1}{2} $$
Multiplying by $2$ gets us:
$$ 2x - 2\lfloor x \rfloor < 1 $$
Now add $2\lfoor x \rfloor$ to both sides:
$$ 2x < 2\lfloor x \rfloor + 1 $$
By definition of floor $\lfoor x \rfloor \leq x$. Thus:
$$ 2\lfloor x \rfloor \leq 2x $$
Putting these two inequalities together shows:
$$ 2\lfloor x \rfloor \leq 2x < 2\lfloor x \rfloor + 1 $$
By definition of floor, this means then that:
$$ \lfloor 2x \rfloor = 2\lfloor x \rfloor $$
Which is what was to be shown.
Q.E.D.
27. For every real number $x$, if $x - \lfloor x \rfloor \geq \dfrac{1}{2}$ then
$\lfloor 2x \rfloor = 2\lfloor x \rfloor + 1$.
**Proof:**
Suppose $x$ is any real number where $x - \lfloor x \rfloor \geq \dfrac{1}{2}$.
By the definition of floor, one can express some integer $n$ such that:
$$ \lfloor x \rfloor = n \Leftrightarrow n \leq x < n + 1 $$
Then subtract n from the inequality:
$$ 0 \leq x - n < 1 $$
We know that $x - \lfloor x \rfloor \geq \dfrac{1}{2}$.
$$ \frac{1}{2} \leq x - n < 1 $$
Multiply all sides by $2$:
$$ 1 \leq 2(x - n) < 2 $$
By substitution:
$$ \lfloor 2(x - n) \rfloor = 1 $$
_[To get to the form we want, we have to express $x$ as a further expression of
$n$]_
We can rewrite $x$ as $x = x + n - n$:
Since $n = \lfloor x \rfloor$:
$$ 2x = 2(n + (x - n)) $$
$$ 2x = 2n + 2(x - n) $$
Then take the floor:
$$ \lfloor 2x \rfloor = \lfloor 2n + 2(x - n) \rfloor $$
Since $2n$ is an integer:
$$ = 2n + \lfloor 2(x - n) \rfloor $$
And we know that $\lfloor 2(x - n) \rfloor = 1$, so substitute:
$$ = 2n + 1 $$
And we know $n = \lfoor x \rfloor$, so substitute again:
$$ = 2\lfoor x \rfloor + 1 $$
Therefore $\lfloor 2x \rfloor = 2\lfloor x \rfloor + 1$.
Q.E.D.
28. For any odd integer $n$,
$$ \left\lfloor \frac{n^2}{4} \right\rfloor = \left(\frac{n - 1}{2}\right)\left(\frac{n + 1}{2}\right) $$
**Proof:**
Suppose $n$ is any odd integer.
Since $n$ is odd, $n = 2k + 1$ for some integer $k$.
Then, by substitution:
$$ \left\lfloor \frac{n^2}{4} \right\rfloor = \left\lfloor \frac{(2k + 1)^2}{4} \right\rfloor $$
$$ = \left\lfloor \frac{(2k + 1)(2k + 1)}{4} \right\rfloor $$
$$ = \left\lfloor \frac{4k^2 + 4k + 1}{4} \right\rfloor $$
$$ = \left\lfloor \frac{4k^2 + 4k}{4} + \frac{1}{4} \right\rfloor $$
$$ = \left\lfloor k^2 + k + \frac{1}{4} \right\rfloor $$
By the definition of floor, since $k^2 + k$ is an integer:
$$ k^2 + k \leq k^2 + k + \frac{1}{4} < k^2 + k + 1 $$
It then follows that:
$$ = \left\lfloor k^2 + k + \frac{1}{4} \right\rfloor = k^2 + k $$
Now, also by substitution:
$$ \left(\frac{n - 1}{2}\right)\left(\frac{n + 1}{2}\right) = \left(\frac{(2k + 1) - 1}{2}\right)\left(\frac{(2k + 1) + 1}{2}\right) $$
$$ = \left(\frac{2k}{2}\right)\left(\frac{2k + 2}{2}\right) $$
$$ = k(k + 1) $$
$$ k^2 + k $$
We have thus shown that the two expressions are equal:
$$ \left\lfloor \frac{n^2}{4} \right\rfloor = \left(\frac{n - 1}{2}\right)\left(\frac{n + 1}{2}\right) $$
Q.E.D.
29. For any odd integer $n$,
$$ \left\lceil \frac{n^2}{4} \right\rceil = \frac{n^2 + 3}{4} $$
**Proof:**
Suppose $n$ is any odd integer.
Since $n$ is an odd integer, $n = 2k + 1$ for some integer $k$.
Then by substitution:
$$ \left\lceil \frac{n^2}{4} \right\rceil = \left\lceil \frac{(2k + 1)^2}{4} \right\rceil $$
$$ = \left\lceil \frac{(2k + 1)(2k + 1)}{4} \right\rceil $$
$$ = \left\lceil \frac{4k^2 + 4k + 1}{4} \right\rceil $$
$$ = \left\lceil k^2 + k + \frac{1}{4} \right\rceil $$
By the definition of ceiling:
$$ k^2 + k < k^2 + k + \frac{1}{4} \leq k^2 + k + 1 $$
It then follows that
$$ \lceil k^2 + k + \frac{1}{4} \rceil = k^2 + k + 1 $$
Then by substution:
$$ \frac{n^2 + 3}{4} = \frac{(2k + 1)^2 + 3}{4} $$
$$ = \frac{(2k + 1)(2k + 1) + 3}{4} $$
$$ = \frac{4k^2 + 4k + 1 + 3}{4} $$
$$ = \frac{4k^2 + 4k + 4}{4} $$
$$ = k^2 + k + 1 $$
Thus we have shown that these two expressions are equal.
$$ \left\lceil \frac{n^2}{4} \right\rceil = \frac{n^2 + 3}{4} $$
Q.E.D.
30. For every integer $n$,
$\left\lfloor \dfrac{n}{2} \right\rfloor + \left\lceil \frac{n}{2} \right\rceil = n$.
**Proof:**
Suppose $n$ is any integer.
_Case where $n$ is even:_
Since $n$ is even, $n = 2k$ for some integer $k$.
By substitution:
$$ \left\lfloor \frac{n}{2} \right\rfloor + \left\lceil \frac{n}{2} \right\rceil = \left\lfloor \frac{2k}{2} \right\rfloor + \left\lceil \frac{2k}{2} \right\rceil $$
$$ = \lfloor k \rfloor + \lceil k \rceil $$
Since $k$ is an integer, then we know that:
$$ \lfloor k \rfloor = k \quad \text{ and } \lceil k \rceil = k $$
So:
$$ = \lfloor k \rfloor + \lceil k \rceil = k + k = 2k = n $$
_Case where $n$ is odd:_
Since $n$ is odd, $n = 2k + 1$ for some integer $k$.
By substitution:
$$ \left\lfloor \frac{n}{2} \right\rfloor + \left\lceil \frac{n}{2} \right\rceil = \left\lfloor \frac{2k + 1}{2} \right\rfloor + \left\lceil \frac{2k + 1}{2} \right\rceil $$
$$ = \left\lfloor \frac{2k}{2} + \frac{1}{2} \right\rfloor + \left\lceil \frac{2k}{2} + \frac{1}{2} \right\rceil $$
$$ = \left\lfloor k + \frac{1}{2} \right\rfloor + \left\lceil k + \frac{1}{2} \right\rceil $$
By the definition of floor:
$$ k \leq k + \frac{1}{2} < k + 1 $$
So:
$$ \left\lfloor k + \frac{1}{2} \right\rfloor = k $$
By the definition of ceiling:
$$ k < k + \frac{1}{2} \leq k + 1 $$
So:
$$ \left\lceil k + \frac{1}{2} \right\rceil = k + 1 $$
Then:
$$ = \left\lfloor k + \frac{1}{2} \right\rfloor + \left\lceil k + \frac{1}{2} \right\rceil = k + k + 1 $$
$$ = 2k + 1 = n $$
In both cases we have shown that the given equation is true. Therefore we have
shown that the given equation is true for every integer $n$.
Q.E.D.
31. For every integer $n$,
$\left\lfloor \dfrac{\left\lceil \dfrac{n}{2} \right\rceil}{3} \right\rfloor = \left\lfloor \dfrac{n}{6} \right\rfloor$.
Omitted.
32. For every integer $n$,
$\left\lceil \dfrac{\left\lceil \frac{n}{2} \right\rceil}{3} \right\rceil = \left\lceil \dfrac{n}{6} \right\rceil$
Omitted.
33. A necessary and sufficient condition for an integer $n$ to be divisible by a
nonzero integer $d$ is that
$n = \left\lfloor \dfrac{n}{d} \right\rfloor \cdot d$. In other words, for
every integer $n$ and nonzero integer $d$,
a. if $d \mid n$, then $n = \left\lfloor \dfrac{n}{d} \right\rfloor \cdot d$.
b. if $n = \left\lfloor \dfrac{n}{d} \right\rfloor \cdot d$ then $d \mid n$.
Omitted.

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chapter_4/notes.md
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@ -658,3 +658,159 @@ It follows, by Theorem T26 of Appendix A, that
$$ |x + y| = (-x) + (-y) \leq |x| + |y| $$
Hence in both cases $|x + y| = |x| + |y|$ _[as was to be shown]._
---
Page 234
**Definition**
Given any real number $x$, the **floor of** $x$, denoted $\lfloor x \rfloor$, is
defined as follows:
$\lfloor x \rfloor =$ that unique integer $n$ such that $n \leq x < n + 1$.
Symbolically, if $x$ is a real number and $n$ is an integer, then
$$ \lfloor x \rfloor = n \Leftrightarrow n \leq x < n + 1 $$
---
Page 235
**Definition**
Given any real number $x$, the **ceiling of** $x$, denoted $\lceil x \rceil$, is
defined as follows:
$\lceil x \rceil =$ that unique integer $n$ such that $n - 1 < x \leq n$.
Symbolically, if $x$ is a real number and $n$ is an integer, then
$$ \lceil x \rceil = n \Leftrightarrow n - 1 < x \leq n $$
---
Page 237
**Theorem 4.6.1**
For every real number $x$ and every integer $m$,
$\lfloor x + m \rfloor = \lfloor x \rfloor + m$.
**Proof:**
Suppose any real number $x$ and any integer $m$ are given. _[We must show that
$\lfloor x + m \rfloor = \lfloor x \rfloor + m$.]_ Let $n = \lfloor x \rfloor$.
By definition of floor, $n$ is an integer and
$$ n \leq x < n + 1 $$
Add $m$ to all three parts to obtain
$$ n + m \leq x + m < n + m + 1 $$
_[since adding a number to both sides of an inequality does not change the
direction of the inequality]._
Now $n + m$ is an integer _[since $n$ and $m$ are integers and a sum of integers
is an integer]_, and so, by definition of floor, the left-hand side of the
equation to be shown is
$$ \lfloor x + m \rfloor = n + m $$
But $n = \lfloor x \rfloor$. Hence, by substitution,
$$ n + m = \lfloor x \rfloor + m $$
which is the right-hand side of the equation to be shown. Thus
$\lfloor x + m \rfloor = \lfloor x \rfloor + m$ _[as was to be shown]._
---
Page 238
**Theorem 4.6.2 The Floor of $\dfrac{n}{2}$**
For any integer $n$,
$$
\left\lfloor \frac{n}{2} \right\rfloor =
\begin{cases}
\dfrac{n}{2} & \text{if } n \text{ is even} \\
\dfrac{n - 1}{2} & \text{if } n \text{ is odd} \\
\end{cases}
$$
**Proof:**
Suppose $n$ is a _[particular but arbitrarily chosen]_ integer. By the quotient
remainder theorem, either $n$ is odd or $n$ is even.
_Case 1 ($n$ is odd):_
In this case, $n = 2k + 1$ for some integer $k$. _[We must show that
$\left\lfloor \dfrac{n}{2} \right\rfloor = \dfrac{(n - 1)}{2}$.]_ But the
left-hand side of the equation to be shown is
$$ \left\lfloor \frac{n}{2} \right\rfloor = \left\lfloor \frac{2k + 1}{2} \right\rfloor = \left\lfloor \frac{2k}{2} + \frac{1}{2} \right\rfloor = \left\lfloor k + \frac{1}{2} \right\rfloor = k $$
because $k$ is an integer and $k \leq k + \dfrac{1}{2} < k + 1$. And the
right-hand side of the equation to be shown is
$$ \frac{n - 1}{2} = \frac{(2k + 1) - 1}{2} = \frac{2k}{2} = k $$
also. So since both the left-hand and right-hand sides equal $k$, they are equal
to each other. That is,
$\left\lfloor \dfrac{n}{2} \right\rfloor = \dfrac{n - 1}{2}$ _[as was to be
shown]._
_Case 2 ($n$ is even):_
In this case, $n = 2k$ for some integer $k$. _[We must show that
$\left\lfloor \dfrac{n}{2} \right\rfloor$]_ The rest of the proof of this case
is left as an exercise.
---
Page 239
**Theorem 4.6.3**
If $n$ is any integer and $d$ is a positive integer, and if
$q = \left\lfloor \frac{n}{d} \right\rfloor$ and
$r = n - d \cdot \left\lfloor \frac{n}{d} \right\rfloor$, then
$$ n = dq + r \quad \text{ and } \quad 0 \leq r < d $$
**Proof:**
Suppose $n$ is any integer, $d$ is a positive integer,
$q = \left\lfloor \dfrac{n}{d} \right\rfloor$, and
$r = n - d \cdot \left\lfloor \frac{n}{d} \right\rfloor$. _[We must show that
$n = dq + r$ and $0 \leq r < d$.]_ By substitution,
$$ dq + r = d \cdot \left\lfloor \frac{n}{d} \right\rfloor + \left(n - d \cdot \left\lfloor \frac{n}{d} \right\rfloor\right) = n $$
So it remains only to show that $0 \leq r < d$. But
$q = \left\lfloor \frac{n}{d} \right\rfloor$. Thus, by definition of floor,
$$ q \leq \frac{n}{d} < q + 1 $$
Then
$$ dq \leq n < dq + d \quad \text{ by multiplying all parts by } d $$
and so
$$ 0 \leq n - dq < d \quad \text{ by subtracting } dq \text{ from all parts.} $$
But
$$ r = n - d\left\lfloor \frac{n}{d} \right\rfloor = n - dq $$
Hence
$$ 0 \leq r < d \quad \text{ by substitution} $$
_[This is what was to be shown.]_

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@ -184,3 +184,19 @@ If $A_1$ then $C$; If $A_2$ then $C$; If $A_3$ then $C$
6. The triangle inequality says that for all real numbers $x$ and $y$, ______.
$|x + 6| \leq |x| + |y|$
---
**Test Yourself**
Page 239
1. Given any real number $x$, the floor of $x$ is the unique integer $n$ such
that ______.
$n \leq x < n + 1$
2. Given any real number $x$, the ceiling of $x$ is the unique integer $n$ such
that ______.
$n - 1 < x \leq n$

2
leftoff.txt
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@ -1 +1 @@
230
234