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@ -3943,23 +3943,250 @@ Page 507
$B$, a student replies, "$A$ and $B$ are one-to-one and onto." What _should_
the student have replied? Why?
Since $A$ and $B$ are sets and not functions, the student's statement is
incorrect, since sets cannot have the properties of being one-to-one and onto,
only a function can have these properties. Instead, the student have claimed
that $A$ and $B$ have the same cardinality if, and only if, there is a function
from $A$ to $B$ that is both one-to-one and onto (a one-to-one correspondence).
2. Show that "there are as many squares as there are numbers" by exhibiting a
one-to-one correspondence from the positive integers, $\mathbb{Z}^+$, to the
set $S$ of all squares of positive integers:
$$ S = \{n \in \mathbb{Z}^+ | n = k^2, \text{ for some positive integer } k\} $$
**Proof:**
Suppose there is a set $S$ that is the set of all squares:
$$ S = \{n \in \mathbb{Z}^+ | n = k^2, \text{ for some positive integer } k\} $$
To prove that $S$ and the set of all positive integers, $\mathbb{Z}^+$, have the
same cardinality, it must be shown that there exists some function, $f$, such
that $f$ is a one-to-one correspondence from $\mathbb{Z}^+$ to $S$. In other
words, it must be shown that there exists some function $f: \mathbb{Z}^+ \to S$
such that $f$ is one-to-one and onto.
_Proof ($f$ is one-to-one):_
Suppose $k_1, k_2 \in \mathbb{Z}^+$ such that $f(k_1) = f(k_2)$.
To prove that $f$ is one-to-one, it must be shown that $k_1 = k_2$.
By definition of $f$:
$$ k_1^2 = k_2^2 $$
By algebra:
$$ \sqrt{k_1^2} = \sqrt{k_2^2} $$
$$ \pm k_1 = \pm k_2 $$
Recall that $k_1, k_2 \in \mathbb{Z}^+$, so it follows that:
$$ k_1 = k_2 $$
This is what was to be shown. Therefore, it can be concluded that $f$ is
one-to-one.
_Proof ($f$ is onto):_
Suppose there is some $m \in S$.
To prove that $f$ is onto, it must be shown that there exists some
$k \in \mathbb{Z}^+$ such that $f(k) = m$.
By definition of $S$:
$$ m = k^2 $$
for some integer $k$.
Then, by definition of $f$:
$$ f(k) = k^2 = m $$
This is what was to be shown. Therefore it can be concluded that $f$ is onto.
_Conclusion:_
Since it has been shown that $f$ is both one-to-one and onto, it follows by the
properties of cardinality that the sets $S$ and $\mathbb{Z}^+$ have the same
cardinality (the same number of elements in them).
Q.E.D.
3. Let
$3\mathbb{Z} = \{n \in \mathbb{Z} | n = 3k, \text{ for some integer } k\}$.
Prove that $\mathbb{Z}$ and $3\mathbb{Z}$ have the same cardinality.
**Proof:**
Suppose there is a set $3\mathbb{Z}$ that represents the set of all integers
divisible by $3$:
$$ 3\mathbb{Z} = \{n \in \mathbb{Z} | n = 3k, \text{ for some integer } k\} $$
Define $f: \mathbb{Z} \to 3\mathbb{Z}$ as $f(n) = 3n$ for some integer $n$.
To prove that $\mathbb{Z}$ and $3\mathbb{Z}$ have the same cardinality, it must
shown that $f$ is a one-to-one correspondence from $\mathbb{Z}$ to 3\mathbb{Z}.
_Proof ($f$ is one-to-one):_
Suppose there exists some $x_1, x_2 \in 3\mathbb{Z}$ such that
$f(x_1) = f(x_2)$.
To prove that $f$ is one-to-one, it must be shown that $x_1 = x_2$.
By definition of $f$:
$$ 3x_1 = 3x_2 $$
By algebra:
$$ x_1 = x_2 $$
This is what was to be shown. Therefore, it can be concluded that $f$ is
one-to-one.
_Proof ($f$ is onto):_
Suppose there is some $m \in 3\mathbb{Z}$.
To prove that $f$ is onto, it must be shown that $f(k) = m$ for some integer
$k$.
By definition of $3\mathbb{Z}$:
$$ f(k) = 3k = m $$
for some integer $k$.
This is what was to be shown. Therefore, it can be concluded that $f$ is onto.
_Conclusion:_
Since it has been shown that $f$ is a one-to-one correspondence from
$\mathbb{Z}$ to $3\mathbb{Z}$, it can be concluded that $\mathbb{Z}$ and
$3\mathbb{Z}$ have the same cardinality.
Q.E.D.
4. Let $\mathbb{O}$ be the set of all odd integers. Prove that $\mathbb{O}$ has
the same cardinality as $2\mathbb{Z}$, the set of all even integers.
**Proof:**
Suppose there is a set $\mathbb{O}$ that represents the set of all odd integers:
$$ \mathbb{O} = \{n \in \mathbb{Z} | n = 2k + 1 \text{ for some integer } k\} $$
Let $f: \mathbb{O} \to 2\mathbb{Z}$ (where $2\mathbb{Z}$ is the set of all even
integers) such that $f(n) = n - 1$ for some integer $n$.
To prove that $\mathbb{O}$ has the same cardinality as $2\mathbb{Z}$, it must be
shown that $f$ is a one-to-one correspondence from $\mathbb{O}$ to
$2\mathbb{Z}$.
_Proof ($f$ is one-to-one):_
Suppose $x_1, x_2 \in \mathbb{O}$ such that $f(x_1) = f(x_2)$.
To prove that $f$ is one-to-one, it must be shown that $x_1 = x_2$.
By the definition of $f$:
$$ x_1 - 1 = x_2 - 1 $$
By algebra:
$$ x_1 = x_2 $$
This is what was to be shown. Therefore, it can be concluded that $f$ is
one-to-one.
_Proof ($f$ is onto):_
Suppose $m \in 2\mathbb{Z}$.
To prove that $f$ is onto, it must be shown that $f(k) = m$ for some integer
$k \in \mathbb{O}$.
Let $k = m + 1$. Since $m \in 2\mathbb{Z}$, $m + 1$ is odd (by definition of
odd). This means that $k$ is odd, so $k \in \mathbb{O}$. Then:
$$ f(k) = k - 1 = (m + 1) - 1 = m $$
This is what was to be shown. Therefore it can be concluded that $f$ is onto.
_Conclusion:_
Since it has been shown that $f$ is a one-to-one correspondence from
$\mathbb{O}$ to $2\mathbb{Z}$, it can be concluded that $\mathbb{O}$ and
$2\mathbb{Z}$ have the same cardinality.
5. Let $25\mathbb{Z}$ be the set of all integers that are multiples of $25$.
Prove that $25\mathbb{Z}$ has the same cardinality as $2\mathbb{Z}$, the set
of all even integers.
**Proof:**
Suppose $25\mathbb{Z}$ represents the set of all integers that are multiples of
$25$.
$$ 25\mathbb{Z} = \{n \in \mathbb{Z} | n = 25k \text{ for some integer } k\} $$
Let $f: 25\mathbb{Z} \to 2\mathbb{Z}$ (where $2\mathbb{Z}$ represents the set of
all even integers), and define $f$ as $f(n) = \dfrac{2}{25}n$ for some integer
$n$.
To prove that $25\mathbb{Z}$ has the same cardinality as $2\mathbb{Z}$, it must
be shown that $f$ is a one-to-one correspondence from
$25\mathbb{Z} \to 2\mathbb{Z}$.
_Proof ($f$ is one-to-one):_
Suppose $x_1, x_2 \in 25\mathbb{Z}$ such that $f(x_1) = f(x_2)$.
To prove that $f$ is one-to-one, it must be shown that $x_1 = x_2$.
By the definition for $f$:
$$ \frac{2}{25}x_1 = \frac{2}{25}x_2 $$
By algebra:
$$ x_1 = x_2 $$
This is what was to be shown. Therefore, it can be concluded that $f$ is
one-to-one.
_Proof ($f$ is onto):_
Let $m \in 2\mathbb{Z}.
To show that $f$ is onto, it must be shown that $f(k) = m$ for some
$k \in 25\mathbb{Z}$.
Let $k = \dfrac{25m}{2}$.
Since $m$ is even, $\dfrac{m}{2}$ is an integer, so $k = 25 \cdot \dfrac{m}{2}$,
which is a multiple of $25$. It follows that $k \in 25\mathbb{Z}$.
Then:
$$ f(k) = \frac{2}{25}\left(\frac{25m}{2}\right) = m $$
This is what was to be shown. Therefore, it can be concluded that $f$ is onto.
_Conclusion:_
Since it has been shown that $f$ is a one-to-one correspondence from
$25\mathbb{Z} \to 2\mathbb{Z}$, it can be concluded that $25\mathbb{Z}$ and
$2\mathbb{Z}$ have the same cardinality.
6. Use the functions $I$ and $J$ defined in the paragraph following Example
7.4.1 to show that even though there is a one-to-one correspondence, $H$,
from $2\mathbb{Z}$ to $\mathbb{Z}$, there is also a function from
@ -3968,33 +4195,476 @@ $$ S = \{n \in \mathbb{Z}^+ | n = k^2, \text{ for some positive integer } k\} $$
words, show that $I$ is one-to-one but not onto, and show that $J$ is onto
but not one-to-one.
_Hint:_ If $m \in 2\mathbb{Z}$, show that $J(m) = J(m + 1) = m$.
Suppose $I: 2\mathbb{Z} \to \mathbb{Z}$, and define $I$ as $I(n) = n$ for some
even integer $n$.
Furthermore, suppose $J: \mathbb{Z} \to 2\mathbb{Z}$, and define $J$ as $J(m) =
2\left\lfloor \dfrac{m}{2} \right\rfloor$ for some integer $m$.
**Proof ($I$ is one-to-one, but not onto):**
_Proof ($I$ is one-to-one):_
Suppose $x_1, x_2 \in 2\mathbb{Z}$ such that $I(x_1) = I(x_2)$. To prove that
$I$ is one-to-one, it must be shown that $x_1 = x_2$.
By definition for $I$:
$$ x_1 = x_2 $$
This is what was to be shown. Therefore it can be concluded that $I$ is
one-to-one.
_Proof ($I$ is not onto):_
Suppose $m \in \mathbb{Z}$.
To prove that $I$ is onto, it would need to be shown that $I(k) = m$, for some
$k \in \mathbb{Z}$.
But since $m \in \mathbb{Z}$, $m$ could be an odd integer. If $m$ is odd, then
$m = 2k + 1$ for some integer $k$, and so $I$ is not onto.
Consider $m = 3$, then there would exist no $k$ for which $I(k) = 3$, since $I$.
Therefore $I$ is not onto.
**Proof ($J$ is onto, but not one-to-one):**
_Proof ($J$ is onto):_
Suppose $m \in 2\mathbb{Z}$.
To prove that $J$ is onto, it must be shown that $J(k) = m$ for some
$k \in \mathbb{Z}$.
Let $k = m$.
By the definition for $J$:
$$ J(k) = 2\lfloor \frac{k}{2} \rfloor $$
Since $k$ is even (since $k = m$ and $m \in 2\mathbb{Z}$), $k = 2p$ for some
integer $p$. By substitution:
$$ J(k) = 2\lfloor \frac{2p}{2} \rfloor $$
$$ = 2\lfloor p \rfloor $$
Then, by definition of floor:
$$ = 2p $$
Since $p$ is an integer, $2p \in 2\mathbb{Z}$, by the definition of even. It
follows that:
$$ J(k) = 2p = k = m $$
This is what was to be shown. Therefore it can be concluded that $J$ is onto.
_Proof ($J$ is not one-to-one):_
Were it to be proven that $J$ is one-to-one, it would have to assumed that for
some $x_1, x_2 \in \mathbb{Z}$, such that $J(x_1) = J(x_2)$, and then shown that
$x_1 = x_2$.
By the definition for $J$:
$$ 2\lfloor \frac{x_1}{2} \rfloor = 2\lfloor \frac{x_2}{2} \rfloor $$
But the floor function does not necessarily have to take the same image to
generate the same elements of its co-domain.
Consider $x_1 = 2$ and $x_2 = 3$, then by definition for $J$:
$$ J(x_1) = 2\lfloor \frac{2}{2} \rfloor = 2\lfloor 1 \rfloor = 2(1) = 2 $$
$$ J(x_2) = 2\lfloor \frac{3}{2} \rfloor = 2(1) = 2 $$
So, $J(x_1) = J(x_2)$, but $x_1 \neq x_2$. Therefore $J$ is not one-to-one.
7.
a. Check that the formula for $F$ given at the end of Example 7.4.2 produces the
correct values for $n = 1, 2, 3, \text{ and } 4$.
The formula for $F$ is as follows:
$$
F(n) =
\begin{cases}
\dfrac{n}{2} & \text{if } n \text{ is an even positive integer} \\
-\dfrac{n - 1}{2} & \text{if } n \text{ is an odd positive integer}
\end{cases}
$$
_Case $n = 1$:_
$$ F(1) = -\frac{(1) - 1}{2} = -\frac{0}{2} = (-1)0 = 0 $$
_Case $n = 2$:_
$$ F(2) = \frac{(2)}{2} = 1 $$
_Case $n = 3$:_
$$ F(3) = -\frac{(3) - 1}{2} = (-1)\left(\frac{2}{2}\right) = (-1)(1) = -1 $$
_Case $n = 4$:_
$$ F(4) = \frac{(4)}{2} = 2 $$
b. Use the floor function to write a formula for $F$ as a single algebraic
expression for each positive integer $n$.
$$ F(n) = (-1)^n\lfloor \frac{n}{2} \rfloor $$
8. Use the result of exercise 3 to prove that $3\mathbb{Z}$ is countable.
Recall that a set is countable if, and only if, it is finite or countably
infinite.
Additionally, recall that a set is countably infinite if, and only if, it has
the same cardinality as the set of positive integers $\mathbb{Z}^+$ (_i.e._
there is a function that is a one-to-one correspondence from the given set to
$\mathbb{Z}^+$).
**Proof:**
Suppose $3\mathbb{Z}$ represents the set of all integers that are multiples of
$3$:
$$ 3\mathbb{Z} = \{n \in \mathbb{Z} | n = 3k, \text{ for some integer } k\} $$
To prove that $3\mathbb{Z}$ is countable, it must be shown that $3\mathbb{Z}$ is
finite or countably infinite.
$3\mathbb{Z}$ is not finite, thus it must be shown that $3\mathbb{Z}$ is
countably infinite.
As was shown in Example 7.4.3, $\mathbb{Z}$ is countably infinite (_i.e._
$\mathbb{Z}$ has the same cardinality as $\mathbb{Z}^+$). Additionally, by
problem 3, it has been shown that $3\mathbb{Z}$ has the same cardinality has
$\mathbb{Z}$. By the transitive property of cardinality, it follows that
$3\mathbb{Z}$ has the same cardinality as $\mathbb{Z}^+$, and therefore
$3\mathbb{Z}$ is countably infinite.
9. Show that the set of all nonnegative integers is countable by exhibiting a
one-to-one correspondence between $\mathbb{Z}^+$ and
$\mathbb{Z}^{\text{nonneg}}$.
Recall that a set is countable if, and only if, it is finite or countably
infinite.
Additionally, recall that a set is countably infinite if, and only if, it has
the same cardinality as the set of positive integers $\mathbb{Z}^+$ (_i.e._
there is a function that is a one-to-one correspondence from the given set to
$\mathbb{Z}^+$).
**Proof:**
Suppose $\mathbb{Z}^+$ represents the set of all positive integers and that
$\mathbb{Z}^{\text{nonneg}}$ represents the set of all nonnegative integers.
To prove that $\mathbb{Z}^{\text{nonneg}}$ is countable, it must be shown that
$\mathbb{Z}^{\text{nonneg}}$ is finite or countably infinite. Since
$\mathbb{Z}^{\text{nonneg}}$ is not finite, it follows that it must be shown
that $\mathbb{Z}^{\text{nonneg}}$ is countably infinite (_i.e._ has the same
cardinality as $\mathbb{Z}^+$).
Let $f: \mathbb{Z}^+ \to \mathbb{Z}^{\text{nonneg}}$ be defined as
$f(n) = n - 1$ for some positive integer $n$.
_Proof ($f$ is one-to-one):_
Suppose $x_1, x_2 \in \mathbb{Z}^+$ such that $f(x_1) = f(x_2)$.
To prove that $f$ is one-to-one, it must be shown that $x_1 = x_2$.
By definition of $f$:
$$ x_1 - 1 = x_2 - 1 $$
By algebra:
$$ x_1 = x_2 $$
This is what was to be shown. Therefore, it can be concluded that $f$ is
one-to-one.
_Proof ($f$ is onto):_
Suppose $m \in \mathbb{Z}^{\text{nonneg}}$.
To prove that $f$ is onto, it must be shown that $f(n) = m$ for some
$n \in \mathbb{Z}^+$.
By definition for $f$:
$$ f(n) = n - 1 $$
Let $n = m + 1$. Since $m \in \mathbb{Z}^{\text{nonneg}}$, it follows that
$n \in \mathbb{Z}^+$ (since adding $1$ to any nonnegative integer is positive).
Then, by substitution:
$$ = (m + 1) - 1 $$
By algebra:
$$ = m $$
This is what was to be shown. Therefore it can be concluded that $f$ is onto.
_Conclusion:_
Since $f$ has been shown to be a one-to-one correspondence, it can be concluded
that $\mathbb{Z}^{\text{nonneg}}$ is countable.
In 10-14 $S$ denotes the set of real numbers strictly between $0$ and $1$. That
is, $s = \{x \in \mathbb{R} | 0 < x < 1\}$.
is, $S = \{x \in \mathbb{R} | 0 < x < 1\}$.
10. Let $U = \{x \in \mathbb{R} | 0 < x < 2\}$. Prove that $S$ and $U$ have the
same cardinality.
**Proof:**
Suppose $S$ represents the set of real numbers strictly between $0$ and $1$:
$$ S = \{x \in \mathbb{R} | 0 < x < 1\} $$
Additionally, suppose $U$ is the set of real numbers strictly between $0$ and
$2$:
$$ U = \{x \in \mathbb{R} | 0 < x < 2\} $$
Let $f: S \to U$, and define $f$ as $f(n) = 2n$ for some $n \in S$.
To prove that $S$ and $U$ have the same cardinality, it must be shown that $f$
is a one-to-one correspondence from $S \to U$.
_Proof ($f$ is one-to-one):_
Suppose $n_1, n_2 \in S$ such that $f(n_1) = f(n_2)$.
To prove that $f$ is one-to-one, it must be shown that $n_1 = n_2$.
By definition of $f$:
$$ 2n_1 = 2n_2 $$
By algebra:
$$ n_1 = n_2 $$
This is what was to be shown. Therefore, it can be concluded that $f$ is
one-to-one.
_Proof ($f$ is onto):_
Let $m \in U$.
To prove that $f$ is onto, it must be shown that $f(n) = m$ for some $n \in S$.
Let $n = \frac{m}{2}$.
Since $m \in U$, $0 < m < 2$. It follows that $0 < \dfrac{m}{2} < 1$. By
substitution, this means that $0 < n < 1$, which means that $n \in S$.
Then, by definition for $f$:
$$ f(n) = 2n $$
By substitution:
$$ = 2\left(\frac{m}{2}\right) $$
By algebra:
$$ = m $$
This is what was to be shown. Therefore, it can be concluded that $f$ is onto.
_Conclusion:_
Since it has been shown that $f$ is a one-to-one correspondence, it can be
concluded that $S$ and $U$ have the same cardinality.
Q.E.D.
11. Let $V = \{x \in \mathbb{R} | 2 < x < 5\}$. Prove that $S$ and $V$ have the
same cardinality.
_Hint:_ Define $h: S \to V$ as follows: $h(x) = 3x + 2$, for every $x \in S$.
**Proof:**
Suppose $S$ represents the set of real numbers strictly between $0$ and $1$:
$$ S = \{x \in \mathbb{R} | 0 < x < 1\} $$
Additionally, suppose $V$ is the set of real numbers strictly between $2$ and
$5$:
$$ V = \{x \in \mathbb{R} | 2 < x < 5\} $$
Let $h: S \to V$, and define $h$ as $h(x) = 3x + 2$ for some $x \in S$.
To prove that $S$ and $V$ have the same cardinality, it must be shown that $h$
is a one-to-one correspondence for $S \to V$.
_Proof ($h$ is one-to-one):_
Suppose $x_1, x_2 \in S$ such that $h(x_1) = h(x_2)$.
To prove $h$ is one-to-one, it must be shown that $x_1 = x_2$.
By the definition of $h$:
$$ 3x_1 + 2 = 3x_2 + 2 $$
By algebra:
$$ 3x_1 = 3x_2 $$
$$ x_1 = x_2 $$
This is what was to be shown. Therefore, it can be concluded that $h$ is
one-to-one.
_Proof ($h$ is onto):_
Let $v \in V$.
To prove that $h$ is onto, it must be shown that $f(x) = v$ for some $x \in S$.
Let $x = \dfrac{v - 2}{3}$.
Since $v \in V$, $2 < v < 5$. It follows that:
$$ 2 - 2 < v - 2 < 5 - 2 $$
$$ 0 < v - 2 < 3 $$
$$ \frac{0}{3} < \frac{v - 2}{3} < \frac{3}{3} $$
$$ 0 < \frac{v - 2}{3} < 1 $$
By substitution, this means that:
$$ 0 < x < 1 $$
This means that $x \in S$.
Then, by definition for $h$:
$$ h(x) = 3x + 2 $$
By substitution:
$$ h(x) = 3\left(\frac{v - 2}{3}\right) + 2 $$
By algebra:
$$ = (v - 2) + 2 $$
$$ = v $$
This is what was to be shown. Therefore it can be concluded that $h$ is onto.
_Conclusion:_
Since it has been shown that $h$ is a one-to-one correspondence for $S \to V$,
it can be concluded that $S$ and $V$ have the same cardinality.
Q.E.D.
12. Let $a$ and $b$ be real numbers with $a < b$, and suppose that
$W = \{x \in \mathbb{R} | a < x < b\}$. Prove that $S$ and $W$ have the same
cardinality.
**Proof:**
Suppose $S$ represents the set of real numbers strictly between $0$ and $1$:
$$ S = \{x \in \mathbb{R} | 0 < x < 1\} $$
Additionally, suppose $a, b \in \mathbb{R}$ such that $a < b$. Then, suppose $W$
is a set defined as:
$$ W = \{x \in \mathbb{R} | a < x < b\} $$
Let $f: S \to W$ be defined as $f(x) = (b - a)x + a$ for some $x \in S$.
To prove that $S$ and $W$ have the same cardinality, it must be shown that $f$
is a one-to-one correspondence for $S \to W$.
_Proof ($f$ is one-to-one):_
Let $x_1, x_2 \in S$ such that $f(x_1) = f(x_2)$.
To prove that $f$ is one-to-one, it must be shown that $x_1 = x_2$.
By the definition of $f$:
$$ (b - a)x_1 + a = (b - a)x_2 + a $$
By algebra:
$$ (b - a)x_1 = (b - a)x_2 $$
$$ x_1 = x_2 $$
This is what was to be shown. Therefore it can be concluded that $f$ is
one-to-one.
_Proof ($f$ is onto):_
Let $w \in W$.
To prove that $f$ is onto, it must be shown that $f(x) = w$ for some $x \in S$.
Let $x = \dfrac{w - a}{b - a}$.
Since $w \in W$, this means that $a < w < b$. It follows that:
$$ a - a < w - a < b - a $$
$$ 0 < w - a < b - a $$
$$ \frac{0}{b - a} < \frac{w - a}{b - a} < \frac{b - a}{b - a} $$
$$ 0 < \frac{w - a}{b - a} < 1 $$
By substitution:
$$ 0 < x < 1 $$
It follows that $x \in S$.
By definition for $f$:
$$ f(x) = (b - a)x + a $$
By substitution:
$$ f(x) = (b - a)\left(\frac{w - a}{b - a}\right) + a $$
$$ = (w - a) + a $$
$$ = w $$
This is what was to be shown. Therefore it can be concluded that $f$ is onto.
_Conclusion:_
Since it has been shown that $f$ is a one-to-one correspondence for $S \to W$,
it can be concluded that $S$ and $W$ have the same cardinality.
Q.E.D.
13. Draw the graph of the function $f$ defined by the following formula:
For each real number $x$ with $0 < x < 1$,
@ -4003,6 +4673,8 @@ $$ f(x) = \tan\left(\pi x - \frac{\pi}{2}\right) $$
Use the graph to explain why $S$ and $\mathbb{R}$ have the same cardinality.
Omitted.
14. Define a function $g$ from the set of real numbers to $S$ by the following
formula:
@ -4014,36 +4686,143 @@ Prove that $g$ is a one-to-one correspondence. (It is possible to prove this
statement either with calculus or without it.) What conclusion can you draw from
this fact?
Omitted.
15. Show that the set of all bit strings (strings of $0$'s and $1$'s) is
countable.
Recall that a set is countable if, and only if, it is finite or countably
infinite.
Additionally, recall that a set is countably infinite if, and only if, it has
the same cardinality as the set of positive integers $\mathbb{Z}^+$ (_i.e._
there is a function that is a one-to-one correspondence from the given set to
$\mathbb{Z}^+$).
**Proof:**
Suppose $B$ is the set of all bit strings (strings of $0$'s and $1$'s).
To prove that $B$ is countable, it must be shown that $B$ is finite or countably
infinite. Since $B$ is not finite, it must be shown that $B$ is countably
infinite.
To show that $B$ is countably infinite, it must be shown that $B$ has the same
cardinality as the set of all positive integers, $\mathbb{Z}^+$.
To show that $B$ and $\mathbb{Z}^+$ have the same cardinality, it must be shown
that there exists some one-to-one correspondence for $B \to \mathbb{Z}^+$.
Consider a function, $f$ that maps the bit strings by their length to some
positive integer. For example, say $f(\lambda) = 1$ (where $\lambda$ represents
the null string). Additional examples would include $f(0) = 2, f(1) = 3$ for bit
strings of length $2$. Further examples for length $3$ include
$f(00) = 4, f(01) = 5, f(10) = 6, f(11) = 7$, and so on.
Generally, for each integer $n \geq 0$, there are $2^n$ bit strings of length
$n$, and $f$ maps them to the positive integers between $2^n$ (inclusive) and
$2^{n + 1} - 1$ (inclusive).
$f$ is one-to-one since two bit strings never map to the same positive integer.
$f$ is onto as every positive integer is in the range for $f$ (namely
$\mathbb{Z}^+$).
Therefore, since $f$ is a one-to-one correspondence for $B \to \mathbb{Z}^+$, it
can be concluded that $B$ and $\mathbb{Z}^+$ have the same cardinality, and
therefore $B$ is countable.
Q.E.D.
16. Show that $\mathbb{Q}$, the set of all rational numbers, is countable.
Omitted.
17. Show that $\mathbb{Q}$, the set of all rational numbers, is dense along the
number line by showing that given any two rational numbers $r_1$ and $r_2$
with $r_2 < r_2$, there exists a rational number $x$ such that
$r_1 < x < r_2$.
18. Must the average of two irrational numbers always be irrational? Prove or
_Hint:_ See the hints for exercises 18 and 19 in Section 4.3.
18. _Hint:_
$$ \frac{\dfrac{a}{b} + \dfrac{c}{d}}{2} = \frac{\dfrac{(ad + bc)}{(bd)}{2} =
\frac{ad + bc}{2bd} $$
19. _Hint:_ If $a < b$ then $a + a < a + b$ (by T19 of Appendix A), or
equivalently, $2a < a + b$. Thus $a < \dfrac{a + b}{2}$ (by T20 of Appendix
A).
**Proof:**
Let $r_1, r_2 \in \mathbb{Q}$ such that $r_1 < r_2$.
To prove $\mathbb{Q}$ is dense along the number line, it must be shown there
exists some $x \in \mathbb{Q}$ such that $r_1 < x < r_2$.
Let $x = \dfrac{r_1 + r_2}{2}$.
Since $r_1, r_2 \in \mathbb{Q}$, it follows that
$\dfrac{r_1 + r_2}{2} \in \mathbb{Q}$, hence $x \in \mathbb{Q}$.
Since $r_1 < r_2$, it follows that $r_1$ is less than their average:
$$ r_1 < \frac{r_1 + r_2}{2} $$
Similarly, since $r_2 > r_1$, it follows that $r_2$ is greater than their
average:
$$ \frac{r_1 + r_2}{2} < r_2 $$
This means that:
$$ r_1 < \frac{r_1 + r_2}{2} < r_2 $$
Now, by substitution:
$$ r_1 < x < r_2 $$
This is what was to be shown. Therefore it can be concluded that $\mathbb{Q}$ is
dense along the number line.
Q.E.D.
20. Must the average of two irrational numbers always be irrational? Prove or
give a counterexample.
19. Show that the set of all irrational numbers is dense along the number line
**Disproof (by counterexample):**
Consider $r_1, r_2 \notin \mathbb{Q}$ where $r_1 = \sqrt{2}$ and
$r_2 = -\sqrt{2}$.
Then, their average would be:
$$ \frac{r_1 + r_2}{2} = \frac{\sqrt{2} + (-\sqrt{2})}{2} = \frac{0}{2} = 0 $$
Now, $0 \in \mathbb{Q}$.
This shows that the average of two irrational numbers is not always irrational.
Q.E.D.
21. Show that the set of all irrational numbers is dense along the number line
by showing that given any two real numbers, there is an irrational number in
between.
20. Give two examples of functions from $\mathbb{Z}$ to $\mathbb{Z}$ that are
22. Give two examples of functions from $\mathbb{Z}$ to $\mathbb{Z}$ that are
one-to-one but not onto.
21. Give two examples of functions from $\mathbb{Z}$ to $\mathbb{Z}$ that are
23. Give two examples of functions from $\mathbb{Z}$ to $\mathbb{Z}$ that are
onto but not one-to-one.
22. Define a function: $g: \mathbb{Z}^+ \times \mathbb{Z}y+ \to \mathbb{Z}^+$ by
24. Define a function: $g: \mathbb{Z}^+ \times \mathbb{Z}y+ \to \mathbb{Z}^+$ by
the formula $g(m, n) = 2^m3^n$ for all
$(m, n) \in \mathbb{Z}^+ \times \mathbb{Z}^+$. Show that $g$ is one-to-one
and use this result to prove that $\mathbb{Z}^+ \times \mathbb{Z}^+$ is
countable.
23.
25.
a. Explain how to use the following diagram to show that
$\mathbb{Z}^{\text{nonneg}} \times \mathbb{Z}^{\text{nonneg}}$ and