diff --git a/chapter_7/exercises.md b/chapter_7/exercises.md index f23ce89..d1c92d6 100644 --- a/chapter_7/exercises.md +++ b/chapter_7/exercises.md @@ -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 diff --git a/chapter_7/notes.md b/chapter_7/notes.md index 9d6f1e3..5fd8c51 100644 --- a/chapter_7/notes.md +++ b/chapter_7/notes.md @@ -484,6 +484,57 @@ cardinality as $B$ or $B$ has the same cardinality as $A$. --- +Page 497 + +**Example 7.4.1** + +**An Infinite Set and a Proper Subset Can Have the Same Cardinality** + +Let $2\mathbb{Z}$ be the set of all even integers. Prove that $2\mathbb{Z}$ and +$\mathbb{Z}$ have the same cardinality. + +_Solution:_ + +Consider the function $H$ from $\mathbb{Z}$ to $2\mathbb{Z}$ defined as follows: + +$$ H(n) = 2n \text{ for each } n \in \mathbb{Z} $$ + +A (partial) arrow diagram for $H$ is shown below. + +(See Page 498 for image). + +To show that $H$ is one-to-one, suppose $H(n_1) = H(n_2)$ for some integers +$n_1$ and $n_2$. Then $2n_1 = 2n_2$ by definition of $H$, and dividing both +sides by $2$ gives $n_1 = n_2$. Hence $h$ is one-to-one. + +To show that $H$ is onto, suppose $m$ is any element of $2\mathbb{Z}$. Then $m$ +is an even integer, and so $m = 2k$ for some integer $k$. It follows that +$H(k) = 2k = m$ . Thus there exists $k$ in $\mathbb{Z}$ with $H(k) = m$, and +hence $H$ is onto. + +Therefore, by definition of cardinality, $\mathbb{Z}$ and $2\mathbb{Z}$ have the +same cardinality. + +In Section 9.4 we will show that a function from one finite set to another set +of the same size is one-to-one if, and only if, it is onto. This result does not +hold for infinite sets. Although it is true that for two infinite sets to have +the same cardinality there must exist a function from one to the other that is +both one-to-one and onto, it is always the case that: + +If $A$ and $B$ are infinite sets with the same cardinality, then there exist +functions from $A$ to $B$ that are one-to-one but not onto and functions from +$A$ to $B$ that are onto but not one-to-one. + +For instance, since the function $H$ in Example 7.4.1 is one-to-one and onto, +$\mathbb{Z}$ and $2\mathbb{Z}$ have the same cardinality. But the "inclusion +function" $I$ from $2\mathbb{Z}$ to $\mathbb{Z}$, given by $I(n) = n$ for all +even integers $n$, is one-to-one but not onto. And the function $J$ from +$\mathbb{Z}$ to $2\mathbb{Z}$ defined by +$J(n) = 2\left\lfloor \dfrac{n}{2} \right\rfloor$, for each integer $n$, is onto +but not one-to-one. (See exercise 6 at the end of this section.) + +--- + Page 499 **Definition** diff --git a/chapter_7/test_yourself.md b/chapter_7/test_yourself.md index 809019b..310a7aa 100644 --- a/chapter_7/test_yourself.md +++ b/chapter_7/test_yourself.md @@ -160,29 +160,57 @@ Page 507 1. A set is finite if, and only if, _____. +it is the empty set or there is a one-to-one correspondence from +$\{1, 2, \dots n\}$ to it, for some positive integer $n$. + 2. To prove that a set $A$ has the same cardinality as a set $B$ you must _____. +show that there is a function one-to-one correspondence from $A$ to $B$. + 3. The reflexive property of cardinality says that given any set $A$, _____. +$A$ has the same cardinality as $A$. + 4. The symmetric property of cardinality says that given any sets $A$ and $B$, _____. +if $A$ has the same cardinality as $B$, then $B$ has the same cardinality as +$A$. + 5. The transitive property of cardinality says that given any sets $A$, $B$, and $C$, _____. +if $A$ has the same cardinality as $B$, and if $B$ has the same cardinality as +$C$, then $A$ has the same cardinality as $C$. + 6. A set is called countably infinite if, and only if, _____. +it has the same cardinality as the set of all positive integers +($\mathbb{Z}^+$). + 7. A set is called countable if, and only if, _____. +it is finite or countably infinite + 8. In each of the following, fill in the blank with the word _countable_ or the word _uncountable_. a. The set of all integers is _____. +countable + b. The set of all rational numbers is _____. +countable + c. The set of all real numbers between $0$ and $1$ is _____. +uncountable + d. The set of all real numbers is _____. +uncountable + 9. The Cantor diagonalization process is used to prove that _____. + +the set of all real numbers between $0$ and $1$ is uncountable