🚧 Fin 8.1

chapter_8/exercises.md

@@ -10,11 +10,174 @@ $$ m E n \Leftrightarrow m - n \text{ is even} $$
 
 a. Is $0 E 0$? Is $5 E 2$? Is $(6, 6) \in E$? Is $(-1, 7) \in E$?
 
+_$0 E 0$:_
+
+Yes, $0 - 0 = 0$, and $0$ is even.
+
+_$5 E 2$:_
+
+No, $5 -  2 = 3$, and $3$ is not even.
+
+_$(6, 6) \in E$:_
+
+Yes, $6 - 6 = 0$, and $0$ is even.
+
+_$(-1, 7) \in E$:_
+
+Yes, $-1 - 7 = -8$, and $-8$ is even.
+
 b. Prove that for any even integer $n$, $n E 0$.
 
+**Proof:**
+
+Suppose $n \in 2\mathbb{Z}$, where $2\mathbb{Z}$ is the set of all even
+integers.
+
+By the definition for even, this means that $n = 2k$ for some integer $k$.
+
+By the definition for $E$, $n E 0$ if, and only if $n - 0$ is even.
+
+By substitution for $E$:
+
+$$ n - 0 = 2k - 0 $$
+
+$$ = 2k $$
+
+By the definition for even, this means that $n - 0$ is even, and therefore
+$n E 0$ is true.
+
+Q.E.D.
+
 2. Prove that for all integers $m$ and $n$, $m - n$ is even if, and only if,
    both $m$ and $n$ are even or both $m$ and $n$ are odd.
 
+_Hint:_ To prove a statement of the form $p \Leftrightarrow (q \vee r)$, you
+need to prove both (1)$p \to (q \vee r)$ and (2) $(q \vee r) \to p$. The easiest
+way to prove $p \to (q \vee r)$ is to prove the logically equivalent statement
+form $(p \wedge \neg q) \to r$. And the easiest way to prove $(q \vee r) \to p$
+is to prove the logically equivalent statement form
+$(q \to p) \wedge (r \to p)$. In this case, suppose $m$ and $n$ are any
+integers, and let $p$ be "$m - n$ is even," let $q$ be "both $m$ and $n$ are
+even," and let $r$ be "both $m$ and $n$ are odd."
+
+**Proof:**
+
+Suppose $m$ and $n$ are any integers.
+
+To prove that for all integers $m$ and $n$, $m - n$ is even if, and only if,
+both $m$ and $n$ are even or both $m$ and $n$ are odd, it must be shown first
+that if $m - n$ is even, then both $m$ and $n$ are even or both $m$ and $n$ are
+odd, then it must be shown second that if both $m$ and $n$ are even or both $m$
+and $n$ are odd, then $m - n$ is even.
+
+_Proof (first):_
+
+Suppose $m - n$ is even. To prove that both $m$ and $n$ must be even or both $m$
+and $n$ must be odd, all cases for where $m$ is even or odd and where $n$ is
+even or odd must be considered.
+
+_Case (both $m$ and $n$ are even):_
+
+Since both $m$ and $n$ are even, this means that $m = 2k$ and $n = 2p$ for some
+integers $k$ and $p$. Then:
+
+$$ m - n = 2k - 2p $$
+
+$$ = 2(k - p) $$
+
+Now, $k - p$ is an integer by the subtraction of integers. Therefore, by the
+definition of even, $m - n$ is even.
+
+_Case (both $m$ and $n$ are odd):_
+
+Since both $m$ and $n$ are odd, this means that $m = 2k + 1$ and $n = 2p + 1$
+for some integers $k$ and $p$. Then:
+
+$$ m - n = (2k + 1) - (2p + 1) $$
+
+$$ = 2k + 1 - 2p - 1 $$
+
+$$ = 2k - 2p $$
+
+$$ = 2(k - p) $$
+
+Now, $k - p$ is an integer by the subtraction of integers. Therefore, by the
+definition of even, $m - n$ is even.
+
+_Case ($m$ is even and $n$ is odd):_
+
+Since $m$ is even and $n$ is odd, $m = 2k$ and $n = 2p + 1$ for some integers
+$k$ and $p$. Then:
+
+$$ m - n = 2k - (2p + 1) $$
+
+$$ = 2k - 2p - 1 $$
+
+$$ = 2(k - p) - 1 $$
+
+Now, $k - p$ is an integer by the subtraction of integers. Thus, by the
+definition of odd, $m - n$ is odd, but by the supposition, $m - n$ is even. This
+is a contradiction.
+
+_Case ($m$ is odd and $n$ is even):_
+
+Since $m$ is odd and $n$ is even, $m = 2k + 1$ and $n = 2p$ for some integers
+$k$ and $p$. Then:
+
+$$ m - n = (2k + 1) - 2p $$
+
+$$ = 2k - 2p + 1 $$
+
+$$ = 2(k - p) + 1 $$
+
+Now, $k - p$ is an integer by the subtraction of integers. Thus, by the
+definition of odd, $m - n$ is odd, but by the supposition, $m - n$ is even. This
+is a contradiction.
+
+_Conclusion:_
+
+It can be concluded based off of all cases that when both $m$ and $n$ are even
+or both $m$ and $n$ are odd, $m - n$ is even.
+
+_Proof (second):_
+
+Suppose both $m$ and $n$ are both even or are both odd.
+
+In order to prove $m - n$ is even, both cases must be considered.
+
+_Case (both $m$ and $n$ are even):_
+
+Since both $m$ and $n$ are even, $m = 2k$ and $n = 2p$ for some integers $k$ and
+$p$. Then:
+
+$$ m - n = 2k - 2p $$
+
+$$ = 2(k - p) $$
+
+Now, $k - p$ is an integer by the subtraction of integers. Therefore, by the
+definition of even, $m - n$ is even.
+
+_Case (both $m$ and $n$ are odd):_
+
+Since both $m$ and $n$ are odd, $m = 2k + 1$ and $n = 2p + 1$ for some integers
+$k$ and $p$. Then:
+
+$$ m - n = (2k + 1) - (2p + 1) $$
+
+$$ = 2k + 1 - 2p - 1 $$
+
+$$ = 2k - 2p $$
+
+$$ = 2(k - p) $$
+
+Now, $k - p$ is an integer by the subtraction of integers. Therefore, by the
+definition of even, $m - n$ is even.
+
+_Conclusion:_
+
+In both cases, $m - n$ is even. Therefore it can be concluded that if both $m$
+and $n$ are even or if both $m$ and $n$ are odd, then $m - n$ is even.
+
 3. The **congruence modulo $3$** relation, $T$, is defined from $\mathbb{Z}$ to
    $\mathbb{Z}$ as follows: For all integers $m$ and $n$,
 
@@ -22,16 +185,53 @@ $$ m T n \Leftrightarrow 3 | (m - n) $$
 
 a. Is $10 T 1$? Is $1 T 10$? Is $(2, 2) \in T$? Is $(8, 1) \in T$?
 
+_$10 T 1$:_
+
+Yes, since $3 | (10 - 1) = 3 | 9 = 3$
+
+_$1 T 10$:_
+
+Yes, since $3 | (1 - 10) = 3 | -9 = -3$
+
+_$(2, 2) \in T$:_
+
+Yes, since $3 | (2 - 2) = 3 | 0 = 0$
+
+_$(8, 1) \in T$:_
+
+No, since $3 | (8 - 1) = 3 \cancel{|} 7$.
+
 b. List five integers $n$ such that $n T 0$.
 
+$3$; $6$, $9$, $12$, $15$
+
 c. List five integers $n$ such that $n T 1$.
 
+$4$; $7$, $10$, $13$, $16$
+
 d. List five integers $n$ such that $n T 2$.
 
+$$ 3 | (n - 2) $$
+
+$5$, $8$, $11$, $14$, $17$
+
 e. Make and prove a conjecture about which integers are related by $T$ to $0$,
 which integers are related to $T$ to $1$, and which integers are related to $T$
 to $2$.
 
+_Hint:_ All integers of the form $3k + 1$, for some integer $k$, are related by
+$T$ to $1$.
+
+**Conjecture:**
+
+All integers of the form $3k$, for some integer $k$, are related by $T$ to $0$.
+
+All integers of the form $3p + 1$, for some integer $p$, are related by $T$ to
+$1$.
+
+All integers of the form $3m + 2$, for some integer $m$, are related to $T$ by
+$2$.
+
 4. Define a relation $P$ on $\mathbb{Z}$ as follows: For every ordered pair
    $(m, n) \in \mathbb{Z} \times \mathbb{Z}$,
 
@@ -39,23 +239,58 @@ $$ m P n \Leftrightarrow m \text{ and } n \text{ have a common prime factor} $$
 
 a. Is $15 P 25$?
 
+Yes, because both $15$ and $25$ are divisible by $5$, which is a prime factor.
+
 b. Is $22 P 27$?
 
+No, because $22$ and $27$ have no common divisors.
+
 c. Is $0 P 5$?
 
+Yes, because both $0$ and $5$ are divisible by $5$, which is a prime factor.
+
 d. Is $8 P 8$?
 
-5. Let $X = \{a, b, c\}$. Recall that $\mathscr{P}(X)$ as follows: For all sets
+Yes, because both $8$ and $8$ are divisible by $2$, which is a prime factor.
+
+5. Let $X = \{a, b, c\}$. Recall that $\mathscr{P}(X)$ is the power set of $X$.
+   Define a relation $\mathbf{S}$ on $\mathscr{P}(X)$ as follows: For all sets
+   $A$ and $B$ in $\mathscr{P}(X)$,
+
+$$ A \mathbf{S}B \Leftrightarrow A \text{ has the same number of elements as } B $$
+
+a. Is $\{a, b\} \mathbf{S} \{b, c\}$?
+
+Yes, since both $\{a, b\}$ and $\{b, c}$ have the same number of elements,
+namely $2$ elements.
+
+b. Is $\{a\} \mathbf{S} \{a, b\}$?
+
+No, since $\{a\}$ has $1$ element and $\{a, b\}$ has $2$ elements, and
+$1 \neq 2$.
+
+c. Is $\{c\} \mathbf{S} \{b\}$?
+
+Yes, since both $\{c\}$ and $\{b\}$ have the same number of elements, namely $1$
+element.
+
+6. Let $X = \{a, b, c\}$. Recall that $\mathscr{P}(X)$ as follows: For all sets
    $A$ and $B$ in $\mathscr{P}(X)$,
 
 $$ A \mathbf{J} B \Leftrightarrow A \cap B \neq \emptyset $$
 
 a. Is $\{a\} \mathbf{J} \{c\}$?
 
+No, since $\{a\} \cap \{\c} = \emptyset$.
+
 b. Is $\{a, b\} \mathbf{J} \{b, c\}$?
 
+Yes, since $\{a, b\} \cap \{b, c\} = \{b\} \neq \emptyset$.
+
 c. Is $\{a, b} \mathbf{J} \{a, b, c\}$?
 
+Yes, since $\{a, b\} \cap \{a, b, c\} = \{a, b\} \neq \emptyset$.
+
 7. Define a relation $R$ on $\mathbb{Z}$ as follows: For all integers $m$ and
    $n$,
 
@@ -63,12 +298,42 @@ $$ m R n \Leftrightarrow 5 | (m^2 - n^2) $$
 
 a. Is $1 R (-9)$?
 
+$$ 5 | ((1)^2 - (-9)^2) $$
+
+$$ 5 | (1 - 81) $$
+
+$$ 5 | (-80) = -16 $$
+
+Yes.
+
 b. Is $2 R 13$?
 
+$$ 5 | ((2)^2 - (13)^2) $$
+
+$$ 5 | (4 - 169) $$
+
+$$ 5 | (-165) = -33 $$
+
+Yes.
+
 c. Is $2 R (-8)$?
 
+$$ 5 | ((2)^2 - (-8)^2) $$
+
+$$ 5 | (4 - (64)) $$
+
+$$ 5 | (-60) = -12  $$
+
+Yes.
+
 d. Is $(-8) R 2$?
 
+$$ 5 | (64 - 4) $$
+
+$$ 5 | 60 = 12 $$
+
+Yes.
+
 8. Let $A$ be the set of all strings of _a_'s and _b_'s of length $4$. Define a
    relation $R$ on $A$ as follows: For every $s, t \in A$,
 
@@ -76,12 +341,23 @@ $$ s R t \Leftrightarrow s \text{ has the same first two characters as } t $$
 
 a. Is _abaa_ $R$ _abba_?
 
+Yes, since _ab_ is the same first two characters of both _abaa_ and _abba_.
+
 b. Is _aabb_ $R$ _bbaa_?
 
+No, since _aa_ is the first two characters of _aabb_ and _bb_ is the first same
+two characters as _bbaa_, it can be concluded that _aabb_ and _bbaa_ do not have
+the same first two characters.
+
 c. Is _aaaa_ $R$ _aaab_?
 
+Yes, since _aa_ is the same first two characters of both _aaaa_ and _aaab_.
+
 d. Is _baaa_ $R$ _abaa_?
 
+No, since _ba_ and _ab_ are the first two characters of _baaa_ and _abaa_
+respectively.
+
 9. Let $A$ be the set of all strings of 0's, 1's, and 2's of length $4$. Define
    a relation $R$ on $A$ as follows: For every $s, t \in A$,
 
@@ -89,12 +365,28 @@ $$ s R t \Leftrightarrow \text{ the same of the characters in } s \text{ equals
 
 a. Is 0121 $R$ 2200?
 
+$$ 0 + 1 + 2 + 1 = 4 =  2 + 2 + 0 + 0 $$
+
+Yes.
+
 b. Is 1011 $R$ 2101?
 
+$$ 1 + 0 + 1 + 1 = 3 = \neq 4 = 2 + 1 + 0 + 1 $$
+
+No.
+
 c. Is 2212 $R$ 2121?
 
+$$ 2 + 2 + 1 + 2 = 7 \neq 6 = 2 + 1 + 2 + 1 $$
+
+No.
+
 d. Is 1220 $R$ 2111?
 
+$$ 1 + 2 + 2 + 0 = 5 = 2 + 1 + 1 + 1 $$
+
+Yes.
+
 10. Let $A = \{3, 4, 5\}$ and $B = \{4, 5, 6\}$ and let $R$ be the "less than"
     relation. That is, for every ordered pair $(x, y) \in A \times B$,
 
@@ -102,6 +394,10 @@ $$ x R y \Leftrightarrow x < y $$
 
 State explicitly which ordered pairs are in $R$ and $R^{-1}$.
 
+$$ R = \{(3, 4), (3, 5), (3, 6), (4, 5), (4, 6), (5, 6) \} $$
+
+$$ R^{-1} = \{(4, 3), (5, 3), (6, 3), (5, 4), (6, 4), (6, 5) \} $$
+
 11. Let $A = \{3, 4, 5\}$ and $B = \{4, 5, 6\}$ and let $S$ be the "divides"
     relation. That is, for every ordered pair $(x, y) \in A \times B$,
 
@@ -109,42 +405,72 @@ $$ x S y \Leftrightarrow x | y $$
 
 State explicitly which ordered pairs are in $S$ and $S^{-1}$.
 
+$$ S = \{(3, 6), (4, 4), (5, 5)\} $$
+
+$$ S^{-1} = \{(6, 3), (4, 4), (5, 5)\} $$
+
 12.
 
 a. Suppose a function $F: X \to Y$ is one-to-one but not onto. Is $F^{-1}$ (the
 inverse relation for $F$) a function? Explain your answer.
 
+No, if $F: X \to Y$ is one-to-one, but not onto, then its inverse relation
+$F^{-1}: Y \to X$ will have some elements in its domain that have not elements
+in the co-domain. More formally:
+
+$$ \exists y \in Y | (y, x) \notin F^{-1} $$
+
+which means $F^{-1}$ does not satisfy property 1 for being a function.
+
 b. Suppose a function $F: X \to Y$ is onto but not one-to-one. Is $F^{-1}$ (the
 inverse relation for $F$) a function? Explain your answer.
 
+No, if $F: X \to Y$ is onto, but not one-to-one, it follows that its inverse
+relation $F^{-1}: Y \to X$ will have at least one
+$y \in Y | (y, x_1) \in F^{-1} \wedge (y, x_2) \in F^{-1}$.
+
+This violates property 2 of the definition of a function.
+
 Draw the directed graphs of the relations defined in 13-18.
 
 13. Define a relation $R$ on $A = \{0, 1, 2, 3\}$ by
     $R = \{(0, 0), (1, 2), (2, 2)\}$.
 
+(Done by hand.)
+
 14. Define a relation $S$ on $B = \{a, b, c, d\}$ by
     $S = \{(a, b), (a, c), (b, c), (d, d)\}$.
 
+(Done by hand.)
+
 15. Let $A = \{2, 3, 4, 5, 6, 7, 8\}$ and define a relation $R$ on $A$ as
     follows: For every $x, y \in A$,
 
 $$ x R y \Leftrightarrow x | y $$
 
+(Done by hand.)
+
 16. Let $A = \{5, 6, 7, 8, 9, 10\}$ and define a relation $S$ on $A$ as follows:
     For every $x, y \in A$,
 
 $$ x S y \Leftrightarrow 2 | (x - y) $$
 
+(Done by hand.)
+
 17. Let $A = \{2, 3, 4, 5, 6, 7, 8\}$ and define a relation $T$ on $A$ as
     follows: For every $x, y \in A$,
 
 $$ x T y \Leftrightarrow 3 | (x - y) $$
 
+(Done by hand.)
+
 18. Let $A = \{0, 1, 3, 4, 5, 6\}$ and define a relation $V$ on $A$ as follows:
     For every $x, y \in A$,
 
 $$ x V y \Leftrightarrow 5 | (x^2 - y^2) $$
 
+(Done by hand.)
+
 Exercises 19-20 refer to unions and intersections of relations. Since relations
 are subsets of Cartesian products, their unions and intersections can be
 calculated as for any subsets. Given two relations $R$ and $S$ from $A$ to $B$,
@@ -161,6 +487,16 @@ $$ x R y \Leftrightarrow x | y \text{ and } x S y \Leftrightarrow y - 4 = x $$
 State explicitly which ordered pairs are in $A \times B$, $R$, $S$, $R \cup S$,
 and $R \cap S$.
 
+$$ A \times B = \{(2, 6), (2, 8), (2, 10), (4, 6), (4, 8), (4, 10)\} $$
+
+$$ R = \{(2, 6), (2, 8), (2, 10), (4, 8)\} $$
+
+$$ S = \{(2, 6), (4, 8)\} $$
+
+$$ R \cup S = \{(2, 6), (2, 8), (2, 10), (4, 8)\} = R $$
+
+$$ R \cap S = \{(2, 6), (4, 8)\} = S $$
+
 20. Let $A = \{-1, 1, 2, 4\}$ and $B = \{1, 2\}$ and define relations $R$ and
     $S$ from $A$ to $B$ as follows: For every $(x, y) \in A \times B$,
 
@@ -169,6 +505,16 @@ $$ x R y \Leftrightarrow |x| = |y| \text{ and } x S y \Leftrightarrow x - y \tex
 State explicitly which ordered pairs are in $A \times B$, $R$, $S$, $R \cup S$,
 and $R \cap S$.
 
+$$ A \times B = \{(-1, 1), (-1, 2), (1, 1), (1, 2), (2, 1), (2, 2), (4, 1), (4, 2)\} $$
+
+$$ R = \{(-1, 1), (1, 1), (2, 2)\} $$
+
+$$ S = \{(-1, 1), (1, 1), (2, 2), (4, 2)\} $$
+
+$$ R \cup S = \{(-1, 1), (1, 1), (2, 2), (4, 2)\} = S $$
+
+$$ R \cap S = \{(-1, 1), (1, 1), (2, 2)\} = R $$
+
 21. Define relations $R$ and $S$ on $\mathbb{R}$ as follows:
 
 $$ R = \{(x, y) \in \mathbb{R} \times \mathbb{R} | x < y\} \text{ and } S = \{(x, y) \in \mathbb{R} \times \mathbb{R} | x = y\}$$
@@ -176,6 +522,8 @@ $$ R = \{(x, y) \in \mathbb{R} \times \mathbb{R} | x < y\} \text{ and } S = \{(x
 That is, $R$ is the "less than" relation and $S$ is the "equals" relation on
 $\mathbb{R}$. Graph $R$, $S$, $R \cup S$, and $R \cap S$ in the Cartesian plane.
 
+Think on this and then see appendix b (Page 975).
+
 22. Define relations $R$ and $S$ on $\mathbb{R}$ as follows:
 
 $$ R = \{(x, y) \in \mathbb{R} \times \mathbb{R} | x^2 + y^2 = 4\} \text{ and } S = \{(x, y) \in \mathbb{R} \times \mathbb{R} | x = y\} $$
@@ -188,6 +536,16 @@ $$ R = \{(x, y) \in \mathbb{R} \times \mathbb{R} | y = |x|\} \text{ and } S = \{
 
 Graph $R$, $S$, $R \cup S$, and $R \cap S$ in the Cartesian plane.
 
+$R$ is a circle about the origin (with intersections along the axis along
+$(-2, 0), (0, 2), (2, 0), (-2, 0)$). $S$ is a straight diagonal line ascending
+from the left to the right, intersecting the origin $(0, 0)$.
+
+$R \cup S$ is just the two graphs drawn together.
+
+$R \cap S$ is only the two points along which the two graphs intersect.
+
+(Done by hand.)
+
 24. In Example 8.1.7 consider the query SELECT Patient_ID#, Name FROM S WHERE
     Primary_Diagnosis = X. The response query is the projection onto the first
     two coordinates of the intersection of the database with the set
@@ -196,5 +554,13 @@ Graph $R$, $S$, $R \cup S$, and $R \cap S$ in the Cartesian plane.
 a. Find the result of the query SELECT Patient_ID#, Name FROM S WHERE
 Primary_Diagnosis = pneumonia.
 
+(574329, Tak Kurosawa),
+
+(011985, John Schmidt)
+
 b. Find the result of the query SELECT Patient_ID#, Name FROM S WHERE
 Primary_Diagnosis = appendicitis.
+
+(466581, Mary Lazars),
+
+(778400, Jamal Baskers)

chapter_8/test_yourself.md

@@ -5,13 +5,23 @@ Page 515
 1. If $R$ is a relation from $A$ to $B$, $x \in A$, and $y \in B$, the notation
    $x R y$ means that ____.
 
+$x$ is related to $y$ by $R$
+
 2. If $R$ is a relation from $A$ to $B$, $x \in A$, and $y \in B$, the notation
    $x \cancel{R} y$ means that ____.
 
+$x$ is not related to $y$ by $R$.
+
 3. If $R$ is a relation from $A$ to $B$, $x \in A$, and $y \in B$, the notation
    $(y, x) \in R^{-1}$ if, and only if, ____.
 
+$$ (x, y) \in R $$
+
 4. A relation on a set $A$ is a relation from ____ to ____.
 
+$A$; $A$
+
 5. If $R$ is a relation on a set $A$, the directed graph of $R$ has an arrow
    from $x$ to $y$ if, and only if, ____.
+
+$x$ is related to $y$ by $R$