tomit4/discrete_mathematics_with_applications
1
0
Fork 0
Code Issues Pull requests Projects Releases Packages Wiki Activity Actions

🚧 Fin 8.1

Browse source
This commit is contained in:
tomit4 2026-08-15 17:12:17 -07:00
parent 91773f401c
commit 5a31dcd06c
2 changed files with 377 additions and 1 deletions
Download patch file Download diff file Expand all files Collapse all files

368
chapter_8/exercises.md
View file

@ -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)

10
chapter_8/test_yourself.md
View file

@ -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 $R$