23 KiB
Page 516
Exercise Set 8.1
- As in Example 8.1.2, the congruence modulo $2$ relation
Eis defined from\mathbb{Z}to\mathbb{Z}as follows: For every ordered pair(m, n) \in \mathbb{Z} \times \mathbb{Z},
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.
- Prove that for all integers
mandn,m - nis even if, and only if, bothmandnare even or bothmandnare 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.
- The congruence modulo $3$ relation,
T, is defined from\mathbb{Z}to\mathbb{Z}as follows: For all integersmandn,
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.
- Define a relation
Pon\mathbb{Z}as follows: For every ordered pair(m, n) \in \mathbb{Z} \times \mathbb{Z},
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?
Yes, because both 8 and 8 are divisible by 2, which is a prime factor.
- Let
X = \{a, b, c\}. Recall that\mathscr{P}(X)is the power set ofX. Define a relation\mathbf{S}on\mathscr{P}(X)as follows: For all setsAandBin\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.
- Let
X = \{a, b, c\}. Recall that\mathscr{P}(X)as follows: For all setsAandBin\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.
- Define a relation
Ron\mathbb{Z}as follows: For all integersmandn,
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.
- Let
Abe the set of all strings of a's and b's of length4. Define a relationRonAas follows: For everys, t \in A,
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.
- Let
Abe the set of all strings of 0's, 1's, and 2's of length4. Define a relationRonAas follows: For everys, t \in A,
s R t \Leftrightarrow \text{ the same of the characters in } s \text{ equals the sum of the characters in } t
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.
- Let
A = \{3, 4, 5\}andB = \{4, 5, 6\}and letRbe the "less than" relation. That is, for every ordered pair(x, y) \in A \times B,
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) \}
- Let
A = \{3, 4, 5\}andB = \{4, 5, 6\}and letSbe the "divides" relation. That is, for every ordered pair(x, y) \in A \times B,
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)\}
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.
- Define a relation
RonA = \{0, 1, 2, 3\}byR = \{(0, 0), (1, 2), (2, 2)\}.
(Done by hand.)
- Define a relation
SonB = \{a, b, c, d\}byS = \{(a, b), (a, c), (b, c), (d, d)\}.
(Done by hand.)
- Let
A = \{2, 3, 4, 5, 6, 7, 8\}and define a relationRonAas follows: For everyx, y \in A,
x R y \Leftrightarrow x | y
(Done by hand.)
- Let
A = \{5, 6, 7, 8, 9, 10\}and define a relationSonAas follows: For everyx, y \in A,
x S y \Leftrightarrow 2 | (x - y)
(Done by hand.)
- Let
A = \{2, 3, 4, 5, 6, 7, 8\}and define a relationTonAas follows: For everyx, y \in A,
x T y \Leftrightarrow 3 | (x - y)
(Done by hand.)
- Let
A = \{0, 1, 3, 4, 5, 6\}and define a relationVonAas follows: For everyx, 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,
R \cup S = \{(x, y) \in A \times B | (x, y) \in R \text{ or } (x, y) \in S\}
R \cap S = \{(x, y) \in A \times B | (x, y) \in R \text{ and } (x, y) \in S\}
- Let
A = \{2, 4\}andB = \{6, 8, 10\}and define relationsRandSfromAtoBas follows: For every(x, y) \in A \times B,
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
- Let
A = \{-1, 1, 2, 4\}andB = \{1, 2\}and define relationsRandSfromAtoBas follows: For every(x, y) \in A \times B,
x R y \Leftrightarrow |x| = |y| \text{ and } x S y \Leftrightarrow x - y \text{ is even}
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
- Define relations
RandSon\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\}
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).
- Define relations
RandSon\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\}
Graph R, S, R \cup S, and R \cap S in the Cartesian plane.
- Define relations
RandSon\mathbb{R}as follows:
R = \{(x, y) \in \mathbb{R} \times \mathbb{R} | y = |x|\} \text{ and } S = \{(x, y) \in \mathbb{R} \times \mathbb{R} | y = 1\}
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.)
- 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
A_1 \times A_2 \times A_3 \times \{X\}.
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)
Page 526
Exercise Set 8.2
In 1-8, a number of relations are defined on the set A = \{0, 1, 2, 3\}. For
each relation:
a. Draw the directed graph.
b. Determine whether the relation is reflexive.
c. Determine whether the relation is symmetric.
d. Determine whether the relation is transitive.
Give a counterexample in each case in which the relation does not satisfy one of the properties.
-
R_1 = \{(0, 0), (0, 1), (0, 3), (1, 1), (1, 0), (2, 3), (3, 3)\} -
R_2= {(0, 0), (0, 1), (1, 1), (1, 2), (2, 2), (2, 3)} -
R_3 = \{(2, 3), (3, 2)\} -
R_4 = \{(1, 2), (2, 1), (1, 3), (3, 1)\} -
R_5 = \{(0, 0), (0, 1), (0, 2), (1, 2)\} -
R_6 = \{(0, 1), (0, 2)\} -
R_7 = \{(0, 3), (2, 3)\} -
R_8 = \{(0, 0), (1, 1)\}
In 9-33, determine whether the given relation is reflexive, symmetric, transitive, or none of these. Justify your answers.
-
Ris the "greater than or equal to" relation on the set of real numbers: For everyx, y \in \mathbb{R},x R y \Leftrightarrow x \geq y. -
Cis the circle relation on the set of real numbers: For everyx, y \in \mathbb{R}, x C y \Leftrightarrow x^2 + y^2 = 1. -
Dis the relation defined on\mathbb{R}as follows: For everyx, y \in \mathbb{R}, x D y \Leftrightarrow xy \geq 0. -
Eis the congruence modulo4relation on\mathbb{Z}: For everym, n \in \mathbb{Z}, m E n \Leftrightarrow 4 | (m - n). -
Fis the congruence modulo5relation on\mathbb{Z}: For everym, n \in \mathbb{Z}, m F n \Leftrightarrow 5 | (m - n). -
Ois the relation defined on\mathbb{Z}as follows: For everym, n \in \mathbb{Z}, m O n \Leftrightarrow m - n \text{ is odd}. -
Dis the "divides" relation on\mathbb{Z}^+: For all positive integersmandn,m D n \Leftrightarrow m | n. -
Ais the "absolute value" relation on\mathbb{R}: For all real numbersxandy,x A y \Leftrightarrow |x| = |y|. -
Recall that a prime number is an integer that is greater than
1and has no positive integer divisors other than1and itself. (In particular,1is not prime.) A relationPis defined on\mathbb{Z}as follows: For everym, n \in \mathbb{Z}, m P n \Leftrightarrow \exists \text{ a prime number } p \text{ such that } p | m \text{ and } p | n. -
Define a relation
Qon\mathbb{R}as follows: For all real numbersxandy,x Q y \Leftrightarrow x - yis rational. -
Define a relation
Ion\mathbb{R}as follows: For all real numbersxandy,x I y \Leftrightarrow x - yis irrational. -
Let
X = \{a, b, c\}and\mathscr{P}(X)be the power set ofX(the set of all subsets ofX). A relation\mathbf{E}is defined on\mathscr{P}(X)as follows: For everyA, B \in \mathscr{P}(X), A \mathbf{E} B \Leftrightarrow \text{ the number of elements in } A \text{ equals the number of elements in } B. -
Let
X = \{a, b, c\}and\mathscr{P}(X)be the power set ofX. A relation\mathbf{L}is defined on\mathscr{P}(X)as follows: For everyA, B \in \mathscr{P}(X), A \mathbf{L} B \Leftrightarrow \text{ the number of elements in } A \text{ is less than the number of elements in } B. -
Let
X = \{a, b, c\}and\mathscr{P}(X)be the power set ofX. A relation\mathbf{N}is defined on\mathscr{P}(X)as follows: For everyA, B \in \mathscr{P}(X), A \mathbf{N} B \Leftrightarrow \text{ the number of elements in } A \text{ is not equal to the number of elements in } B. -
Let
Xbe a nonempty set and\mathscr{P}(X)the power set ofX. Define the "subset" relation\mathbf{S}on\mathscr{P}(X)as follows: For everyA, B \in \mathscr{P}(X), A \mathbf{S} B \Leftrightarrow A \subseteq B. -
Let
Xbe a nonempty set and\mathscr{P}(X)the power set ofX. Define the "not equal to" relation\mathbf{U}on\mathscr{P}(X)as follows: For everyA, B \in \mathscr{P}(X), A \mathbf{U} B \Leftrightarrow A \neq B. -
Let
Abe the set of all strings of a's and b's of length4. Define a relationRonAas follows: For everys, t \in A, s R t \Leftrightarrow s \text{ has the same first two characters as } t. -
Let
Abe the set of all strings of 0's, 1's, and 2's that have length 4 and for which the sum of the characters in the string is less than or equal to 2. Define a relationRonAas follows: For everys, t \in A, s R t \Leftrightarrow \text{ the sum of the characters of } s \text{ equals the sum of the characters of } t. -
Let
Abe the set of all English statements. A relation\mathbf{I}is defined onAas follows: For everyp, q \in A,
p \mathbf{I} q \Leftrightarrow p \to q \text{ is true}
- Let
A = \mathbb{R} \times \mathbb{R}. A relation\mathbf{S}is defined onAas follows: For every(x_1, y_1)and(x_2, y_2)inA,
(x_1, y_2) \mathbf{S} (x_2, y_2) \Leftrightarrow x_1 = x_2
- Let
A = \mathbb{R} \times \mathbb{R}. A relation\mathbf{S}is defined onAas follows: For every(x_1, y_1)and(x_2, y_2)inA,
(x_1, y_2) \mathbf{S} (x_2, y_2) \Leftrightarrow y_1 = y_2
-
Let
Abe the "punctured plane"; that is,Ais the set of all points in the Cartesian plane except the origin(0, 0). A relationRis defined onAas follows: For everyp_1andp_2inA,p_1 R p_2 \Leftrightarrow p_1 \text{ and } p_2 \text{ lie on the same half line emanating from the origin}. -
Let
Abe the set of people living in the world today. A relationRis defined onAas follows: For all peoplepandqinA,
p R q \Leftrightarrow p \text{ lives within 100 miles of } q
-
Let
Abe the set of all lines in the plane. A relationRis defined onAas follows: For everyl_1andl_2inA,l_1 R l_2 \Leftrightarrow l_1 \text{ is parallel to } l_2. (Assume that a line is parallel to itself.) -
Let
Abe the set of all lines in the plane. A relationRis defined onAas follows: For everyl_1andl_2inA,
l_1 R l_2 \Leftrightarrow l_1 \text{ is perpendicular to } l_2
In 34-36, assume that R is a relation on a set A. Prove or disprove each
statement.
-
If
Ris reflexive, thenR^{-1}is reflexive. -
If
Ris symmetric, thenR^{-1}is symmetric. -
If
Ris transitive, thenR^{-1}is transitive.
In 37-42, assume that R and S are relations on a set A. Prove or disprove
each statement.
-
If
RandSare reflexive, isR \cap Sreflexive? Why? -
If
RandSare symmetric, isR \cap Ssymmetric? Why? -
If
RandSare transitive, isR \cap Stransitive? Why? -
If
RandSare reflexive, isR \cup Sreflexive? Why? -
If
RandSare symmetric, isR \cup Ssymmetric? Why? -
If
RandSare transitive, isR \cup Stransitive? Why?
In 43-50, the following definitions are used: A relation on a set A is defined
to be
irreflexive if, and only if, for every x \in A, x \cancel{R} x;
asymmetric if, and only if, for every x, y \in A if x R y then
y \cancel{R} x;
intransitive if, and only if, for every x, y, z \in A, if x R y and y R z
then x \cancel{R} z.
For each of the relations in the referenced exercise, determine whether the relation is irreflexive, asymmetric, intransitive, or none of these.
-
Exercise 1
-
Exercise 2
-
Exercise 3
-
Exercise 4
-
Exercise 5
-
Exercise 6
-
Exercise 7
-
Exercise 8
In 51-53, R, S, and T are relations defined on A = \{0, 1, 2, 3\}.
- Let
R = \{(0, 1), (0, 2), (1, 1), (1, 3), (2, 2), (3, 0)\}.
Find R^t, the transitive closure of R.
- Let
S = \{(0, 0), (0, 3), (1, 0), (1, 2), (2, 0), (3, 2)\}.
Find S^t, the transitive closure of S.
- Let
T = \{(0, 2), (1, 0), (2, 3), (3, 1)\}.
Find T^t, the transitive closure of T.
- Write a computer algorithm to test whether a relation
Rdefined on a finite setAis reflexive, where
A = \{a[1], a[2], \dots, a[n]\}
- Write a computer algorithm to test whether a relation
Rdefined on a finite setAis symmetric, where
A = \{a[1], a[2], \dots, a[n]\}
- Write a computer algorithm to test whether a relation
Rdefined on a finite setAis transitive, where
A = \{a[1], a[2], \dots, a[n]\}