discrete_mathematics_with_applications/chapter_8/exercises.md

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Exercise Set 8.1

1. As in Example 8.1.2, the congruence modulo $2$ relation E is 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.

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,

 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},

 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.

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,

 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,

 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,

 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.

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,

 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,

 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.

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,

 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\} 

19. Let A = \{2, 4\} and B = \{6, 8, 10\} and define relations R and S from A to B as 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 

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,

 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 

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\}

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\} 

Graph R, S, R \cup S, and R \cap S in the Cartesian plane.

23. Define relations R and S on \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.)

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 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)

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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.

1. R_1 = \{(0, 0), (0, 1), (0, 3), (1, 1), (1, 0), (2, 3), (3, 3)\}

a. Draw the directed graph.

(Done by hand.)

b. Determine whether the relation is reflexive.

No, 2 \cancel{R_1} 2.

c. Determine whether the relation is symmetric.

No, 0 R_1 3, but 3 \cancel{R_1} 0.

d. Determine whether the relation is transitive.

No, 1 R_1 0 and 0 R_1 3, but 1 \cancel{R_1} 3

2. R_2 = {(0, 0), (0, 1), (1, 1), (1, 2), (2, 2), (2, 3)}

a. Draw the directed graph.

(Done by hand.)

b. Determine whether the relation is reflexive.

No, since 3 \cancel{R_2} 3.

c. Determine whether the relation is symmetric.

No, 0 R_2 1, but 1 \cancel{R_2} 0.

d. Determine whether the relation is transitive.

No, 0 R_2 1 and 1 R_2 2, but 0 \cancel{R_2} 2.

3. R_3 = \{(2, 3), (3, 2)\}

a. Draw the directed graph.

(Done by hand.)

b. Determine whether the relation is reflexive.

No, 2 \cancel{R_3} 2.

c. Determine whether the relation is symmetric.

Yes, 2 R_3 3 and 3 R_3 2.

d. Determine whether the relation is transitive.

No, 2 R_3 3 and 3 R_3 2, but 2 \cancel{R_3} 2.

4. R_4 = \{(1, 2), (2, 1), (1, 3), (3, 1)\}

a. Draw the directed graph.

(Done by hand.)

b. Determine whether the relation is reflexive.

No, 1 \cancel{R_4} 1.

c. Determine whether the relation is symmetric.

Yes, 1 R_4 2 and 2 R_4 1 and 1 R_4 3 and 3 R_4 1.

d. Determine whether the relation is transitive.

No, 1 R_4 2 and 2 R_4 1, but 1 \cancel{R_4} 1.

5. R_5 = \{(0, 0), (0, 1), (0, 2), (1, 2)\}

a. Draw the directed graph.

(Done by hand.)

b. Determine whether the relation is reflexive.

No, 1 \cancel{R_5} 1.

c. Determine whether the relation is symmetric.

No, 0 R_5 1, but 1 \cancel{R_5} 0.

d. Determine whether the relation is transitive.

Yes, 0 R_5 1 and 1 R_5 2, and 0 R_5 2.

6. R_6 = \{(0, 1), (0, 2)\}

a. Draw the directed graph.

(Done by hand.)

b. Determine whether the relation is reflexive.

No, 0 \cancel{R_6} 0.

c. Determine whether the relation is symmetric.

No, 0 R_6 1, but 1 \cancel{R_6} 0.

d. Determine whether the relation is transitive.

Yes, vacuously.

7. R_7 = \{(0, 3), (2, 3)\}

a. Draw the directed graph.

(Done by hand.)

b. Determine whether the relation is reflexive.

No, 0 \cancel{R_7} 0.

c. Determine whether the relation is symmetric.

No, 0 R_7 3, but 3 \cancel{R_7} 0.

d. Determine whether the relation is transitive.

Yes, vacuously.

8. R_8 = \{(0, 0), (1, 1)\}

a. Draw the directed graph.

(Done by hand.)

b. Determine whether the relation is reflexive.

Yes, both 0 R_8 0 and 1 R_8 1.

c. Determine whether the relation is symmetric.

Yes, since 0 R_8 0 and 0 R_8 0, and also 1 R_8 1 and 1 R_8 1.

d. Determine whether the relation is transitive.

Yes, vacuously.

In 9-33, determine whether the given relation is reflexive, symmetric, transitive, or none of these. Justify your answers.

9. R is the "greater than or equal to" relation on the set of real numbers: For every x, y \in \mathbb{R}, x R y \Leftrightarrow x \geq y.

a. Is R reflexive?

Yes, since \forall x \in \mathbb{R}, x = x, it follows that \forall x \in \mathbb{R}, x \geq x.

b. Is R symmetric?

No, since \forall x, y \in \mathbb{R}, x \geq y \to y \geq x cannot be true. Consider the example that x = 5 and y = 4, then x \geq y, but y \cancel{\geq} x.

c. Is R transitive?

Yes, since \forall x, y, z \in \mathbb{R}, (x \geq y \wedge y \geq z) \to x \geq z is true by the transitive law of greatness (See appendix A, T18).

10. C is the circle relation on the set of real numbers: For every x, y \in \mathbb{R}, x C y \Leftrightarrow x^2 + y^2 = 1.

a. Is C reflexive?

No, C is not reflexive. The statement claims that \forall x \in \mathbb{R}, x C x \Leftrightarrow x^2 + x^2 = 1, but consider x = 0, then 0^2 + 0^2 = 1, but 0 \neq 1, this is a contradiction.

b. Is C symmetric?

Yes, C is symmetric. The statement claims that x, y \in \mathbb{R}, (x^2 + y^2 = 1) \to (y^2 + x^2 = 1). This is true by the commutative laws of addition.

c. Is C transitive?

No, C is not transitive. The statement claims that x, y, z \in \mathbb{R}, [(x^2 + y^2 = 1) \wedge (y^2 + z^2 = 1)] \to x^2 + z^2 = 1. Consider x = 1, y = 0, and z = 1, then x^2 + y^2 = (1)^2 + (0)^2 = 1 and y^2 + z^2 = (0)^2 + (1)^2 = 1, but x^2 + z^2 = (1)^2 + (1)^2 = 2 \neq 1.

11. D is the relation defined on \mathbb{R} as follows: For every x, y \in \mathbb{R}, x D y \Leftrightarrow xy \geq 0.

a. Is D reflexive?

Yes, D is reflexive. \forall x \in \mathbb{R} x \cdot x \geq 0 is a true statement, as even if x is negative, any negative number times itself will always be positive, and so x \geq 0 is true. If x = 0, then x \geq 0 is a true statement. If x is positive, then any positive number times itself will be positive, and so x \geq 0 is true.

b. Is D symmetric?

Yes, D is symmetric, \forall x, y \in \mathbb{R}, (xy \geq 0) \to (yx \geq 0) is true by the commutative laws of multiplication since xy = yx.

c. Is D transitive?

No, D is not transitive. The statement claims \forall x, y, z \in \mathbb{R}, [(xy \geq 0) \wedge (yz \geq 0)] \to (xz \geq 0). This is not true, consider x = 1, y = 0, and z = -1, then xy = (1)(0) = 0 \geq 0, and yz = (0)(-1) = 0 \geq 0, but xz = (1)(-1) = -1 \cancel{\geq} 0.

12. E is the congruence modulo 4 relation on \mathbb{Z}: For every m, n \in \mathbb{Z}, m E n \Leftrightarrow 4 | (m - n).

a. Is E reflexive?

Yes, E is reflexive. The statement claims \forall m \in \mathbb{Z}, 4 | (m - m). Since any integer subtracted from itself is 0, this means that:

 4 | (m - m) = 4 | 0 

Which is true since 4 = 4 \cdot 0.

b. Is E symmetric?

Yes, E is symmetric. The statement claims \forall m, n \in \mathbb{Z}, [4 | (m - n)] \to [4 | (n - m)].

Since 4 | (m - n), this means that m - n = 4k for some integer k. It follows then that:

 n - m = -1(m - n) 

 = -1(4k) 

 = 4(-k) 

Now, -k is an integer by the multiplication of integers. It follows then that 4 | (n - m). This is what was to be shown.

c. Is E transitive?

Yes, E is transitive. The statement claims that \forall m, n, p \in \mathbb{Z}, [(4 | (m - n)) \wedge (4 | (n - p))] \to (4 | (m - p)).

Since 4 | (m - n) and 4 | (n - p), it can be said that m - n = 4r and n - p = 4s for some integers r and s. It follows by addition of these two terms, and substitution, that:

 (m - n) + (n - p) = 4r + 4s 

and also that:

 (m - n) + (n - p) = m - p 

Then, setting the substitution equal to the evaluation/simplification:

 4r + 4s = m - p 

Then, by algebra:

 4(r + s) = m - p 

Now, r + s is an integer by the sum of integers. It follows that 4 | (m - p). This is what was to be shown.

13. F is the congruence modulo 5 relation on \mathbb{Z}: For every m, n \in \mathbb{Z}, m F n \Leftrightarrow 5 | (m - n).

a. Is F reflexive?

Yes, F is reflexive. The statement claims that \forall m \in \mathbb{Z}, 5 | (m - m). This is true since m - m = 0, and 5 | 0 is true since 5 = 5 \cdot 0.

b. Is F symmetric?

Yes, F is symmetric. The statement claims that \forall m, n \in \mathbb{Z}, (5 | (m - n)) \to (5 | (n - m)).

Since 5 | m - n, it can be said that m - n = 5k for some integer k. Then, consider:

 m - n = -1(n - m) 

By substitution then:

 5k = -1(5k) 

 5k = 5(-k) 

Now, -k is an integer by the multiplication of integers. It follows that 5 | (n - m). This is what was to be shown.

c. Is F transitive?

Yes, F is transitive. The statement claims that \forall m, n, p \in \mathbb{Z}, [(5 | (m - n)) \wedge (5 | (n - p))] \to [5 | (m - p)].

Since 5 | (m - n) and 5 | (n - p), it can be said that m - n = 5r and n - p = 5s for some integers r and s. Adding m - n and n - p gives m - p:

 (m - n) + (n - p) = m - p 

Then, by substitution:

 5r + 5s = m - p 

Then, by algebra:

 5(r + s) = m - p 

Now, r + s is an integer by the sum of integers. It follows that 5 | (m - p). This is what was to be shown.

14. O is the relation defined on \mathbb{Z} as follows: For every m, n \in \mathbb{Z}, m O n \Leftrightarrow m - n \text{ is odd}.

a. Is O reflexive?

No, O is not reflexive. The statement claims that \forall m \in \mathbb{Z}, m - m \text{ is odd}. Since m - m = 0, and 0 is even (since 0 = 2(0)), by the definition of even, m - m cannot be odd. Therefore O is not reflexive.

b. Is O symmetric?

Yes, O is symmetric. The statement claims that \forall m, n \in \mathbb{Z}, (m - n \text{ is odd}) \to (n - m \text{ is odd}).

Since m - n is odd, it can be said that m - n = 2k + 1 for some integer k. Consider that:

 m - n = -1(n - m) 

Then, by substitution:

 2k + 1 = -1(n - m) 

By algebra:

 -1(2k + 1) = n - m 

 -2k - 1 = n - m 

 2(-k - 1) + 1 = n - m 

Now, -k - 1 is an integer by the multiplication and sum of integers. Therefore n - m is odd. This is what was to be shown.

c. Is O transitive?

No, O is not transitive. The statement claims that \forall m, n, p \in \mathbb{Z} [(m - n \text{ is odd}) \wedge (n - p \text{ is odd})] \to [m - p \text{ is odd}]. This is not true for all integers. Consider m = 2, n = 1, and p = 0. Then m - n = 2 - 1 = 1 \text{ is odd}, and n - p = 1 - 0 = 1 \text{ is odd}, but m - p = 2 - 0 = 2 \text{ is even}. Therefore 0 is not transitive.

15. D is the "divides" relation on \mathbb{Z}^+: For all positive integers m and n, m D n \Leftrightarrow m | n.

a. Is D reflexive?

Yes, D is reflexive. The statement claims \forall m \in \mathbb{Z}^+, m | m. This is true since any integer divides itself by the definition of divisibility.

b. Is D symmetric?

No, D is not symmetric. The statement claims \forall m, n \in \mathbb{Z}^+, (m | n) \to (n | m), but this is not true for all positive integers. Consider m = 2 and n = 4, then 2 | 4 is true since 2 = 2 \cdot 2 = 4, but 4 \cancel{|} 2 since 4 \neq 4k = 2 for some integer k.

c. Is D transitive?

Yes, D is transitive. The statement claims \forall m, n, p \in \mathbb{Z}^+, [(m | n) \wedge (n | p)] \to [m | p]. This is true by the transitivity of divisibility (see Theorem 4.4.3).

16. A is the "absolute value" relation on \mathbb{R}: For all real numbers x and y, x A y \Leftrightarrow |x| = |y|.

a. Is A reflexive?

Yes, A is reflexive. The statement claims \forall x \in \mathbb{R}, |x| = |x|. This is trivially true.

b. Is A symmetric?

Yes, A is symmetric. The statement claims that \forall x, y \in \mathbb{R}, (|x| = |y|) \to (|y| = |x|). This is true by the definition of equality.

c. Is A transitive?

Yes, A is transitive. The statement claims that \forall x, y, z \in \mathbb{R}, [(|x| = |y|) \wedge (|y| = |z|)] \to |x| = |z|

This is true by the transitivity of equality (since |x| = |y| = |z|).

17. Recall that a prime number is an integer that is greater than 1 and has no positive integer divisors other than 1 and itself. (In particular, 1 is not prime.) A relation P is defined on \mathbb{Z} as follows: For every m, n \in \mathbb{Z}, m P n \Leftrightarrow \exists \text{ a prime number } p \text{ such that } p | m \text{ and } p | n.

a. Is P reflexive?

No, P is not reflexive. The statement claims \forall m \in \mathbb{Z}, \exists \text{ a prime number } p \text{ such that } p | m. Consider m = 1 (note that 1 \in \mathbb{Z}), then there is no such prime number p that divides m.

b. Is P symmetric?

Yes, P is symmetric. The statement claims \forall m, n \in \mathbb{Z}, \exists \text{ some prime number } p \text{ such that } p | m \wedge p | n \to p | n \wedge p | m.

Since there is a prime number p that divides m and n, it is trivially true that p divides n and m.

c. Is P transitive?

No, P is not transitive. The statement claims that:

 \forall m, n, o \in \mathbb{Z}, [\exists \text{ some prime } p_1, p_1 | m \wedge p_1 | n] \wedge [\exists \text{ some prime } p_2, p_2 | n \wedge p_2 | o] \to [\exists \text{ some prime } p_3, p_3 | m \wedge p_3 | o] 

But this is not true for all integers m, n, and o.

Consider m = 6, n = 15, o = 35.

Then there exists the prime number p_1 = 3 such that 3 | m since 3 | 6 since 6 = 3 \cdot 2. Additionally, 3 | n since 3 | 15 since 15 = 3 \cdot 5, so the first term of the supposition is true.

Next, there exists the prime number p_2 = 5 such that 5 | n since 5 | 15 since 15 = 5 \cdot 3. Additionally 5 | o since 5 | 35 since 35 = 5 \cdot 7, so the second term of the supposition is true.

Then, the conclusion claims that there exists some prime p_3 such p_3 | m and p_3 | o, but the only prime numbers that divide m are 3 and 2 since m = 6, and the only prime numbers that divide o are 7 and 5 since o = 35. None of these primes are equal to each other, and so p_3 does not exist. Therefore P is not transitive.

18. Define a relation Q on \mathbb{R} as follows: For all real numbers x and y, x Q y \Leftrightarrow x - y is rational.

Hint: Q is reflexive, symmetric, and transitive.

a. Is Q reflexive?

Yes, Q is reflexive. The statement claims that \forall x \in \mathbb{R}, x - x \text{ is rational}. This is true since x - x = 0, and 0 is rational since 0 = \dfrac{0}{1}.

b. Is Q symmetric?

Yes, Q is symmetric. The statement claims that \forall x, y \in \mathbb{R}, (x - y \text{ is rational }) \to (y - x \text{ is rational}).

Since x - y is rational, it can be said that x - y = \dfrac{a}{b}, where a is some integer and b is some integer with b \neq 0. Now, consider that:

 x - y = -1(y - x) 

 -1(x - y) = y - x 

Then, by substitution:

 -1\left(\frac{a}{b}\right) = y - x 

Now, -1\left(\dfrac{a}{b}\right) is a rational number (since -1 multiplied by a rational number is a rational number). Therefore y - x is rational. This is what was to be shown.

c. Is Q transitive?

Yes, Q is transitive. The statement claims that \forall x, y, z \in \mathbb{R}, [(x - y \text{ is rational}) \wedge (y - z \text{ is rational})] \to x - z \text{ is rational}.

Since x - y is rational and y - z is rational, it can be said that x - y = \dfrac{a}{b} and y - z = \dfrac{c}{d}, where a, b, c, d \in \mathbb{Z} with b \neq 0 and d \neq 0.

Then, consider the addition of x - y and y - z:

 (x - y) + (y - z) = x - z 

Then, by substitution:

 x - z = \frac{a}{b} + \frac{c}{d} 

 = \frac{ad + cb}{bd}

Now, ad + cb is an integer by the product and sum of integers, and bd is an integer by the product of integers and bd \neq 0 (since b \neq 0 and d \neq 0). Thus \dfrac{ad + cb}{bd} is a rational number, and therefore x - z is rational. This is what was to be shown.

19. Define a relation I on \mathbb{R} as follows: For all real numbers x and y, x I y \Leftrightarrow x - y is irrational.

a. Is I reflexive?

No, I is not reflexive. The statement claims that \forall x \in \mathbb{R}, x - x \text{ is irrational}. Since x - x = 0, and 0 = \dfrac{0}{1}, it follows that x - x is rational. Therefore I is not reflexive.

b. Is I symmetric?

Yes, I is symmetric. The statement claims \forall x, y \in \mathbb{R}, (x - y \text{ is irrational}) \to (y - x \text{ is irrational}).

Consider that:

 x - y = -1(y - x) 

 -1(x - y) = y - x 

Now, the product of -1 and an irrational number (x - y) is irrational. It follows that y - x is irrational. This is what was to be shown.

c. Is I transitive?

The statement claims that \forall x, y, z \in \mathbb{R}, [(x - y \text{ is irrational}) \wedge (y - z \text{ is irrational})] \to x - z \text{ is irrational}. But this is not true for all integers x, y, and z.

Consider x = \sqrt{2}, y = 0, and z = \sqrt{2}.

Then x - y = \sqrt{2} - 0 = \sqrt{2}, which is irrational. Additionally, y - z = 0 - \sqrt{2} = -\sqrt{2}, which is irrational. Thus the supposition is true.

Then x - z = \sqrt{2} - \sqrt{2} = 0, which is rational (since 0 = \dfrac{0}{1}). Therefore I is not transitive.

20. Let X = \{a, b, c\} and \mathscr{P}(X) be the power set of X (the set of all subsets of X). A relation \mathbf{E} is defined on \mathscr{P}(X) as follows: For every A, 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.

a. Is E reflexive?

Yes, E is reflexive. The statement claims that \forall A \in \mathscr{P}(X), \text{ the number of elements in } A \text{ equals the number of elements in } A.

This is trivially true.

b. Is E symmetric?

Yes, E is symmetric. The statement claims that \forall A, B \in \mathscr{P}(X), (\text{the number of elements in } A \text{ equals the number of elements in } B) \to (\text{the number of elements in } B \text{ equals the number of elements in } A).

This is trivially true (by the commutative laws of equality).

c. Is E transitive?

Yes, E is transitive. The statement claims that $\forall A, B, C \in \mathscr{P}(X), [(\text{ the number of elements in } A \text{ equals the number of elements in } B) \wedge (\text{ the number of elements in } B \text{ equals the number of elements in } C)] \to \text{the number of elements in } A \text{ equals the number of elements in } C$.

This is trivially true (by the transitivity of equality).

21. Let X = \{a, b, c\} and \mathscr{P}(X) be the power set of X. A relation \mathbf{L} is defined on \mathscr{P}(X) as follows: For every A, 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.

a. Is L reflexive?

No, L is not reflexive. The statement claims \forall A \in \mathscr{P}(X), \text{ the number of elements in } A \text{ is less than the number of elements in } A.

This cannot be true, since the number of elements in A will always equal the number of elements in A.

b. Is L symmetric?

No, L is not symmetric. The statement claims that \forall A, B \in \mathscr{P}(X), (\text{the number of elements in } A \text{ is less than the number of elements in } B) \to (\text{the number of elements in } B \text{ is less than the number of elements in } A).

Let x= \text{ the number of elements in } A and y = \text{ the number of elements in } B. Then, by the supposition, x < y. By the definition of inequality, this means that y \cancel{<} x. Therefore L is not symmetric.

c. Is L transitive?

Yes, L is transitive. The statement claims that \forall A, B, C \in \mathscr{P}(X), [(\text{the number of elements in } A \text{ is less than the number of elements in } B) \wedge (\text{the number of elements in } B \text{ is less than the number of elements in } C)] \to \text{ the number of elements in } A \text{ is less than the number of elements in } C.

Let x = \text{ the number of elements in } A, y = \text{ the number of elements in } B, and z = \text{ the number of elements in } C.

Then, by the supposition, x < y and y < z. Since x < y < z (by the transitivity of inequality), it follows that x < z. This is what was to be shown. Therefore L is transitive.

22. Let X = \{a, b, c\} and \mathscr{P}(X) be the power set of X. A relation \mathbf{N} is defined on \mathscr{P}(X) as follows: For every A, 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.

a. Is \mathbf{N} reflexive?

No, \mathbf{N} is not reflexive. The statement claims \forall A \in \mathscr{P}(X), \text{ the number of elements in } A \text{ is not equal to the number of elements in } A.

This is trivially false.

b. Is \mathbf{N} symmetric?

Yes, \mathbf{N} is symmetric. The statement claims \forall A, B \in \mathscr{P}(X), (\text{the number of elements in } A \text{ is not equal to the number of elements in } B) \to (\text{ the number of elements in } B \text{ is not equal to the number of elements in } A).

This is true.

Let x = \text{ the number of elements in } A, y = \text{ the number of elements in } B. Then, by the supposition, x \neq y. It follows by the definition of inequality that y \neq x.

Therefore \mathbf{N} is symmetric.

c. Is \mathbf{N} transitive?

No, \mathbf{N} is not transitive. The statement claims \forall A, B, C \in \mathscr{P}(X), [(\text{the number of elements in } A \text{ is not equal to the number of elements in } B) \wedge (\text{the number of elements in } B \text{ is not equal to the number of elements in } C)] \to \text{the number of elements in } A \text{ is not equal to the number of elements in } C. But this is not true for all subsets A, B, and C.

Consider A = \{a\}, B = \{a, b\}, and C = \{c\}.

Then, by the supposition, the number of elements in A does not equal the number of elements in B, and the number of elements in B does not equal the number of elements in C, but the number of elements in A is equal to the number of elements in C.

Therefore, \mathbf{N} is not transitive.

23. Let X be a nonempty set and \mathscr{P}(X) the power set of X. Define the "subset" relation \mathbf{S} on \mathscr{P}(X) as follows: For every A, B \in \mathscr{P}(X), A \mathbf{S} B \Leftrightarrow A \subseteq B.

a. Is \mathbf{S} reflexive?

Yes, \mathbf{S} is reflexive. The statement claims \forall A \in \mathscr{P}(X), A \subseteq A. By the definition of subset, this is true.

b. Is \mathbf{S} symmetric?

No, \mathbf{S} is not symmetric. The statement claims \forall A, B \in \mathscr{P}(X), (A \subseteq B) \to (B \subseteq A).

Consider X = \{1, 2, 3\}, A = \{1\}, B = \{1, 2\}. Then, by the supposition A, B \in \mathscr{P}(X), and A \subseteq B, but B \nsubseteq A. Therefore \mathbf{S} is not symmetric.

c. Is \mathbf{S} transitive?

Yes, \mathbf{S} is transitive. The statement claims that \forall A, B, C \in \mathscr{P}(X), [(A \subseteq B) \wedge (B \subseteq C)] \to [A \subseteq C].

By the supposition A \subseteq B and B \subseteq C, it follows by the transitivity property of subset that A \subseteq B \subseteq C, and thus A \subseteq C. Therefore \mathbf{S} is transitive.

24. Let X be a nonempty set and \mathscr{P}(X) the power set of X. Define the "not equal to" relation \mathbf{U} on \mathscr{P}(X) as follows: For every A, B \in \mathscr{P}(X), A \mathbf{U} B \Leftrightarrow A \neq B.

a. Is \mathbf{U} reflexive?

No, \mathbf{U} is not reflexive. The statement claims \forall A \in \mathscr{P}(X), A \neq A. This is trivially false.

b. Is \mathbf{U} symmetric?

Yes, \mathbf{U} is symmetric. The statement claims \forall A, B \in \mathscr{P}, (A \neq B) \to (B \neq A). This is true by the definition of inequality.

c. Is \mathbf{U} transitive?

No, \mathbf{U} is not transitive. The statement claims \forall A, B, C \in \mathscr{P}, [(A \neq B) \wedge (B \neq C)] \to [A \neq C].

Let X = \{1, 2, 3\}, A = \{1\}, B = \{2\}, and C = \{1\}. Then, by the supposition, A, B, C \in \mathscr{P}(X), A \neq B and B \neq C, but A = C.

Therefore \mathbf{U} is not transitive.

25. 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, s R t \Leftrightarrow s \text{ has the same first two characters as } t.

a. Is R reflexive?

Yes, R is reflexive. The statement claims \forall s \in A, s \text{ has the same first two characters as } s. This is trivially true.

b. Is R symmetric?

Yes, R is symmetric. The statement claims \forall s, t \in A, (s \text{ has the same first two characters as } t) \to (t \text{ has the same first two characters as} s).

This is trivially true.

c. Is R transitive?

Yes, R is transitive. The statement claims \forall s, t, u \in A, [(s \text{ has the same first two characters as } t) \wedge (t \text{ has the same first two characters as } u)] \to s \text{ has the same first two characters as } u.

This is true by the transitivity of equality, since s and t have the same first two characters, and t and u have the same first two characters, it follows that s and u have the same first two characters. Therefore R is transitive.

26. Let A be 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 relation R on A as follows: For every s, t \in A, s R t \Leftrightarrow \text{ the sum of the characters of } s \text{ equals the sum of the characters of } t.

a. Is R reflexive?

Yes, R is reflexive. The statement claims \forall s \in A, \text{ the sum of the characters of } s \text{ equals the sum of the characters of } s. This is trivially true.

b. Is R symmetric?

Yes, R is symmetric. The statement claims \forall s, t \in A, (\text{ the sum of the characters of} s \text{ equals the sum of the characters of } t) \to (\text{ the sum of the characters of } t \text{ equals the sum of the characters of } s).

Let x = \text{ the sum of the characters of } s and y = \text{ the sum of the characters of } t. Then, by the supposition, x = y. It follows by symmetry of equality that y = x. This is what was to be shown. Therefore R is symmetric.

c. Is R transitive?

Yes, R is transitive. The statement claims \forall s, t, u \in A, [(\text{ the sum of the characters of } s \text{ equals the sum of the characters of } t) \wedge (\text{ the sum of the characters of } t \text{ equals the sum of the characters of } u)] \to \text{ the sum of the characters of } s \text{ equals the sum of the characters of } u.

Let x = \text{ the sum of the characters of } s, y = \text{ the sum of the characters of } t, and z = \text{ the sum of the characters of } u.

By the supposition x = y and y = z. By the transitivity of equality, x = y = z, and it follows that x = z. This is what was to be shown. Therefore R is transitive.

27. Let A be the set of all English statements. A relation \mathbf{I} is defined on A as follows: For every p, q \in A,

 p \mathbf{I} q \Leftrightarrow p \to q \text{ is true} 

a. Is \mathbf{I} reflexive?

Yes \mathbf{I} is reflexive. The statement claims \forall p \in A, p \to p \text{ is true}. This is true by the law of identity (tautology).

b. Is \mathbf{I} symmetric?

No, \mathbf{I} is not symmetric. The statement claims \forall p, q \in A, (p \to q) \to (q \to p).

Consider p is the statement "All pigs can fly", and q is the statement "The sky is blue". Then, by the supposition p, q \in A, and p \to q is vacuously true. But, q \to p is false, since q is true and p is false.

Therefore \mathbf{I} is not symmetric.

c. Is \mathbf{I} transitive?

Yes, \mathbf{I} is transitive. The statement claims \forall p, q, r \in A, [(p \to q) \wedge (q \to r)] \to (p \to r).

This is true, since p \to q and q \to r is true, it follows that p \to q \to r, and that p \to r is true.

28. Let A = \mathbb{R} \times \mathbb{R}. A relation \mathbf{F} is defined on A as follows: For every (x_1, y_1) and (x_2, y_2) in A,

 (x_1, y_2) \mathbf{F} (x_2, y_2) \Leftrightarrow x_1 = x_2 

a. Is \mathbf{F} reflexive?

Yes, \mathbf{F} is reflexive. The statement claims \forall (x_1, y_1) \in A, x_1 = x_1. This is trivially true.

b. Is \mathbf{F} symmetric?

Yes, \mathbf{F} is symmetric. The statement claims \forall (x_1, y_1), (x_2, y_2) \in A, (x_1 = x_2) \to (x_2 = x_1).

This is true by the symmetry of equality.

c. Is \mathbf{F} transitive?

The statement claims \forall (x_1, y_1), (x_2, y_2), (x_3, y_3) \in A, [(x_1 = x_2) \wedge (x_2 = x_3)] \to x_1 = x_3.

This is true by the transitivity of equality.

29. Let A = \mathbb{R} \times \mathbb{R}. A relation \mathbf{S} is defined on A as follows: For every (x_1, y_1) and (x_2, y_2) in A,

 (x_1, y_2) \mathbf{S} (x_2, y_2) \Leftrightarrow y_1 = y_2 

a. Is \mathbf{S} reflexive?

Yes, \mathbf{S} is reflexive. The statement claims \forall (x_1, y_1) \in A, y_1 = y_1. This is trivially true.

b. Is \mathbf{S} symmetric?

Yes, \mathbf{S} is symmetric. The statement claims \forall (x_1, y_1), (x_2, y_2) \in A, (y_1 = y_2) \to (y_2 = y_1).

This is true by the symmetry of equality.

c. Is \mathbf{S} transitive?

Yes, \mathbf{S} is transitive. The statement claims \forall (x_1, y_1), (x_2, y_2), (x_3, y_3) \in A, [(y_1 = y_2) \wedge (y_2 = y_3)] \to y_1 = y_3.

This is true by the transitivity of equality.

30. Let A be the "punctured plane"; that is, A is the set of all points in the Cartesian plane except the origin (0, 0). A relation R is defined on A as follows: For every p_1 and p_2 in A, p_1 R p_2 \Leftrightarrow p_1 \text{ and } p_2 \text{ lie on the same half line emanating from the origin}.

a. Is reflexive?

b. Is symmetric?

c. Is transitive?

31. Let A be the set of people living in the world today. A relation R is defined on A as follows: For all people p and q in A,

 p R q \Leftrightarrow p \text{ lives within 100 miles of } q 

a. Is reflexive?

Omitted.

b. Is symmetric?

Omitted.

c. Is transitive?

Omitted.

32. Let A be the set of all lines in the plane. A relation R is defined on A as follows: For every l_1 and l_2 in A, l_1 R l_2 \Leftrightarrow l_1 \text{ is parallel to } l_2. (Assume that a line is parallel to itself.)

a. Is reflexive?

Omitted.

b. Is symmetric?

Omitted.

c. Is transitive?

Omitted.

33. Let A be the set of all lines in the plane. A relation R is defined on A as follows: For every l_1 and l_2 in A,

 l_1 R l_2 \Leftrightarrow l_1 \text{ is perpendicular to } l_2 

a. Is reflexive?

Omitted.

b. Is symmetric?

Omitted.

c. Is transitive?

Omitted.

In 34-36, assume that R is a relation on a set A. Prove or disprove each statement.

34. If R is reflexive, then R^{-1} is reflexive.

35. If R is symmetric, then R^{-1} is symmetric.

36. If R is transitive, then R^{-1} is transitive.

In 37-42, assume that R and S are relations on a set A. Prove or disprove each statement.

37. If R and S are reflexive, is R \cap S reflexive? Why?

38. If R and S are symmetric, is R \cap S symmetric? Why?

39. If R and S are transitive, is R \cap S transitive? Why?

40. If R and S are reflexive, is R \cup S reflexive? Why?

41. If R and S are symmetric, is R \cup S symmetric? Why?

42. If R and S are transitive, is R \cup S transitive? 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.

43. Exercise 1

44. Exercise 2

45. Exercise 3

46. Exercise 4

47. Exercise 5

48. Exercise 6

49. Exercise 7

50. Exercise 8

In 51-53, R, S, and T are relations defined on A = \{0, 1, 2, 3\}.

51. Let R = \{(0, 1), (0, 2), (1, 1), (1, 3), (2, 2), (3, 0)\}.

Find R^t, the transitive closure of R.

52. Let S = \{(0, 0), (0, 3), (1, 0), (1, 2), (2, 0), (3, 2)\}.

Find S^t, the transitive closure of S.

53. Let T = \{(0, 2), (1, 0), (2, 3), (3, 1)\}.

Find T^t, the transitive closure of T.

54. Write a computer algorithm to test whether a relation R defined on a finite set A is reflexive, where

 A = \{a[1], a[2], \dots, a[n]\} 

55. Write a computer algorithm to test whether a relation R defined on a finite set A is symmetric, where

 A = \{a[1], a[2], \dots, a[n]\} 

56. Write a computer algorithm to test whether a relation R defined on a finite set A is transitive, where

 A = \{a[1], a[2], \dots, a[n]\}