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.

Proof:

Suppose R is any relation on a set A, such that R is reflexive.

By the definition of reflexive, this means that \forall x \in A, (x, x) \in R, or \forall x \in A, x R x. Then, by definition of an inverse relation, it follows that (x, x) \in R^{-1}, or x R^{-1} x.

Therefore, R^{-1} is reflexive.

Q.E.D.

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

Proof:

Suppose R is any relation on a set A, such that R is symmetric.

By the definition of symmetric, this means that \forall (x, y) \in A, (x, y) \in R \to (y, x) \in R. Since (y, x) \in R, it follows, by definition of inverse relation, that (x, y) \in R^{-1}. Furthermore, since (x, y) \in R, it follows that (y, x) \in R^{-1}.

Therefore R^{-1} is symmetric.

Q.E.D.

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

Proof:

Suppose R is any relation on a set A such that R is transitive.

By the definition of transitive, this means that \forall x, y, z \in A, [(x, y) \in R \wedge (y, z) \in R] \to (x, z) \in R.

Since (x, y), (y, z), (x, z) \in R, it follows by the definition of inverse that (y, x), (z, y), (z, x) \in R^{-1}. This means that \forall x, y, z \in A, [(z, y) \in R^{-1} \wedge (y, x) \in R^{-1}] \to (z, x) \in R^{-1}.

Therefore R^{-1} is transitive.

Q.E.D.

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?

R \cap S is reflexive.

Proof:

Suppose R and S are any relations on some set A such that R and S are reflexive.

By the definition of reflexive, this means that \forall x \in A, (x, x) \in R, and \forall x \in A, (x, x) \in S.

Since (x, x) \in R and (x, x) \in S, it follows (by the definition of intersection), that (x, x) \in R \cap S.

Therefore R \cap S is reflexive.

Q.E.D.

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

R \cap S is symmetric.

Proof:

Suppose R and S are any relations on a set A such that R and S are symmetric.

By the definition of symmetric, this means that \forall x, y \in A, (x, y) \in R \to (y, x) \in R. Similarly, \forall x, y \in A, (x, y) \in S \to (y, x) \in S.

Since (x, y) \in R, (y, x) \in R, (x, y) \in S, (y, x) \in S, it follows by the definition of intersection that (x, y) \in R \cap S and (y, x) \in R \cap S.

Therefore R \cap S is symmetric.

Q.E.D.

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

R \cap S is transitive.

Proof:

Suppose R and S are any relations on a set A such that R and S are transitive.

By the definition of transitive, this means that \forall x, y, z \in A, [(x, y) \in R \wedge (y, z) \in R] \to (x, z) \in R. Similarly, \forall x, y, z \in A, [(x, y) \in S \wedge (y, z) \in S] \to (x, z) \in S.

Since (x, y), (y, z), (x, z) \in R and (x, y), (y, z), (x, z) \in S, it follows by the definition of intersection that (x, y), (y, z), (x, z) \in (R \cap S). Furthermore, this means that \forall x, y, z \in A, [(x, y) \in (R \cap S) \wedge (y, z) \in (R \cap S)] \to (x, z) \in (R \cap S).

Therefore, by the definition of transitive, R \cap S is transitive.

Q.E.D.

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

R \cup S is reflexive.

Proof:

Suppose R and S are any relations on some set A such that R and S are reflexive.

By the definition of reflexive, this means that \forall x \in A, (x, x) \in R, and \forall x \in A, (x, x) \in S.

Since (x, x) \in R and (x, x) \in S, it follows (by the definition of union), that (x, x) \in R \cup S (since in order to satisfy the definition of union, (x, x) \in R or (x, x) \in S).

Therefore R \cup S is reflexive.

Q.E.D.

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

R \cup S is symmetric.

Proof:

Suppose R and S are any relations on a set A such that R and S are symmetric.

By the definition of symmetric, this means that \forall x, y \in A, (x, y) \in R \to (y, x) \in R. Similarly, \forall x, y \in A, (x, y) \in S \to (y, x) \in S.

Since (x, y) \in R, (y, x) \in R, (x, y) \in S, (y, x) \in S, it follows by the definition of union that (x, y) \in R \cup S and (y, x) \in R \cup S (since in order to satisfy the definition of union, (x, y) \in R and (y, x) \in R or (x, y ) \in S and (y, x) \in S).

Therefore R \cup S is symmetric.

Q.E.D.

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

Disproof (by counterexample):

Let A = \{a, b, c, d\}, R = {(a, b), (b, c), (a, c)}, and S = \{(b, c), (c, d), (b, d)\}. Note that R and S are transitive. However, when we take the union, R \cup S:

 (R \cup S) = \{(a, b), (b, c), (a, c), (c, d), (b, d)\} 

Note that (a, b), (b, d) \in (R \cup S), but (a, d) \notin (R \cup S). By the definition of transitive, it follows that R \cup S is not transitive.

Q.E.D.

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

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

a. Irreflexive?:

No, since 0 R_1 0, R_1 is not irreflexive.

b. Asymmetric?:

No, since (0, 1) \in R_1 and (1, 0) \in R_1, R_1 is not asymmetric.

c. Intransitive?:

No, since (0, 1), (1, 0), (0, 0) \in R_1, R_1 is not intransitive.

44. Exercise 2

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

a. Irreflexive?:

No, since 0 R_2 0, R_2 is not irreflexive.

b. Asymmetric?:

No, since (0, 0) \in R_2 and (0, 0) \in R_2, R_2 is not asymmetric.

c. Intransitive?:

No, since (1, 1), (1, 2), (2, 2) \in R_2, R_2 is not intransitive.

45. Exercise 3

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

a. Irreflexive?:

Yes, R_3 is irreflexive.

b. Asymmetric?:

No, since (2, 3), (3, 2) \in R_3, R_3 is not asymmetric.

c. Intransitive?:

Yes, since (2, 3), (3, 2) \in R_3, but (2, 2) \notin R_3, R_3 is intransitive.

46. Exercise 4

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

a. Irreflexive?:

Yes, R_4 is irreflexive.

b. Asymmetric?:

No, since (1, 2), (2, 1) \in R_4, R_4 is not asymmetric.

c. Intransitive?:

Yes, R_4 is intransitive.

 (1, 2), (2, 1) \in R_4, \text{ but } (1, 1) \notin R_4 

 (2, 1), (1, 3) \in R_4, \text{ but } (2, 3) \notin R_4 

 (1, 3), (3, 1) \in R_4 , \text{ but } (1, 1) \notin R_4 

etc. (note that a more rigorous proof would check all examples.)

47. Exercise 5

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

a. Irreflexive?:

No, since (0, 0) \in R_5

b. Asymmetric?:

No, since (0, 0) \in R_5 and (0, 0) \in R_5.

c. Intransitive?:

No, since (0, 1), (1, 2), (0, 2) \in R_5.

48. Exercise 6

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

a. Irreflexive?:

Yes.

b. Asymmetric?:

Yes.

c. Intransitive?:

Yes, since there is no (1, x) for some element x, nor is there (2, y) for some element y, the supposition is always false, and is therefore the if/then proposition is vacuously true.

49. Exercise 7

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

a. Irreflexive?:

Yes.

b. Asymmetric?:

Yes.

c. Intransitive?:

Yes (see Exercise 48 for vacuous truth explanation, which applies here as well.)

50. Exercise 8

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

a. Irreflexive?:

No, since (0, 0) \in R_8.

b. Asymmetric?:

No, since (0, 0) \in R_8.

c. Intransitive?:

No, since (0, 0), (0, 0), (0, 0) \in R_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.

First, by definition of the transitive closure, R \subseteq R^t, so (building R^t, i.e. not finished):

 R^t = \{(0, 1), (0, 2) (1, 1), (1, 3), (2, 2), (3, 0)\} 

Since R^t must be transitive, every ordered pair triple must have a transitive "third":

 (0, 1), (1, 1) \to (0, 1) 

 (0, 1), (1, 3) \to (0, 3) 

 (0, 2), (2, 2) \to (0, 2) 

 (1, 1), (1, 3) \to (1, 3) 

 (1, 3), (3, 0) \to (1, 0) 

 (3, 0), (0, 1) \to (3, 1) 

 (3, 0), (0, 2) \to (3, 2) 

Now, add all missing ordered pairs to R^t:

 R^t = \{(0, 1), (0, 2), (0, 3), (1, 0), (1, 1), (1, 3), (2, 2), (3, 0), (3, 1), (3, 2)\} 

Now, check to be sure all ordered triples yields:

 (3, 1), (1, 3) \to (3, 3) 

 \boxed{R^t = \{(0, 1), (0, 2), (0, 3), (1, 0), (1, 1), (1, 3), (2, 2), (3, 0), (3, 1), (3, 2), (3, 3)\}} 

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

Find S^t, the transitive closure of S.

 S^t = \{(0, 0), (0, 3), (1, 0), (1, 2), (2, 0), (3, 2)\} 

Then:

 (0, 0), (0, 3) \to (0, 3) 

 (0, 3), (3, 2) \to (0, 2) 

 (1, 0), (0, 0) \to (1, 0) 

 (1, 0), (0, 3) \to (1, 3) 

 (1, 2), (2, 0) \to (1, 0) 

 (2, 0), (0, 0) \to (2, 0) 

 (2, 0), (0, 3) \to (2, 3) 

 (3, 2), (2, 0) \to (3, 0) 

New:

 S^t = \{(0, 0), (0, 2), (0, 3), (1, 0), (1, 2), (1, 3), (2, 0), (2, 3), (3, 0), (3, 2)\} 

Check again:

 (2, 0), (0, 2) \to (2, 2) 

 (3, 0), (0, 3) \to (3, 3) 

Finally:

 S^t = \{(0, 0), (0, 2), (0, 3), (1, 0), (1, 2), (1, 3), (2, 0), (2, 2), (2, 3), (3, 0), (3, 2), (3, 3)\} 

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

Find T^t, the transitive closure of T.

 $T^t = \{(0, 2), (1, 0), (2, 3), (3, 1)\}. 

 (0, 2), (2, 3) \to (0, 3) 

 (1, 0), (0, 2) \to (1, 2) 

 (2, 3), (3, 1) \to (2, 1) 

 (3, 1), (1, 0) \to (3, 0) 

Now:

 $T^t = \{(0, 2), (0, 3), (1, 0), (1, 2), (2, 1), (2, 3), (3, 0), (3, 1)\}. 

Furthermore:

 (0, 2), (2, 1) \to (0, 1) 

 (0, 3), (3, 0) \to (0, 0) 

 (1, 0), (0, 3) \to (1, 3) 

 (1, 2), (2, 1) \to (1, 1) 

 (2, 3), (3, 0) \to (2, 0) 

Now:

$T^t = {(0, 0), (0, 1), (0, 2), (0, 3), (1, 0), (1, 1), (1, 2), (1, 3), (2, 0), (2, 1), (2, 3), (3, 0), (3, 1)}.

And:

 (2, 1), (1, 2) \to (2, 2) 

 (3, 1), (1, 2) \to (3, 2) 

 (3, 1), (1, 3) \to (3, 3) 

So:

$T^t = {(0, 0), (0, 1), (0, 2), (0, 3), (1, 0), (1, 1), (1, 2), (1, 3), (2, 0), (2, 1), (2, 2), (2, 3), (3, 0), (3, 1), (3, 2), (3, 3)}.

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]\} 

Omitted.

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]\} 

Omitted.

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]\} 

Omitted.

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Page 543

Exercise Set 8.3

1. Suppose that S = \{a, b, c, d, e\} and R is a relation on S such that a R b, b R c, and d R e. List all of the following that must be true if R is (a) reflexive (but not symmetric or transitive), (b) symmetric (but not reflexive or ransitive), c transitive (but not reflexive or symmetric), and (d) an equivalence relation.

 c R b \quad c R c  \quad a R c \quad b R a 

 a R d \quad e R a \quad e R d \quad c R a 

a. reflexive

c R c

b. symmetric

b R a, c R b, e R d

c. transitive

a R c

d. equivalence relation

c R c, b R a, c R b, e R d, a R c, c R a

2. Each of the following partitions of \{0, 1, 2, 3, 4\} induces a relation R on \{0, 1, 2, 3, 4\}. In each case, find the ordered pairs in R.

a. \{0, 2\}, \{1\}, \{3, 4\}

 R = \{(0, 0), (0, 2), (2, 0), (2, 2), (1, 1), (3, 3), (3, 4), (4, 3) , (4, 4)\} 

b. \{0\}, \{1, 3, 4\}, \{2\}

 R = \{(0, 0), (1, 1), (1, 3), (1, 4), (2, 2), (3, 1), (3, 3), (3, 4), (4, 1), (4, 3), (4, 4)\} 

c. \{0\}, \{1, 2, 3, 4\}

 R = \{(0, 0), (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4)\} 

In each of 3-6, the relation R is an equivalence relation on A. As in example 8.3.5, first find the specified equivalence classes. Then state the number of distinct equivalence classes for R and list them.

 A = \{0, 1, 2, 3, 4\} 

 R = \{(0, 0), (0, 4), (1, 1), (1, 3), (2, 2), (3, 1), (3, 3), (4, 0), (4, 4)\} 

equivalence classes: [0], [1], [2], [3]

 [0] = \{x \in A | x R 0\} = \{0, 4\} 

 [1] = \{x \in A | x R 1\} = \{1, 3\} 

 [2] = \{x \in A | x R 2\} = \{2\} 

 [3] = \{x \in A | x R 3\} = \{1, 3\} 

The distinct number of classes is 3. List:

 [0] = \{0, 4\}, [1] = \{1, 3\} = [3], [2] = \{2\} 

 A = \{a, b, c, d\} 

 R = \{(a, a), (b, b), (b, d), (c, c), (d, b), (d, d)\} 

equivalence classes: [a], [b], [c], [d]

 [a] = \{x \in A | x R a\} = \{a\} 

 [b] = \{x \in A | x R b\} = \{b, d\} 

 [c] = \{x \in A | x R c\} = \{c\} 

 [d] = \{x \in A | x R d\} = \{b, d\} 

The number of distinct classes is 3. List:

 [a] = \{a\}, [b] = \{b, d\} = [d], [c] = \{c\} 

 A = \{1, 2, 3, 4, \dots, 20\} 

R is defined on A as follows:

 \text{For all } x, y \in A, x R y \Leftrightarrow 4 | (x - y) 

equivalence classes: [1], [2], [3], [4], [5]

 [1] = \{1, 5, 9, 13, 17\} 

 [2] = \{2, 6, 10, 14, 18\} 

 [3] = \{3, 7, 11, 15, 19\} 

 [4] = \{4, 8, 12, 16, 20\} 

 [5] = \{1, 5, 9, 13, 17\} 

There are 4 distinct classes:

 [1] = \{1, 5, 9, 13, 17\} = [5], [2] = \{2, 6, 10, 14, 18\}, [3] = \{3, 7, 11, 15, 19\}, [4] = \{4, 8, 12, 16, 20\} 

 A = \{-4, -3, -2, -1, 0, 1, 2, 3, 4, 5\} 

R is defined on A as follows:

 \text{For all } x, y \in A, x R y \Leftrightarrow 3 | (x - y) 

equivalence classes: [0], [1], [2], [3]

 [0] = \{-3, 0, 3\} 

 [1] = \{-2, 1, 4\} 

 [2] = \{-4, -1, 2, 5\} 

 [3] = \{-3, 0, 3\} 

There are 3 distinct equivalence classes:

 [0] = \{-3, 0, 3\} = [3], [1] = \{-2, 1, 4\}, [2] = \{-4, -1, 2, 5\} 

In each of 7-14, the relation R is an equivalence relation on the set A. Find the distinct equivalence classes of R.

7. A = \{(1, 3), (2, 4), (-4, -8), (3, 9), (1, 5), (3, 6)\}. R is defined on A as follows: For every (a, b), (c, d) \in A,

 (a, b) R (c, d) \Leftrightarrow ad = bc 

 \{(1, 3), (3, 9)\}, \{(2, 4), (-4, -8), (3, 6)\}, \{(1, 5)\} 

8. X = \{a, b, c\} and A = \mathscr{P}(X). R is defined on A as follows: For all sets u and v in \mathscr{P}(X),

 u R v \Leftrightarrow N(u) = N(v) 

(That is, the number of elements in u equals the number of elements in v.)

 \mathscr{P}(X) = \{\emptyset, \{a\}, \{b\}, \{c\}, \{a, b\}, \{a, c\}, \{b, c\}, \{a, b, c\}\} 

 \{\emptyset\}, \{\{a\}, \{b\}, \{c\}\}, \{\{a, b\}, \{a, c\}, \{b, c\}\}, \{\{a, b, c\}\} 

9. X = \{-1, 0, 1\} and A = \mathscr{P}(X). R is defined on \mathscr{P}(X) as follows: For all sets s and t in \mathscr{P}(X),

 s R t \Leftrightarrow \text{ the sum of the elements in } s \text{ equals the sum of the elements in } t 

 \mathscr{P}(X) = \{\emptyset, \{-1\}, \{0\}, \{1\}, \{-1, 0\}, \{-1, 1\}, \{0, 1\}, \{-1, 0, 1\}\} 

 \{\{\emptyset\}, \{-1\}, \{-1, 0\}\}, \{\{0\}, \{-1, 1\}, \{-1, 0, 1\}\}, \{\{1\}, \{0, 1\}\} 

10. A = \{-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5\}. R is defined on A as follows: For all m, n \in \mathbb{Z},

 m R n \Leftrightarrow 3 |(m^2 - n^2) 

 \{-5, -4, -2, -1, 1, 2, 4, 5\}, \{-3, 0, 3\} 

11. A = \{-4, -3, -2< -1, 0, 1, 2, 3, 4\}. R is defined on A as follows: For every (m, n) \in A,

 m R n \Leftrightarrow 4 | (m^2 - n^2) 

 [0] = \{x \in A | 4 | (x^2 - 0^2)\} = \{x \in A | 4 | x^2\} 

 = \{-4, -2, 0, 2, 4\} 

 [1] = \{x \in A | 4 | (x^2 - 1^2) = \{x \in A | 4 | (x^2 - 1)\}\} 

 = \{-3, -1, 1, 3\} 

12. A = \{-4, -3, -2, -1, 0, 1, 2, 3, 4\}. R is defined on A as follows: For all (m, n) \in A,

 m R n \Leftrightarrow 5 | (m^2 - n^2) 

 [0] = \{x \in A | 5 | (x^2 - 0^2)\} = \{x \in A | 5 | x^2\} 

 = \{0\} 

 [1] = \{x \in A | 5 | (x^2 - 1^2)\} = \{x \in A | 5 | (x^2 - 1) \} 

 = \{-4, -1, 1, 4\} 

 [2] = \{x \in A | 5 | (x^2 - 2^2)\} = \{x \in A | 5 | (x^2 - 4)\} 

 = \{-3, -2, 2, 3\} 

13. A is the set of all strings of length 4 in a's and b's. R is defined on A as follows: For all strings s and t in A,

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

 A = \{aaaa, aaab, aabb, aaba, abbb, abba, abaa, abab, bbbb, bbba, bbaa, bbab, baaa, baab, babb, baba\} 

 \{aaaa, aaab, aabb, aaba\}, \{abbb, abba, abaa, abab\}, \{bbbb, bbba, bbaa, bbab\}, \{baaa, baab, babb, baba\} 

14. A is 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. R is defined 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 

 \{0000\}, \{0001, 0010, 0100, 1000\}, \{0011, 0101, 1001, 1010, 1100, 0002, 0020, 0200, 2000\}

15. Determine which of the following congruence relations are true and which are false.

 m \equiv n (\mod d) \Leftrightarrow d | (m - n) 

a. 17 \equiv 2 (\mod 5)

 5 | 17 - 2 

 5 | 15 

Yes, this congruence relation is true, because 15 = 5 \cdot 3, therefore 5 | 15.

b. 4 \equiv -5 (\mod 7)

 7 | 4 - (-5) 

 7 | 9 

No, this congruence relation is not true, since 7 \cancel{|} 9.

c. -2 \equiv -8 (\mod 3)

 3 | -2 - (-8) 

 3 | 6 

Yes, this congruence relation is true, since 6 = 3 \dot 2, therefore 3 | 6.

d. -6 \equiv 22 (\mod 2)

 2 | -6 - 22 

 2 | -28 

Yes, this congruence relation is true, since -28 = 2 \cdot -14, therefore 2 | -28.

a. Let R be the relation of congruence modulo 3. Which of the following equivalence classes are equal?

 [7], [-4], [-6], [17], [4], [27], [19] 

 7 \mod 3 = 1, -4 \mod 3 = 2, -6 \mod 3 = 0, 17 \mod 3 = 2, 4 \mod 3 = 1, 27 \mod 3 = 0, 19 \mod 3 = 1 

 [7] = [4] = [19], [-4] = [17], [-6] = [27] 

b. Let R be the relation of congruence modulo 7. Which of the following equivalence classes are equal?

 [35], [3], [-7], [12], [0], [-2], [17] 

 35 \mod 7 = 0, 3 \mod 7 = 3, -7 \mod 7 = 0, 12 \mod 7 = 5, 0 \mod 7 = 0, -2 \mod 7 = 5, 17 \mod 7 = 3 

 [35] = [-7] = [0], [12] = [-2], [3] = [17] 

a. Prove that for all integers m and n, m \equiv n (\mod 3) if, and only if, m \mod 3 = n \mod 3.

Proof:

To prove that m \equiv n (\mod 3) \Leftrightarrow m \mod 3 = n \mod 3, it must be shown that m \equiv n (\mod 3) \to m \mod 3 = n \mod 3, and it must also be shown that m \mod 3 = n \mod 3 \to m \equiv n (\mod 3).

Proof (m \equiv n (\mod 3)\to m \mod 3 = n \mod 3):

Suppose m \in \mathbb{Z} and n \in \mathbb{Z}, such that m \equiv n (\mod 3).

It is to be shown that m \mod 3 = n \mod 3.

Since m \equiv n (\mod 3), by the definition of congruence, this means that:

 3 | (m - n) 

By the definition of divisiblity:

 m - n = 3a 

For some integer a.

Let r = m \mod 3.

Then, by the definition of modulo:

 m = 3b + r 

for some integer b.

Since m - n = 3a, it follows by substitution that:

 m - n = (3b + r) - n = 3a 

Equivalently (by algebra):

 (3b + r) - n = 3a 

 -n = 3a - (3b + r) 

 n = (3b + r) - 3a 

 n = 3b + r - 3a 

 n = 3b - 3a + r 

 n = 3(b - a) + r 

Now, b - a is an integer (by the difference of integers), and 0 \leq r < 3. So, by definition of \mod, n \mod 3 = r, which equals m \mod 3.

This is what was to be shown.

Q.E.D.

Proof (m \mod 3 = n \mod 3 \to m \equiv n (\mod 3)):

Suppose m \in \mathbb{Z} and n \in \mathbb{Z} such that m \mod 3 = n \mod 3.

It must be shown that m \equiv n (\mod 3).

Let r = m \mod 3 = n \mod 3.

Then, by definition of \mod, m = 3p + r and n = 3q + r for some integers p and q.

By substitution:

 m - n = (3p + r) - (3q + r) 

 = 3p + r - 3q - r 

 = 3p - 3q 

 = 3(p - q) 

Now, p - q is an integer (by the difference of integers). It follows by the definition of divisibility, that 3 | (m - n). Therefore, by the definition of congruence, m \equiv n (\mod 3).

This is what was to be shown.

Q.E.D.

Conclusion:

Since it has been shown that m \equiv n (\mod 3) \to m \mod 3 = n \mod 3 and it has also been shown that m \mod 3 = n \mod 3 \to m \equiv n (\mod 3), it can be concluded that m \equiv n (\mod 3) \Leftrightarrow m \mod 3 = n \mod 3.

b. Prove that for all integers m and n and any positive integer d, m \equiv n (\mod d) if, and only if, m \mod d = n \mod d.

Proof:

To prove that m \equiv n (\mod d) \Leftrightarrow m \mod d = n \mod d, it must be shown that m \equiv n (\mod d) \to m \mod d = n \mod d, and it must also be shown that m \mod d = n \mod d \to m \equiv n (\mod d).

Proof (m \equiv n (\mod d)\to m \mod d = n \mod d):

Suppose m \in \mathbb{Z}, n \in \mathbb{Z}, and d \in \mathbb{Z}^+ such that m \equiv n (\mod d).

It is to be shown that m \mod d = n \mod d.

Since m \equiv n (\mod d), by the definition of congruence, this means that:

 d | (m - n) 

By the definition of divisiblity:

 m - n = da 

For some integer a.

Let r = m \mod d.

Then, by the definition of modulo:

 m = db + r 

for some integer b.

Since m - n = da, it follows by substitution that:

 m - n = (db + r) - n = da 

Equivalently (by algebra):

 (db + r) - n = da 

 -n = da - (db + r) 

 n = (db + r) - da 

 n = db + r - da 

 n = db - da + r 

 n = d(b - a) + r 

Now, b - a is an integer (by the difference of integers), and 0 \leq r < d. So, by definition of \mod, n \mod d = r, which equals m \mod d.

This is what was to be shown.

Q.E.D.

Proof (m \mod d = n \mod d \to m \equiv n (\mod d)):

Suppose m \in \mathbb{Z}, n \in \mathbb{Z}, d \in \mathbb{Z}^+ such that m \mod d = n \mod d.

It must be shown that m \equiv n (\mod d).

Let r = m \mod d = n \mod d.

Then, by definition of \mod, m = dp + r and n = dq + r for some integers p and q.

By substitution:

 m - n = (dp + r) - (dq + r) 

 = dp + r - dq - r 

 = dp - dq 

 = d(p - q) 

Now, p - q is an integer (by the difference of integers). It follows by the definition of divisibility, that d | (m - n). Therefore, by the definition of congruence, m \equiv n (\mod d).

This is what was to be shown.

Q.E.D.

Conclusion:

Since it has been shown that m \equiv n (\mod d) \to m \mod d = n \mod d and it has also been shown that m \mod d = n \mod d \to m \equiv n (\mod d), it can be concluded that m \equiv n (\mod d) \Leftrightarrow m \mod d = n \mod d.

a. Give an example of two sets that are distinct but not disjoint.

Consider \{1, 2, 3\}, \{2\}, then they are distinct since \{1, 2, 3\} \neq \{2\}, but they are not disjoint since \{1, 2, 3\} \cap \{2\} = \{2\}.

b. Find sets A_1 and A_2 and elements x, y, and z such that x and y are in A_1 and y and z are in A_2 but x and z are not both in either of the sets A_1 or A_2.

 A_1 = \{x, y\} 

 A_2 = \{y, z\} 

In 19-31, (1) prove that the relation is an equivalence relation, and (2) describe the distinct equivalence classes of each relation.

19. A is the set of all students at your college.

a. R is the relation defined on A a follows: For every x and y in A,

 x R y \Leftrightarrow x \text{ has the same major (or double major) as } y 

(Assume "undeclared" is a major.)

(1) Prove that the relation is an equivalence relation.

Proof:

Suppose A is the set of all students at my college, and R is a relation defined on A defined as follows:

\forall x, y \in A, x R y \Leftrightarrow x \text{ has the same major (or double major) as } y

It must be shown that R is an equivalence relation.

To prove that R is an equivalence relation, it must be shown that R is reflexive, symmetric, and transitive.

Proof (that R is reflexive):

Let x \in A.

To prove that R is reflexive, it must be shown that x R x. It is true that x has the same major as x. Therefore R is reflexive. This is what was to be shown.

Proof (that R is symmetric):

Let x, y \in A.

To prove that R is symmetric, it must be shown that (x, y) \in R \to (y, x) \in R.

Suppose (x, y) \in R. Then, by definition of R, this means that x has the same major as y. By symmetric property of equality, this means that y has the same major as x. Therefore (y, x) \in R.

This is what was to be shown.

Proof (that R is transitive):

Let x, y, z \in A.

To prove that R is transitive, it must be shown that (x, y) \in R \wedge (y, z) \in R \to (x, z) \in R.

Suppose (x, y) \in R and (y, z) \in R. Then, by definition of R, this means that x has the same major as y, and y has the same major as z. By the transitive property of equality, it follows that x has the same major as z.

Therefore (x, z) \in R, and it can be concluded that R is transitive.

This is what was to be shown.

Conclusion:

Since it has been shown that R is reflexive, symmetric, and transitive, it can be concluded that R is an equivalence relation.

This is what was to be shown.

Q.E.D.

(2) Describe the distinct equivalence classes of each relation.

There is one equivalence class for each major and double major at the college. Each class consists of all students with that major (or double major).

b. S is the relation defined on A as follows: For every x, y \in A,

 x S y \Leftrightarrow x \text{ is the same age as } y 

(1) Prove that the relation is an equivalence relation.

Proof:

Suppose A is the set of all students at my college, with S being a relation defined on A as follows:

 \forall x, y \in A, x S y \Leftrightarrow x \text{ is the same age as } y 

To prove that S is an equivalence relation, it must be shown that S is reflexive, symmetric, and transitive.

Proof (that S is reflexive):

Let x \in A.

To prove that S is reflexive, it must be shown that x S x. It is true that x is the same age as x. Thus x S x, and therefore S is reflexive.

This is what was to be shown.

Proof (that S is symmetric):

Let x, y \in A.

To prove that S is symmetric, it must be shown that (x, y) \in S \to (y, x) \in S.

Suppose x S y. By the definition of S, this means that x is the same age as y. By the symmetric property of equality, this means that y is the same age as x. It follows that y R x, and therefore S is symmetric.

This is what was to be shown.

Proof (that S is transitive):

Let x, y, z \in A.

To prove that S is transitive, it must be shown that (x, y) \in S \wedge (y, z) \in S \to (x, z) \in S.

Suppose x S y and y S z. Then, by the definition of S, this means that x is the same age as y and y is the same age as z. By the transitive property of equality, this means that x is the same age as z.

It follows that (x, z) \in S, and therefore S is transitive.

This is what was to be shown.

Conclusion:

Since S has been shown to be reflexive, symmetric, and transitive, it can be concluded that S is an equivalence relation.

This is what was to be shown.

Q.E.D.

(2) Describe the distinct equivalence classes of each relation.

There is one equivalence class for each student age (by year) at the college. Each class consists of all students with that age.

20. E is the relation defined on \mathbb{Z} as follows:

 \text{For every } m, n \in \mathbb{Z}, m E n \Leftrightarrow 4 | (m - n) 

(1) Prove that the relation is an equivalence relation.

Proof:

Suppose m \in \mathbb{Z} and n \in \mathbb{Z}. Let E be a relation defined on \mathbb{Z} as follows:

 \forall m, n \in \mathbb{Z}, m E n \Leftrightarrow 4 | (m - n) 

To prove that E is an equivalence relation, it must be shown that E is reflexive, symmetric, and transitive.

Proof (E is reflexive):

Let m \in \mathbb{Z}.

To prove that E is reflexive, it must be shown that (m, m) \in E.

By the definition for E, this means that:

 4 | (m - m) 

Since m - m = 0, this means that:

 4 | 0 

This is true, since 0 = 4 \cdot 0. It follows that (m, m) \in E, and therefore E is reflexive.

Proof (E is symmetric):

Let m \in \mathbb{Z} and n \in \mathbb{Z}.

To prove that E is symmetric, it must be shown that (m, n) \in E \to (n, m) \in E.

Since (m, n) \in E, this means that:

 4 | (m - n) 

By the definition of divisibility, this means that:

 m - n = 4k 

for some integer k.

Now, consider:

 -1(m - n) = -1(4k) 

 n - m = 4(-k) 

Now, -k is an integer (by the product of integers). It follows (by the definition of divisibility), that:

 4 | (n - m) 

This means that (n, m) \in E, and therefore E is symmetric.

Proof (E is transitive):

Let m \in \mathbb{Z}, n \in \mathbb{Z}, and p \in \mathbb{Z}.

To prove that E is transitive, it must be shown that (m, n) \in E \wedge (n, p) \in E \to (m, p) \in E.

Suppose (m, n) \in E and (n, p) \in E. By definition of E, this means that:

 4 | (m - n) 

and

 4 | (n - p) 

By the definition of divisibility, this means that:

 m - n = 4k 

 n - p = 4q 

for some integers k and q.

Subtracting the two yields:

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

And then by substitution this is:

 m - p = 4k - 4q 

By algebra:

 = 4(k - q) 

Now, k - q is an integer (by the difference of integers). It follows that 4 | (m - p), and thus (m, p) \in E. Therefore, it can be concluded that E is transitive.

Conclusion:

Since it has been shown that E is reflexive, symmetric, and transitive, it can be concluded that E is an equivalence relation. This is what was to be shown.

Q.E.D.

(2) Describe the distinct equivalence classes of each relation.

Observe that for any a \in \mathbb{Z}, the equivalence class of a, ([a]), is:

 [a] = \{x \in \mathbb{Z} | x E a\} = \{x \in \mathbb{Z} | 4 | x - a\} 

By definition of divisiblity:

 = \{x \in \mathbb{Z} | x - a = 4k \text{ for some integer } k\} 

By algebra:

 = \{x \in \mathbb{Z} | x = 4k + a \} 

So, our equivalence classes are defined as follows:

 \{x \in \mathbb{Z} | x = 4k \}, \{x \in \mathbb{Z} | x = 4k + 1 \}, \{x \in \mathbb{Z} | x = 4k + 2 \}, \{x \in \mathbb{Z} | x = 4k + 3 \} 

21. R is the relation defined on \mathbb{Z} as follows:

 \text{For every } m, n \in \mathbb{Z}, m R n \Leftrightarrow 7m - 5n \text{ is even} 

(1) Prove that the relation is an equivalence relation.

Proof:

Suppose m \in \mathbb{Z} and n \in \mathbb{Z}, such that R is a relation on \mathbb{Z} defined as follows:

 \forall m, n \in \mathbb{Z}, m R n \Leftrightarrow 7m - 5n \text{ is even} 

It must be shown that R is an equivalence relation.

To show that R is an equivalence relation, it must be shown that R is reflexive, symmetric, and transitive.

Proof (R is reflexive):

Let m \in \mathbb{Z}.

To prove that R is reflexive, it must be shown that (m, m) \in R. By the definition of R, it then must be shown that:

 7m - 5m \text{ is even} 

Consider that:

 7m - 5m = 2m 

Since m is an integer (by the supposition), it follows that 7m - 5m is even (by the definition of even, since 7m - 5m = 2m).

It follows that (m, m) \in R, and therefore R is reflexive.

Proof (R is symmetric):

Let m, n \in \mathbb{Z}.

To prove that R is symmetric, it must be shown that (m, n) \in R \to (n, m) \in R.

Suppose (m, n) \in R. By definition of R, this means that:

 7m - 5n \text{ is even} 

By definition of even, this means that:

 7m - 5n = 2k 

for some integer k.

Then, consider:

 7n - 5m = (12 - 5)n - (12 - 7)m 

 = 12n - 5n - 12m + 7m 

 = 12n - 12m + (7m - 5n) 

 = 12n - 12m + 2k 

 = 2(6n - 6m + k) 

Now, 6n - 6m + k is an integer (by the product, sum, and difference of integers). It follows that 7n - 5m is even (by the definition of even). Therefore (n, m) \in R, and therefore R is symmetric.

Proof (R is transitive):

Let m, n, p \in \mathbb{Z}.

To prove that R is transitive, it must be shown that (m, n) \in R \wedge (n, p) \in R \to (m, p) \in R.

Suppose (m, n) \in R and (n, p) \in R. By definition of R, this means that:

 7m - 5n \text{ is even} 

and

 7n - 5p \text{ is even}  

By the definition of even, this means that:

 7m - 5n = 2r  

and

 7n - 5p = 2s 

for some integers r and s.

It must be shown that 7m - 5p \text{ is even}. Consider:

 7m - 5p = (7m - 5n + 5n) + (7n - 7n - 5p) 

 = ((7m - 5n) + 5n) + (7n - (7n - 5p)) 

 = (2r + 5n) + (7n - 2s) 

 = 2r + 5n + 7n - 2s 

 = 2r + 12n - 2s 

 = 2(r + 6n - s) 

Now, r + 6n - s is an integer (by the product, sum, and difference of integers). By the definition of even, this means that 7m - 5p is even. It follows that (m, p) \in R, and therefore R is transitive.

Conclusion:

Since it has been shown that R is reflexive, symmetric, and transitive, it can be concluded that R is an equivalence relation.

This is what was to be shown.

Q.E.D.

(2) Describe the distinct equivalence classes of each relation.

 \forall m, n \in \mathbb{Z}, m R n \Leftrightarrow 7m - 5n \text{ is even} 

Consider $a \in \mathbb{Z}, then, by the definition of r, this means that:

 \{x \in \mathbb{Z} | x R a \} 

By the definition of R:

 \{x \in \mathbb{Z} | 7x - 5a \text{ is even} \} 

Since 7x - 5a is even, this means that both 7x and 5a are even, or both 7x and 5a are odd. Since 7 and 5 are both odd (and odd times odd is odd, and odd times even is even), this means that 7x and 5a have the same parity.

Thus there are two equivalency cases, one the set of all even integers, and the other the set of all odd integers.

22. Let A be the set of all statement forms in three variables p, q, and r. \mathbf{R} is the relation defined on A as follows: For all P and Q in A,

 P \mathbf{R} Q \Leftrightarrow P \text{ and } Q \text{ have the same truth table} 

(1) Prove that the relation is an equivalence relation.

Proof:

Suppose A is the set of all statement forms in three variables p, q, and r. Let \mathbf{R} be a relation on the set A defined as follows:

 P \mathbf{R} Q \Leftrightarrow P \text{ and } Q \text{ have the same truth table} 

To prove that \mathbf{R} is an equivalence relation, it must be shown that \mathbf{R} is reflexive, symmetric, and transitive.

Proof (\mathbf{R} is reflexive):

Let P \in A.

To prove that \mathbf{R} is reflexive, it must be shown that (P, P) \in \mathbf{R}. By the definition of \mathbf{R}, this means that P and P have the same truth table.

It is true that P has the same truth table as itself.

It follows that (P, P) \in \mathbf{R}, and therefore \mathbf{R} is reflexive.

Proof (\mathbf{R} is symmetric):

Let P, Q \in A.

To prove that \mathbf{R} is symmetric, it must be shown that (P, Q) \in \mathbf{R} \to (Q, P) \in \mathbf{R}.

Suppose (P, Q) \in \mathbf{R}, by the definition for \mathbf{R}, this means that P and Q have the same truth tables.

It follows by the symmetric property of equality that Q and P have the same truth tables.

Thus (Q, P) \in \mathbf{R}, and therefore \mathbf{R} is symmetric.

Proof (\mathbf{R} is transitive):

Let P, Q, S \in A.

To prove that \mathbf{R} is transitive, it must be shown that (P, Q) \in \mathbf{R} \wedge (Q, S) \in \mathbf{R} \to (P, S) \in \mathbf{R}.

Suppose (P, Q) \in \mathbf{R} and (Q, S) \in \mathbf{R}. By the definition of \mathbf{R}, this means that P and Q have the same truth tables, and that Q and S have the same truth tables.

It follows, by the transitive property of equality, that P and S have the same truth tables.

Thus (P, S) \in \mathbf{R}, and therefore \mathbf{R} is transitive.

Conclusion:

Since it has been shown that \mathbf{R} is reflexive, symmetric, and transitive, it can be concluded that \mathbf{R} is an equivalence relation.

Q.E.D.

(2) Describe the distinct equivalence classes of each relation.

There is an equivalence class corresponding to every possible truth table in 3 variables, p, q, r. There are 8 lines in every truth table, and each line has 2 options (true or false), so there are 2^8 equivalence classes.

23. Let P be a set of parts shipped to a company from various suppliers. S is the relation defined on P as follows: For every x, y \in P,

 x S y \Leftrightarrow  x \text{ has the same part number and is shipped from the same supplier as } y 

(1) Prove that the relation is an equivalence relation.

Proof:

Suppose P is the set of all parts shipped to a company from various suppliers. Let S be a relation defined on P as follows:

 \forall x, y \in P, x S y \Leftrightarrow x \text{ has the same part number and is shipped from the same supplier as } y 

To prove that S is an equivalence relation, it must be shown that S is reflexive, symmetric, and transitive.

Proof (S is reflexive):

Let x \in P.

To prove that S is reflexive, it must be shown that (x, x) \in S. By the definition for S, this means it must be shown that x has the same part number and is shipped from the same supplier as x.

It is true that x has the same part number as x and that x is shipped from the same supplier as x.

Thus (x, x) \in S, and therefore S is reflexive.

Proof (S is symmetric):

Let x, y \in P.

To prove that S is symmetric, it must be shown that (x, y) \in S \to (y, x) \in S.

Suppose (x, y) \in S. By the definition for S, this means that x has the same part number as y and x is shipped from the same supplier as y.

It follows by the symmetry of equality that y has the same part number as x and y is shipped from the same supplier as x.

Thus (y, x) \in S, and therefore S is symmetric.

Proof (S is transitive):

Let x, y, z \in P.

To prove that S is transitive, it must be shown that (x, y) \in S \wedge (y, z) \in S \to (x, z) \in S.

Suppose (x, y) \in S and (y, z) \in S. By the definition for S, this means that:

x has the same part number and is shipped from the same supplier as y.

and that:

y has the same part number and is shipped from the same supplier as z.

By the definition of the transitivity of equality, this means that x has the same part number and is shipped from the same supplier as z.

Thus (x, z) \in S, and therefore S is transitive.

Conclusion:

Since it has been shown that S is reflexive, symmetric, and transitive, it can be concluded that S is an equivalence relation. This is what was to be shown.

Q.E.D.

(2) Describe the distinct equivalence classes of each relation.

The number of distinct equivalence classes is grouped based off of parts that all have the same part number and are shipped from the same supplier (i.e. the equivalence classes are sets of all parts with the same part number and supplier.)

24. Let A be the set of identifiers in a computer program. It is common for identifiers to be used for only a short part of the execution time of a program and not to be used again to execute other parts of the program. In such cases, arranging for identifiers to share memory locations makes efficient use of a computer's memory capacity. Define a relation R on A as follows: For all identifiers x and y,

 x R y \Leftrightarrow \text{ the values of } x \text{ and } y \text{ are stored in the same memory location during execution of the program} 

(1) Prove that the relation is an equivalence relation.

Proof:

Suppose A is the set of identifiers in a computer program. Let R be a relation on the set A such that it is defined as follows:

 \forall x, y \in A, x R y \Leftrightarrow \text{ the values of } x \text{ and } y \text{ are stored in the same memory location during execution of the program} 

To prove that R is an equivalence relation, it must be shown that R is reflexive, symmetric, and transitive.

Proof (R is reflexive):

Let x \in A.

To prove that R is reflexive, it must be shown that (x, x) \in R.

By definition of R, this means that it must be shown that the values of x and x are stored in the same memory location during execution of the program.

It is true that x and x are stored in the same memory location during execution of the program (since x is the same identifier as x.)

Thus (x, x) \in R and therefore R is reflexive.

Proof (R is symmetric):

Let x, y \in A.

To prove that R is symmetric, it must be shown that (x, y) \in R \to (y, x) \in R.

Suppose (x, y) \in R. Then, by definition of R, this means that the values of x and y are stored in the same memory location during the execution of the program.

By the symmetric property of equality, this means that the values of y and x are stored in the same memory location during the execution of the program.

Thus, (y, x) \in R, and therefore R is symmetric.

Proof (R is transitive):

Let x, y, z \in A.

To prove that R is transitive, it must be shown that (x, y) \in R \wedge (y, z) \in R \to (x, z) \in R.

Suppose (x, y) \in R and (y, z) \in R. By the definition for R, this means that:

The values of x and y are stored in the same memory location during execution of the program.

and that:

The values of y and z are stored in the same memory location during execution of the program.

By the transitive property of equality, this means that the values of x and z are stored in the same memory location during execution of the program.

Thus (x, z) \in R, and therefore R is transitive.

Conclusion:

Since it has been shown that R is reflexive, symmetric, and transitive, it can be concluded that R is an equivalence relation.

(2) Describe the distinct equivalence classes of each relation.

The number of equivalence classes is based off the number of identifiers in a computer program that are stored in the same memory location during execution of the program.

25. A is the "absolute value" relation defined on \mathbb{R} as follows:

 \text{For every } x, y \in \mathbb{R}, x A y \Leftrightarrow |x| = |y| 

(1) Prove that the relation is an equivalence relation.

Proof:

Suppose A is the "absolute value" relation on \mathbb{R}, defined as follows:

 \forall x, y \in \mathbb{R}, x A y \Leftrightarrow |x| = |y| 

To prove that A is an equivalence relation, it must be shown that A is reflexive, symmetric, and transitive.

Proof (A is reflexive):

Let x \in \mathbb{R}.

To prove that A is reflexive, it must be shown that (x, x) \in A.

By definition for A, this means that it must be proved that:

 |x| = |x| 

It is trivially true that |x| = |x|.

Thus (x, x) \in A, and therefore A is reflexive.

Proof (A is symmetric):

Let x, y \in \mathbb{R}.

To prove that A is symmetric, it must be shown that (x, y) \in A \to (y, x) \in A.

Suppose (x, y) \in A. By the definition for A, this means that:

 |x| = |y| 

By the symmetric property of equality, it follows that:

 |y| = |x| 

Thus (y, x) \in A, and therefore A is symmetric.

Proof (A is transitive):

Let x, y, z \in \mathbb{R}.

To prove that A is transitive, it must be shown that (x, y) \in A \wedge (y, z) \in A \to (x, z) \in A.

Suppose (x, y) \in A and (y, z) \in A. By the definition for A, this means that:

 |x| = |y| 

and that:

 |y| = |z| 

It follows, by the transitive property of equality that |x| = |z|.

Thus (x, z) \in A, and therefore A is transitive.

Conclusion:

Since it has been shown that A is reflexive, symmetric, and transitive, it can be concluded that A is an equivalence relation. This is what was to be shown.

Q.E.D.

(2) Describe the distinct equivalence classes of each relation.

Let a \in \mathbf{R}, then by the definition of absolute value:

 |-a| = |a| 

with the exception of 0, since 0 \in \mathbb{R}, but there is no -0.

Thus the equivalence classes are all sets of all real numbers and their corresponding negative counterpart, and also the set \{0\}.

26. D is the relation defined on \mathbb{Z} as follows: For every m, n \in \mathbb{Z},

 m D n \Leftrightarrow 3 | (m^2 - n^2) 

(1) Prove that the relation is an equivalence relation.

Proof:

Suppose D is a relation on \mathbb{Z} defined as follows:

 \forall m, n \in \mathbb{Z}, m D n \Leftrightarrow 3 | (m^2 - n^2) 

To prove that D is an equivalence relation, it must be shown that D is reflexive, symmetric, and transitive.

Proof (D is reflexive):

Let x \in \mathbb{Z}.

To prove that D is reflexive, it must be shown that (x, x) \in D. By the definition for D, this means it must be shown that:

 3 | (x^2 - x^2) 

Now, x^2 - x^2 = 0, and it is true that 3 | 0, since 0 = 3 \cdot 0. Thus (x, x) \in D, and it can be concluded that D is reflexive.

Proof (D is symmetric):

Let x, y \in \mathbb{Z}.

To prove that D is symmetric, it must be shown that (x, y) \in D \to (y, x) \in D. By definition of D, this means it must be shown that:

 [3 | (x^2 - y^2)] \to [3 | (y^2 - x^2)] 

Suppose 3 | (x^2 - y^2). By the definition of divisibility, this means that:

 x^2 - y^2 = 3k 

for some integer k.

Now, consider that:

 y^2 - x^2 = -1(x^2 - y^2) 

Then, by substitution:

 = -1(3k) 

 = 3(-k) 

Now, -k is an integer (by the product of integers), thus 3 | (y^2 - x^2), and hence (y, x) \in D, and therefore D is symmetric.

Proof (D is transitive):

Let x, y, z \in \mathbb{Z}.

To prove that D is transitive, it must be shown that [(x, y) \in D \wedge (y, z) \in D] \to [(x, z) \in D].

Suppose (x, y) \in D and (y, z) \in D. Then, by the definition for D, this means:

 3 | (x^2 - y^2) 

and also:

 3 | (y^2 - z^2) 

(It must be shown that 3 | (x^2 - z^2).)

By the definition of divisibility, this means that:

 x^2 - y^2 = 3k 

and also that:

 y^2 - z^2 = 3p 

for some integers k and p.

Now, if one adds x^2 - y^2 and y^2 - z^2, this yields:

 x^2 - y^2 + y^2 - z^2 = x^2 - z^2 

Then, by substitution:

 x^2 - z^2 = (3k) + (3p) 

 = 3(k + p) 

Now, k + p is an integer (by the sum of integers). Thus 3 | (x^2 - z^2) (by the definition of divisibility). It follows that (x, z) \in D, and therefore D is transitive.

Conclusion:

Since D has been shown to be reflexive, symmetric, and transitive, it follows that D is an equivalence relation. This is what was to be shown.

Q.E.D.

(2) Describe the distinct equivalence classes of each relation.

There are two distinct equivalence classes:

 [0] = \{\dots, -6, -3, 0, 3, 6, \dots\}, [1] = \{\dots, -5, -4, -2, -1, 1, 2, 4, 5, \dots\} 

27. R is the relation defined on \mathbb{Z} as follows: For every (m, n) \in \mathbb{Z},

 m R n \Leftrightarrow 4 | (m^2 - n^2) 

(1) Prove that the relation is an equivalence relation.

Proof:

Suppose R is a relation defined on \mathbb{Z} as follows:

 \forall m, n \in \mathbb{Z}, m R n \Leftrightarrow 4 | (m^2 - n^2) 

To prove that R is an equivalence relation, it must be shown that R is reflexive, symmetric, and transitive.

Proof (R is reflexive):

Let x \in \mathbb{Z}.

To prove that R is reflexive, it must be shown that (x, x) \in R. By the definition for R, this means it must be shown that:

 4 | (x^2 - x^2) 

Since x^2 - x^2 = 0, this means it must be shown that 4 | 0. Now, 4 | 0 because 0 = 4 \cdot 0. Therefore (x, x) \in R, and it can be concluded that R is reflexive.

Proof (R is symmetric):

Let x, y \in \mathbb{Z}.

To prove that R is symmetric, it must be shown that (x, y) \in R \to (y, x) \in R.

Suppose (x, y) \in R, then, by definition for R, this means:

 4 | (x^2 - y^2) 

By the definition of divisibility, this means that:

 x^2 - y^2 = 4k 

for some integer k.

Now, consider that:

 y^2 - x^2 = -1(x^2 - y^2) 

Then, by substitution:

 y^2 - x^2 = -1(4k) 

 = 4(-k) 

Now, -k is an integer (by the product of integers). Hence 4 | (y^2 - x^2), and it follows that (y, x) \in R, and therefore R is symmetric.

Proof (R is transitive):

Let x, y, z \in \mathbb{Z}.

To prove that R is transitive, it must be shown that [(x, y) \in R \wedge (y, z) \in R] \to (x, z) \in R.

Suppose (x, y) \in R and (y, z) \in R. By the definition for R, this means that:

 4 | (x^2 - y^2) 

and also that:

 4 | (y^2 - z^2) 

Now, by the definition for divisibility, this means that:

 x^2 - y^2 = 4k 

and also that:

 y^2 - z^2 = 4p 

for some integers k and p.

Now, consider that:

 x^2 - z^2 = x^2 - y^2 + y^2 - z^2 

Then, by substitution:

 x^2 - z^2 = 4k + 4p 

 x^2 - z^2 = 4(k + p) 

Now, k + p is an integer (by the sum of integers), and so it follows that 4 | (x^2 - z^2). This means that (x, z) \in R, and therefore R is transitive.

Conclusion:

Since it has been shown that R is reflexive, symmetric, and transitive, it can be concluded that R is an equivalence relation. This is what was to be shown.

Q.E.D.

(2) Describe the distinct equivalence classes of each relation.

There are two distinct equivalence classes:

 [0] = \{\dots, -8, -4, -2, 0, 2, 4, 8, \dots\} = \text{ the set of all even integers } 

 [1] = \{\dots, -9, -5, -1, 1, 5, 9\dots\} = \text{ the set of all odd integers } 

28. I is the relation defined on \mathbb{R} as follows:

 \text{For every } x, y \in \mathbb{R}, m I n \Leftrightarrow x - y \text{ is an integer} 

(1) Prove that the relation is an equivalence relation.

Proof:

Suppose I is a relation defined on \mathbb{R} as follows:

 \forall x, y \in \mathbb{R}, m I n \Leftrightarrow (x - y) \in \mathbb{Z} 

To prove that I is an equivalence relation, it must be shown that I is reflexive, symmetric, and transitive.

Proof (I is reflexive):

Let x \in \mathbb{R}.

To prove that I is reflexive, it must be shown that (x, x) \in I. By the definition for I, this means it must be shown that:

 (x - x) \in \mathbb{Z} 

Now, x - x = 0, and 0 \in \mathbb{Z}. Thus (x, x) \in I, and therefore I is reflexive.

Proof (I is symmetric):

Let x, y \in \mathbb{R}.

To prove that I is symmetric, it must be shown that (x, y) \in I \to (y, x) \in I.

Suppose (x, y) \in I, by the definition for I, this means that:

 (x - y) \in \mathbb{Z} 

Now, consider:

 y - x = -1(x - y) 

Now, -1(x - y) is an integer (by the product of integers), and thus (y - x) \in \mathbb{Z}. Thus (y, x) \in I, and therefore I is symmetric.

Proof (I is transitive):

Let x, y, z \in \mathbb{R}.

To prove that I is transitive, it must be shown that [(x, y) \in I \wedge (y, z) \in I] \to (x, z) \in I.

Suppose (x, y) \in I and (y, z) \in I. By the definition for I, this means that:

 (x - y) \in \mathbb{Z} 

and also that:

 (y - z) \in \mathbb{Z} 

Now, consider that:

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

Thus, (x - z) \in \mathbb{Z} (by the sum of integers). It follows that (x, z) \in I, and therefore I is transitive.

Conclusion:

Since it has been shown that I is reflexive, symmetric, and transitive, it can be concluded that I is an equivalence relation. This is what was to be shown.

Q.E.D.

(2) Describe the distinct equivalence classes of each relation.

There is one class for each real number x with 0 \leq x < 1. The distinct classes are all sets of the form [x] = y \in \mathbb{R}, | y = n + x \text{ for some integer } n, where x is a real number such that 0 \leq x < 1.

29. Define P on the set \mathbb{R} \times \mathbb{R} of ordered pairs of real numbers as follows: For every (w, x), (y, z) \in \mathbb{R} \times \mathbb{R},

 (w, x) P (y, z) \Leftrightarrow w = y 

(1) Prove that the relation is an equivalence relation.

Proof:

Suppose P is a relation on \mathbb{R} \times \mathbb{R}, defined as:

 \forall (w, x), (y, z) \in \mathbb{R} \times \mathbb{R}, (w, x) P (y, z) \Leftrightarrow w = y 

To prove that P is an equivalence relation, it must be shown that P is reflexive, symmetric, and transitive.

Proof (P is reflexive):

Let (w, x) \in \mathbb{R} \times \mathbb{R}.

To prove that P is reflexive, it must be shown that [(w, x), (w, x)] \in P. By the definition for P, this means it must be shown that:

 w = w 

This is trivially true. Thus [(w, x), (w, x)] \in P, and therefore P is reflexive.

Proof (P is symmetric):

Let (w, x), (y, z) \in \mathbb{R} \times \mathbb{R}.

TO prove that P is symmetric, it must be shown that [(w, x), (y, z)] \in P \to [(y, z), (w, x)] \in P.

Suppose [(w, x), (y, z)] \in P. By definition for P, this means that:

 w = y 

This means that y = w, by the symmetric property of equality. This means that [(y, z), (w, x)] \in P], and therefore P is symmetric.

Proof (P is transitive):

Let (w, x), (y, z), (a, b) \in \mathbb{R} \times \mathbb{R}.

To prove that P is transitive, it must be shown that [[(w, x), (y, z)] \in P \wedge [(y, z), (a, b)] \in P \to [(w, x), (a, b)] \in P.

Suppose [(w, x), (y, z)] \in P and [(y, z), (a, b)] \in P. By the definition for P, this means that:

 w = y 

And also that:

 y = a 

By the transitive property of equality, this means that w = a. It follows that [(w, x), (a, b)] \in P, and therefore P is transitive.

Conclusion:

Since it has been shown that P is reflexive, symmetric, and transitive, it follows that P is an equivalence relation. This is what was to be shown.

Q.E.D.

(2) Describe the distinct equivalence classes of each relation.

There is one equivalence class for each real number. The distinct equivalence classes are all sets of ordered pairs (x, y) \in \mathbb{R} \times \mathbb{R}, | x = a for each real number a.

30. Define Q on the set \mathbb{R} \times \mathbb{R} as follows: For every (w, x), (y, z) \in \mathbb{R} \times \mathbb{R},

 (w, x) Q (y, z) \Leftrightarrow x = z 

(1) Prove that the relation is an equivalence relation.

Omitted.

(2) Describe the distinct equivalence classes of each relation.

Omitted.

31. Let P be the set of all points in the Cartesian plane except the origin. R is the relation defined on P as follows: For every p_1 and p_2 in P,

 p_1 R p_2 \Leftrightarrow p_1 \text{ and } p_2 \text{ lie on the same half-line emanating from the origin} 

(1) Prove that the relation is an equivalence relation.

Omitted.

(2) Describe the distinct equivalence classes of each relation.

Omitted.

32. Let A be the set of all straight lines in the Cartesian plane. Define a relation \mid \mid on A as follows: For every l_1 and l_2 in A,

 l_1 \parallel l_2 \Leftrightarrow l_1 \text{ is parallel to } l_2 

Then \parallel is an equivalence relation on A. Describe the equivalence classes of this relation.

Every possible slope is an equivalence class, including vertical lines (undefined).

33. Let A be the set of points in the rectangle with x and y coordinates between 0 and 1. That is,

 A = \{(x, y) \in \mathbb{R} \times \mathbb{R} | 0 \leq x \leq 1 \text{ and } 0 \leq y \leq 1\} 

Define a relation R on A as follows: For all $(x_1, y_1) and (x_2, y_2) in A,

 (x_1, y_1) R (x_2, y_2) \Leftrightarrow (x_1, y_1) = (x_2, y_2) 

or

 x_1 = 0 \text{ and } x_2 = 1 \text{ and } y_1 = y_2 

or

 x_1 = 1 \text{ and } x_2 = 0 \text{ and } y_1 = y_2 

or

 y_1 = 0 \text{ and } y_2 = 1 \text{ and } x_1 = x_2 

or

 y_1 = 1 \text{ and } y_2 = 0 \text{ and } x_1 = x_2 

In other words, all points along the top edge of the rectangle are related to the points along the bottom edge directly beneath them, and all points directly opposite each other along the left and right edges are related to each other. The points in the interior of the rectangle are not related to anything other than themselves. Then R is an equivalence relation on A. Imagine gluing together all the points that are in the same equivalence class. Describe the resulting figure.

Gluing the top and the bottom edges of the rectangle together forms a cylinder, and then gluing the left and right edges of the rectangle together forms a doughnut shape (a torus).

34. The documentation for the computer language Java recommends that when an "equals method" is defined for an object, it be an equivalence relation. That is, if R is defined as follows:

 x R y \Leftrightarrow \text{x.equals}(y) \text{ for all objects in the class} 

then R should be an equivalence relation. Suppose that in trying to optimize some of the mathematics of a graphics application, a programmer creates an object called a point, consisting of two coordinates in the plane. The programmer defines an equals method as follows: If p and q are any points, then

 \text{p.equals}(q) \Leftrightarrow \text{ the distance from } p \text{ to } q \text{ is less than or equal to } c 

where c is a small positive number that depends on the resolution of the computer display. Is the programmer's equals method an equivalence relation? Justify your answer.

No. If points p, q, and r all lie on a straight line with q in the middle, and if p is c units from q and q is c units from r, then p is more than c units from r. In other words, the programmer's equals method is not an equivalence relation because it is not transitive.

35. Find an additional representative circuit for the input/output table of Example 8.3.9.

Omitted.

Let R be an equivalence relation on a set A. Prove each of the statements in 36-41 directly from the definitions of equivalence relation and equivalence class without using the results of Lemma 8.3.2, Lemma 8.3.3, or Theorem 8.3.4.

36. For every a in a, a \in [a].

Proof:

Suppose R is an equivalence relation on a set A, and let a \in A.

Since R is an equivalence relation, this means that R is reflexive, or, in other words, every element in A is related to itself by R. In particular, a R a, and hence, by definition of an equivalence class, a \in [a]. This is what was to be shown.

Q.E.D.

37. For every a and b in A, if b \in [a] then a R b.

Proof:

Suppose R is an equivalence relation on a set A, and let a, b \in A.

Let b \in [a].

By definition of class, this means that:

 b \in [a] \Leftrightarrow b R a 

Since R is an equivalence relation, R is symmetric. By the definition of symmetry, this means that:

 a R b 

This is what was to be shown.

Q.E.D.

38. For every a, b, and c in A, if b R c and c \in [a] then b \in [a].

Proof:

Suppose R is an equivalence relation on a set A, and let a, b, c \in A.

Let b R c and let c \in [a].

We must show that b \in [a].

By the definition of class, since c \in [a], this means that c R a. Since R is an equivalence relation, and therefore transitive, and also since b R c, it follows, by the definition of transitive, that b R c and c R a. In other words b R a, and by definition of class, this means that b \in [a]. This is what was to be shown.

Q.E.D.

39. For every a and b in A, if [a] = [b] then a R b.

Proof:

Suppose R is an equivalence relation on a set A, and let a, b \in A.

Let [a] = [b].

Since R is reflexive (by the definition of equivalence relation), it follows that a \in [a] and b \in [b].

By the supposition, [a] = [b], and so it follows that a \in [b]. By the definition of class, this means that a R b. This is what was to be shown.

Q.E.D.

40. For every a, b, and x in A, if a R b and x \in [a] then x \in [b].

Proof:

Suppose R is an equivalence relation on a set A, and let a, b, x \in A.

Let a R b and x \in [a].

It must be shown that x \in [b].

Since x \in [a], by the definition of equivalence class, this means that x R a. Since a R b, by the definition of transitivity (since R is an equivalence relation and therefore transitive), it follows that x R b. By the definition of equivalence class, this means that x \in [b]. This is what was to be shown.

Q.E.D.

41. For every a and b in A, if a \in [b] then [a] = [b].

Proof:

Suppose R is an equivalence relation on a set A, and let a, b \in A.

Let a \in [b].

To prove that [a] = [b], it must be shown that [a] \subseteq [b], and that [b] \subseteq [a].

Proof ([a] \subseteq [b]):

Let x \in [a].

By the definition of equivalence class, this means that x R a. Since a \in [b], this means that a R b. Since R is transitive (because R is an equivalence relation), this means that x R b. By the definition of class, this means that x \in [b]. It follows that [a] \subseteq [b]. This is what was to be shown.

Proof ([b] \subseteq [a]):

Let x \in [b].

By the definition of equivalence, class this means that x R b. Since a \in [b], this means that a R b. Since R is symmetric (because R is an equivalence relation), this means that b R a. Then, since R is transitive (again, because R is an equivalence relation), it follows that x R a. By the definition of equivalence class, this means that x \in [a]. It follows that [b] \subseteq [a]. This is what was to be shown.

Conclusion:

Since it has been shown that [a] \subseteq [b] and also that [b] \subseteq [a], it follows (by the definition for subset), that [a] = [b]. This is what was to be shown.

Q.E.D.

42. Let R be the relation defined in Example 8.3.12.

a. Prove that R is reflexive.

Proof:

Suppose A is the set of all ordered pairs of integers for which the second element of the pair is nonzero:

 A = \mathbb{Z} \times (\mathbb{Z} - \{0\}) 

Then, define a relation R on A as follows:

 \forall (a, b), (c, d) \in A, (a, b) R (c, d) \Leftrightarrow ad = bc 

Let (x, y) \in A.

To prove that R is reflexive, it must be shown that [(x, y), (x, y)] \in R. By the definition of R, this means it must be shown that:

 xy = yx 

By the commutative law of product, this is true. Therefore [(x, y), (x, y)] \in R, and R is reflexive.

Q.E.D.

b. Prove that R is symmetric.

Proof:

Suppose A is the set of all ordered pairs of integers for which the second element of the pair is nonzero:

 A = \mathbb{Z} \times (\mathbb{Z} - \{0\}) 

Then, define a relation R on A as follows:

 \forall (a, b), (c, d) \in A, (a, b) R (c, d) \Leftrightarrow ad = bc 

Let (x, y), (z, a) \in A.

To prove that R is symmetric, it must be shown that [(x, y), (z, a)] \in R \to [(z, a), (x, y)] \in R.

Suppose [(x, y), (z, a)] \in R. By the definition for R, this means that:

 xa = yz 

(It must be shown that zy = ax.)

By the commutative property for product, and the symmetric property of equality, xa = yz can be rewritten as:

 zy = ax 

It follows that [(z, a), (x, y)] \in R, and therefore R is symmetric. This is what was to be shown.

Q.E.D.

c. List four distinct elements in [(1, 3)].

 (2, 6), (-2, -6), (3, 9), (-3, -9) 

d. List four distinct elements in [(2, 5)].

 (4, 10), (6, 15), (8, 20), (10, 25) 

43. In Example 8.3.12, define operations of addition (+) and multiplication (\cdot) as follows: For every (a, b), (c, d) \in A,

 [(a, b)] + [(c, d)] = [(ad + bc, bd)] 

 [(a, b)] \cdot [(c, d)] = [(ac, bd)] 

a. Prove that this addition is well defined. That is, show that if [(a, b)] = [(a', b')] and [(c, d)] = [(c', d')], then [(ad + bc), bd] = [(a'd' + b'c', b'd')].

Omitted.

b. Prove that this multiplication is well defined. That is, show that if [(a, b)] = [(a', b')] and [(c, d)] = [(c', d')], then [(ac, bd)] = [(a'c', b'd')].

Omitted.

c. Show that [(0, 1)] is an identity element for addition. That is, show that for any (a, b) \in A,

 [(a, b)] + [(0, 1)] = [(0, 1)] + [(a, b)] = [(a, b)] 

Omitted.

d. Find an identity element for multiplication. That is, find (i, j) in A so that for every (a, b) in A, [(a, b)] \cdot [(i, j)] = [(i, j)] \cdot [(a, b)] = [(a, b)].

Omitted.

e. For any (a, b) \in A, show that [(-a, b)] is an inverse for [(a, b)] for addition. That is, show that [(-a, b)] + [(a, b)] = [(a, b)] + [(-a, b)] = [(0, 1)].

Omitted.

f. Given any (a, b) \in A with a \neq 0, find an inverse for [(a, b)] for multiplication. That is, find (c, d) in A so that [(a, b)] \cdot [(c, d)] = [(c, d)] \cdot [(a, b)] = [(i, j)], where [(i, j)] is the identity element you found in part (d).

Omitted.

44. Let A = \mathbb{Z}^+ \times \mathbb{Z}^+. Define a relation R on A as follows: For every (a, b) and (c, d) in A,

 (a, b) R (c, d) \Leftrightarrow a + d = c + b 

a. Prove that R is reflexive.

Omitted.

b. Prove that R is symmetric.

Omitted.

c. Prove that R is transitive.

Omitted.

d. List five elements in [(1, 1)].

Omitted.

e. List five elements in [(3, 1)].

Omitted.

f. List five elements in [(1, 2)].

Omitted.

g. Describe the distinct equivalence classes of R.

45. The following argument claims to prove that the requirement that an equivalence relation be reflexive is redundant. In other words, it claims to show that if a relation is symmetric and transitive, then it is reflexive. Find the mistake in the argument.

"Proof: Let R be a relation on a set A and suppose R is symmetric and transitive. For any two elements x and y in A, if x R y then y R x since R is symmetric. Thus it follows by transitivity that x R x, and hence R is reflexive."

The mistake in the argument is that just because R is symmetric and transitive does not necessarily mean it is reflexive. Recall that for R to be reflexive, \forall x \in A, x R x. Consider, however, a set where the relation is both symmetric and transitive, but not reflexive:

 A = \{1, 2\} 

 R = \{(1, 1)\} 

Now, R is symmetric, since (1, 1) \to (1, 1), and R is transitive, since (1, 1) \wedge (1, 1) \to (1, 1), and while (1, 1) is reflexive, there is no ordered pair in the set where 2 R 2, so therefore R is not reflexive, even though R is symmetric and transitive.

46. Let R be a relation on a set A and suppose R is symmetric and transitive. Prove the following: If for every x in A there is a y in A such that x R y, then R is an equivalence relation.

Omitted.

47. Refer to the quote at the beginning of this section to answer the following questions.

a. What is the name of the Knight's song called?

Omitted.

b. What is the name of the Knight's song?

Omitted.

c. What is the Knight's song called?

Omitted.

d. What is the Knight's song?

Omitted.

e. What is your (full, legal) name?

Omitted.

f. What are you called?

Omitted.

g. What are you? (Do not answer this on paper; just think about it.)

Omitted.