discrete_mathematics_with_applications/chapter_8/exercises.md

Page 516

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)