discrete_mathematics_with_applications/chapter_7/exercises.md

Page 458

Exercise Set 7.1

1. Let X = \{1, 3, 5\} and Y = \{s, t, u, v\}. Define f: X \to Y by the following arrow diagram.

(See page 458 for image)

a. Write the domain of f and the co-domain of f.

Domain: \{1, 3, 5\}

Co-domain: \{s, t, u, v\}

b. Find f(1), f(3), and f(5).

f(1) = v, f(3) = s, f(5) = v

c. What is the range of f?

\{s, v\}

d. Is 3 an inverse image of s? Is 1 an inverse image of u?

yes; no

e. What is the inverse image of s? of u? of v?

\{3\};$\emptyset$;${1, 5}$

f. Represent f as a set of ordered pairs.

\{(1, v), (3, s), (5, v)\}

2. Let X = \{1, 3, 5\} and Y = \{a, b, c, d\}. Define g: X \to Y by the following arrow diagram.

(See page 459 for image)

a. Write the domain of g and the co-domain of g.

Domain: \{1, 3, 5\}

Co-domain: \{a, b, c, d\}

b. Find g(1), g(3), and g(5).

g(1) = b, g(3) = b, g(5) = b

c. What is the range of g?

\{b\}

d. Is 3 an inverse image of a? Is 1 an inverse image of b?

no;yes

e. What is the inverse image of b? of c?

\{1, 3, 5\}, \emptyset

f. Represent g as a set of ordered pairs.

 \{(1, b), (3, b), (5, b)\} 

3. Indicate whether the statements in parts (a)-(d) are true or false for all functions. Justify your answers.

a. If two elements in the domain of a function are equal, then their images in the co-domain are equal.

True. The definition of a function states that every input element in the domain must have an output element in the co-domain. Since two elements in the domain of the function are equal, then their outputs in the co-domain must be equal by this definition.

b. If two elements in the co-domain of a function are equal, then their preimages in the domain are also equal.

This is false. A function can have the same output for two different inputs.

c. A function can have the same output for more than one input.

True, the definition of a function only states that every input to the function must have an output, not necessarily unique outputs.

d. A function can have the same input for more than one output.

This is false. A single input can only map to a single output, not multiple outputs.

a. Find all functions from X = \{a, b\} to Y = \{u, v\}.

 f(a) = u, f(a) = v, f(b) = u, f(b) = v 

b. Find all functions from X = \{a, b, c\} to Y = \{u\}.

 f(a) = u, f(b) = u, f(c) = u 

c. Find all functions from X = \{a, b, c\} to Y = \{u, v\}.

 f(a) = u, f(a) = v, f(b) = u, f(b) = v, f(c) = u, f(c) k v 

5. Let I_{\mathbb{z}} bee the identity function defined on the set of all integers, and suppose that e, b_i^{jk}, K(t), and u_{kj} all represent integers. Find the following:

a. I_{\mathbb{Z}}(e)

 I_{\mathbb{Z}}(e) = e 

b. I_{\mathbb{Z}}\left(b_i^{jk}\right)

 I_{\mathbb{Z}}\left(b_i^{jk}\right) = b_i^{jk}\right 

c. I_{\mathbb{Z}}(K(t))

 I_{\mathbb{Z}}(K(t)) = K(t) 

d. I_{\mathbb{Z}}(u_{kj})

 I_{\mathbb{Z}}(u_{kj}) = u_{kj} 

6. Find functions defined on the set of nonnegative integers that can be used to define the sequences whose first six terms are given below.

a. 1, -\dfrac{1}{3}, \dfrac{1}{5}, -\dfrac{1}{7}, \dfrac{1}{9}, -\dfrac{1}{11}

 f: \mathbb{Z}^{\text{nonneg}} \to \mathbb{R} 

 f(n) = \frac{(-1)^n}{2n + 1} 

b. 0, -2, 4, -6, 8, -10

 f: \mathbb{Z}^{\text{nonneg}} \to \mathbb{R} 

 f(n) = (-1)^n \cdot 2n 

7. Let A = \{1, 2, 3, 4, 5\}, and define a function F: \mathscr{P}(A) \to \mathbb{Z} as follows: For each set X in \mathscr{P}(A),

F(x) = \begin{cases} 0& \text{if } X \text{ has an even number of elements} \ 1 & \text{if } X \text{ has an odd number of elements} \end{cases}

Find the following:

a. F(\{1, 3, 4\})

 F(\{1, 3, 4\}) = 1 

because \{1, 3, 4\} has an odd number of elements.

b. F(\emptyset)

 F(\emptyset) = 0 

because \emptyset has an even number of elements.

c. F(\{2, 3\})

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

because \{2, 3\} has an even number of elements.

d. F(\{2, 3, 4, 5\})

 F(\{2, 3, 4, 5\}) = 0 

because \{2, 3, 4, 5\} has an even number of elements.

8. Let J_5 = \{0, 1, 2, 3, 4\}, and define a function F: J_5 \to J_5 as follows: For each x \in J_5, F(x) = (x^3 + 2x + 4) \mod 5.

Find the following:

a. F(0)

 F(0) = ((0)^3 + 2(0) + 4) \mod 5 

 = (0 + 0 + 4) \mod 5 

 = 4 \mod 5 

 = 4 

b. F(1)

 F(1) = ((1)^3 + 2(1) + 4) \mod 5 

 = (1 + 2 + 4) \mod 5 

 = 7 \mod 5 

 = 2 

c. F(2)

 F(2) = ((2)^3 + 2(2) + 4) \mod 5 

 = (8 + 4 + 4) \mod 5 

 = 16 \mod 5 

 = 1 

d. F(3)

 F(3) = ((3)^3 + 2(3) + 4) \mod 5 

 = (27 + 6 + 4) \mod 5 

 = 37 \mod 5 

 = 2 

e. F(4)

 F(4) = ((4)^3 + 2(4) + 4) \mod 5 

 = (64 + 8 + 4) \mod 5 

 = 76 \mod 5 

 = 1 

9. Define a function S: \mathbb{Z}^+ \to \mathbb{Z}^+ as follows: For each positive integer n,

 S(n) = \text{ the sum of the positive divisors of } n 

Find the following:

a. S(1)

 S(1) = 1 

b. S(15)

 S(15) = 1 + 3 + 5 + 15 = 24 

c. S(17)

 S(17) = 1 + 17 = 18 

d. S(5)

 S(5) = 1 + 5 = 6 

e. S(18)

 S(18) = 1 + 2 + 3 + 6 + 9 + 18  = 39 

f. S(21)

 S(21) = 1 + 3 + 7 + 21 = 32 

10. Let D be the set of all finite subsets of positive integers.

Define a function T: \mathbb{Z}^+ \to D as follows: For each positive integer n, T(n) = the set of positive divisors of n.

Find the following:

a. T(1)

 T(1) = \{1\} 

b. T(15)

 T(15) = \{1, 3, 5, 15\} 

c. T(17)

 T(17) = \{1, 17\} 

d. T(5)

 T(5) = \{1, 5\} 

e. T(18)

 T(18) = \{1, 2, 3, 6, 9, 18\} 

f. T(21)

 T(21) = \{1, 3, 7, 21\} 

11. Define F: \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z} \times \mathbb{Z} as follows: For every ordered pair (a, b) of integers, F(a, b) = (2a + 1, 3b - 2).

Find the following:

a. F(4, 4)

 F(4, 4) = (2(4) + 1, 3(4) - 2) 

 = (8 + 1, 12 - 2) 

 = (9, 10) 

b. F(2, 1)

 F(2, 1) = (2(2) + 1, 3(1) - 2) 

 = (4 + 1, 3 - 2) 

 = (5, 1) 

c. F(3, 2)

 F(3, 2) = (2(3) + 1, 3(2) - 2) 

 = (6 + 1, 6 - 2) 

 = (7, 4) 

d. F(1, 5)

 F(1, 5) = (2(1) + 1, 3(5) - 2) 

 = (2 + 1, 15 - 2) 

 = (3, 13) 

12. Let J_5 = \{0, 1, 2, 3, 4\}, and define G: J_5 \times J_5 \to J_5 \times J_5 as follows: For each (a, b) \in J_5 \times J_5,

 G(a, b) = ((2a + 1) \mod 5, (3b - 2) \mod 5) 

Find the following:

a. G(4, 4)

 G(4, 4) = ((2(4) + 1) \mod 5, (3(4) - 2) \mod 5) 

 = ((8 + 1) \mod 5, (12 - 2) \mod 5) 

 = (9 \mod 5, 10 \mod 5) 

 = (4, 0) 

b. G(2, 1)

 G(2, 1) = ((2(2) + 1) \mod 5, (3(1) - 2) \mod 5) 

 = ((4 + 1) \mod 5, (3 - 2) \mod 5) 

 = (5 \mod 5, 1 \mod 5) 

 = (0, 1) 

c. G(3, 2)

 G(3, 2) = ((2(3) + 1) \mod 5, (3(2) - 2) \mod 5) 

 = ((6 + 1) \mod 5, (6 - 2) \mod 5) 

 = (7 \mod 5, 4 \mod 5) 

 = (2, 4) 

d. G(1, 5)

 G(1, 5) = ((2(1) + 1) \mod 5, (3(5) - 2) \mod 5) 

 = ((2 + 1) \mod 5, (15 - 2) \mod 5) 

 = (3 \mod 5, 13 \mod 5) 

 = (3, 3) 

13. Let J_5 = \{0, 1, 2, 3, 4\}, and define functions f: J_5 \to J_5 and g: J_5 \to J_5 as follows: For each x \in J_5,

 f(x) = (x + 4)^2 \mod 5 \quad \text{ and } \quad g(x) = (x^2 + 3x + 1) \mod 5 

Is f = g? Explain.

x	f(x)	g(x)
0	1	1
1	0	0
2	1	1
3	4	4
4	4	4

The table shows that f(x) = g(x) for every x \in J_5. Therefore f = g by definition of equality of functions.

14. Define functions H and K from \mathbb{R} to \mathbb{R} by the following formulas:

For every x \in \mathbb{R},

 H(x) = \lfloor x \rfloor + 1 \quad \text{ and } \quad K(x) = \lceil x \rceil 

Does H = K? Explain.

No. For example say x = 0, then H(0) = \lfloor 0 \rfloor + 1 = 0 + 1 = 1 and K(0) = \lceil 0 \rceil = 0. Therefore it cannot be said that for every x \in \mathbb{R} that H(x) = K(x), and thus H \neq K.

15. Let F and G be functions from the set of all real numbers to itself. Define the product functions F \cdot G: \mathbb{R} \to \mathbb{R} and G \cdot F: \mathbb{R} \to \mathbb{R} as follows: For every x \in \mathbb{R},

 (F \cdot G)(x) = F(x) \cdot G(x) 

 (G \cdot F)(x) = G(x) \cdot F(x) 

Does F \cdot G = G \cdot F? Explain.

Yes, by the commutative law of multiplication of Real numbers:

 (F \cdot G)(x) = F(x) \cdot G(x) = G(x) \cdot F(x) = (G \cdot F)(x) 

Therefore, since (F \cdot G)(x) = (G \cdot F)(x) for all x \in \mathbb{R}, it can be concluded that F \cdot G = G \cdot F by the definition of equality of functions.

16. Let F and G be function sfrom the set of all real numbers to itself. Define new functions F - G: \mathbb{R} \to \mathbb{R} and G - F: \mathbb{R} \to \mathbb{R} as follows: For every x \in \mathbb{R},

 (F - G)(x) = F(x) - G(x) 

 (G - F)(x) = G(x) - F(x) 

Does F - G = G - F? Explain.

No. Consider the definition of the difference of sets:

 (F - G)(x) = F(x) - G(x) = F(x) 

and:

 (G - F)(x) = G(x) - F(x) = G(x) 

Since F(x) \neq G(x) for all x \in \mathbb{R}, it can be concluded that F - G \neq G - F by the definition of the equality of functions.

17. Use the definition of logarithm to fill in the blanks below.

a. \log_28 = 3 because _____.

 2^3 = 8 

b. \log_5\left(\dfrac{1}{25}\right) = -2 because _____.

 5^{-2} = \frac{1}{5^2} = \frac{1}{25} 

c. \log_44 = 1 because _____.

 4^1 = 4 

d. \log_3(3^n) = n because _____.

 3^n = 3^n 

e. \log_41 = 0 because _____.

 4^0 = 1 

18. Find exact values for each of the following quantities without using a calculator.

a. \log_{3}81

 3^{\text{?}} = 81 

 \log_{3}81 = 4 

b. \log_{2}1024

 2^{\text{?}} = 1024 

 \log_{2}1024 = 10 

c. \log_{3}\left(\dfrac{1}{27}\right)

 \log_{3}\left(\frac{1}{27}\right) = -3 

d. \log_{2}1

 \log_{2}1 = 0 

e. \log_{10}\left(\dfrac{1}{10}\right)

 \log_{10}\left(\dfrac{1}{10}\right) = -1 

f. \log_{3}3

 \log_{3}3 = 1 

g. \log_{2}(2^k)

\log_{2}(2^k) = k 

19. Use the definition of logarithm to prove that for any positive real number b with b \neq 1, \log_{b}b = 1.

Proof:

Let b be any positive real number with b \neq 1. Since b^1 = b, then \log_{b}b = 1 by definition of logarithm.

Q.E.D.

20. Use the definition of logarithm to prove that for any positive real number b with b \neq 1, \log_{b}1 = 0.

Proof:

Let b be any positive real number with b \neq 1. Since b^0 = 1, then \log_{b}1 = 0 by definition of logarithm.

Q.E.D.

21. If b is any positive real number with b \neq 1 and x is any real number, b^{-x} is defined as follows:

b^{-x} = \dfrac{1}{b^x}. Use this definition and the definition of logarithm to prove that \log_{b}\left(\dfrac{1}{u}\right) = -\log_{b}u for all positive real numbers u and b, with b \neq 1.

Proof:

Let b be any positive real number with b \neq 1. Let u be any positive real number.

Let v = \log_{b}\left(\dfrac{1}{u}\right). By the definition of logarithm, this means that b^v = \dfrac{1}{u}. It follows by algebra that:

 b^v = \frac{1}{u} 

 u \cdot b^v = 1 

 u = \frac{1}{b^v} 

 u = b^{-v} 

Hence, by the definition of logarithm:

 -v = \log_{b}(u) 

and by algebra:

 v = -\log_{b}(u) 

Since v = \log_{b}\left(\dfrac{1}{u}\right) and v = -\log_{b}(u), it follows by the definition of equality that:

 \log_{b}\left(\frac{1}{u}\right) = -\log{b}(u) 

This is what was to be shown.

Q.E.D.

22. Use the unique factorization for the integers theorem (Section 4.4) and the definition of logarithm to prove that \log_{3}(7) is irrational.

Hint: Use a proof by contradiction. Suppose \log_{3}7 is rational. Then \log_{3}7 = \dfrac{a}{b} for some integers a and b with b \neq 0.

Apply the definition of logarithm and rewrite \log_{3}7 = \dfrac{a}{b} in exponential form.

Proof (by contradiction):

Suppose \log_{3}(7) is rational, that is \log_{3}(7) = \dfrac{a}{b} for some integers a and b where b \neq 0.

By the definition of logarithm, this would mean that:

 3^{\frac{a}{b}} = 7 

Then by algebra:

 3^a = 7^b 

Since b \neq 0, we know that 7^b \neq 1, and by equality it follows that 3^a \neq 1. Additionally, by the definition of exponentiation, it is known that 7^b > 0 and 3^a > 0 (they are both positive numbers).

But, by the unique factorization for integers theorem, this means that 7^b and 3^a are two different prime factorizations of the same positive integer. This is only possible if the positive integer is equal to 1.

Hence 3^a = 7^b = 1, but it has already been established that 3^a = 7^b \neq 1. This is a contradiction.

Therefore the supposition is false, and \log_{3}(7) is irrational.

Q.E.D.

23. If b and y are positive real numbers such that \log_{b}y = 3, what is \log_{\frac{1}{b}}y? Explain.

Proof:

Suppose b and y are positive real numbers such that \log_{b}y = 3.

By the definition of logarithm, this means that:

 b^3 = y 

To find \log_{\frac{1}{b}}y, first, replace y by substitution:

 \log_{\frac{1}{b}}y 

 = \log_{\frac{1}{b}}(b^3) 

Then notice that \dfrac{1}{b} = b^{-1}, and then substitute:

 = \log_{b^{-1}}(b^3) 

By the definition of logarithm, this means that:

 (b^{-1})^x = b^3 

Where x is \log_{\frac{1}{b}}y, or our answer. By the multiplication of exponents, this means that:

 b^{-1 \cdot x} = b^3 

And by multiplication of negative numbers:

 b^{-1 \cdot -3} = b^3 

Therefore x = -3, or:

 \log_{\frac{1}{b}}y = -3 

This is what was to be found.

Q.E.D.

24. If b and y are positive real numbers such that \log_{b}y = 2, what is \log_{b^2}(y)? Explain.

Proof:

Suppose b and y are positive real numbers such that \log_{b}y = 2. By the definition of logarithm, this means that:

 \log_{b}y = 2 

 b^2 = y 

To find \log_{b^2}(y), first substitute in for y:

 \log_{b^2}(b^2) 

By the definition of logarithm, this means that:

 \log_{b^2}(b^2) = 1 

because (b^2)^1 = b^2.

This is what was to be found.

Q.E.D.

25. Let A = \{2, 3, 5\} and B = \{x, y\}. Let p_1 and p_2 be the projections of A \times B onto the first and second coordinates. That is, for each pair (a, b) \in A \times B, p_1(a, b) = a and p_2(a, b) = b.

a. Find p_1(2, y) and p_1(5, x). What is the range of p_1?

 p_1(2, y) = 2 

 p_1(5, x) = 5 

Range of p_1:

 \{2, 3, 5\} 

b. Find p_2(2, y) and p_2(5, x). What is the range of p_2?

 p_2(2, y) = y 

 p_2(5, x) = x 

Range of p_2:

 \{x, y\} 

26. Observe that \mod and \text{div} can be defined as functions from \mathbb{Z}^{\text{nonneg}} \times \mathbb{Z}^+$ to \mathbb{Z}. For each ordered pair (n, d) consisting of a nonnegative integer n and a positive integer d, let

\mod(n, d) = n \mod d (the nonnegative remainder obtained when n is divided by d).

\text{div}(n, d) = n \text{ div } d (the integer quotient obtained when n is divided by d).

Find each of the following:

a. \mod(67, 10) and \text{div}(67, 10)

 \mod(67, 10) = 7 

 \text{div}(67, 10) = 6 

b. \mod(59, 8) and \text{div}(59, 8)

 \mod(59, 8) = 3 

 \text{div}(59, 8) = 7 

c. \mod(30, 5) and \text{div}(30, 5)

 \mod(30, 5) = 0 

 \text{div}(30, 5) = 6 

27. Let S be the set of all strings of $a$'s and $b$'s.

a. Define f: S \to \mathbb{Z} as follows: For each string s in S

f(s) = \begin{cases} & \text{ the number of b's to the left-most a in s} \ 0 & \text{if s contains no a's} \end{cases}

Find f(aba), f(bbab), and f(b). What is the range of f?

 f(aba) = 0 

 f(bbab) = 2 

 f(b) = 0 

The range of f: \mathbb{Z}^{\text{nonneg}}

b. Define g: S \to S as follows: For each string s in S,

 g(s) = \text{ the string obtained by writing the characters of s in reverse order} 

Find g(aba), g(bbab), and g(b). What is the range of g?

 g(aba) = aba 

 g(bbab) = babb 

The range of g is S.

28. Consider the coding and decoding functions E and D defined in Example 7.1.9.

a. Find E(0110) and D(111111000111).

 E(0110) = 000111111000 

 D(111111000111) = 1101 

b. Find E(1010) and D(000000111111).

 E(1010) = 111000111000 

 D(000000111111) = 0011 

29. Consider the Hamming distance function defined in Example 7.1.10.

a. Find H(10101, 00011).

 H(10101, 00011) = 3 

b. Find H(00110, 10111).

 H(00110, 10111) = 2 

30. Draw arrow diagrams for the Boolean functions defined by the following input/output tables.

a.

Input	Intput	Output
P	Q	R
-------	-
1	1	0
1	0	1
0	1	0
0	0	1

Omitted.

b.

Input	Intput	Input	Output
P	Q	R	S
-	-	-	-
1	1	1	1
1	1	0	0
1	0	1	1
1	0	0	1
0	1	1	0
0	1	0	0
0	0	1	0
0	0	0	1

Omitted.

31. Fill in the following table to show the values of all possible two-place Boolean functions.

Input	Input	f_1	f_2	f_3	f_4	f_5	f_6	f_7	f_8	f_9	f_{10}	f_{11}	f_{12}	f_{13}	f_{14}	f_{15}	f_{16}
1	1	0	0	0	0	0	0	0	0	1	1	1	1	1	1	1	1
1	0	0	0	0	0	1	1	1	1	0	0	0	0	1	1	1	1
0	1	0	0	1	1	0	0	1	1	0	0	1	1	0	0	1	1
0	0	0	1	0	1	0	1	0	1	0	1	0	1	0	1	0	1

32. Consider the three-place Boolean function f defined by the following rule: For each triple (x_1, x_2, x_3) of $0$'s and $1$'s,

 f(x_1, x_2, x_3) = (4x_1 + 3x_2 + 2x_3) \mod 2 

a. Find f(1, 1, 1) and f(0, 0, 1).

 f(1, 1, 1) = (4(1) + 3(1) + 2(1)) \mod 2 

 f(1, 1, 1) = (4 + 3 + 2) \mod 2 

 f(1, 1, 1) = 9 \mod 2 

 f(1, 1, 1) = 1 

 f(0, 0, 1) = (4(0) + 3(0) + 2(1)) \mod 2 

 f(0, 0, 1) = (0 + 0 + 2) \mod 2 

 f(0, 0, 1) = 2 \mod 2 

 f(0, 0, 1) = 0 

b. Describe f using an input/output table.

x_1	x_2	x_3	f(x_1, x_2, x_3)
0	0	0	0
0	0	1	0
0	1	0	1
0	1	1	1
1	0	0	0
1	0	1	0
1	1	0	1
1	1	1	1

33. Student A tries to define a function g: \mathbb{Q} \to \mathbb{Z} by the rule

g\left(\dfrac{m}{n}\right) = m - n, for all integers m and n with n \neq 0.

Student B claims that g is not well defined. Justify student B's claim.

Suppose \dfrac{m}{n} = \dfrac{1}{2}, this would mean that g\left(\dfrac{m}{n}\right) = 1 - 2 = -1.

Since \dfrac{m}{n} = \dfrac{1}{2}, this means that \dfrac{m}{n} = \dfrac{1}{2} = \dfrac{2}{4}. Since they are equivalent, this means that g\left(\dfrac{1}{2}\right) = g\left(\dfrac{2}{4}\right) = 2 - 4 = -2.

But notice that:

 g\left(\frac{1}{2}\right) = -1 \neq -2 = g\left(\frac{2}{4}\right) 

Since the function g gives two different outputs for the same input, the function g is not well defined.

34. Student C tries to define a function h: \mathbb{Q} \to \mathbb{Q} by the rule

h\left(\dfrac{m}{n}\right) = \dfrac{m^2}{n}, for all integers m and n with n \neq 0.

Student D claims that h is not well defined. Justify student D's claim.

Suppose \dfrac{m}{n} = \dfrac{2}{3}, then h\left(\dfrac{2}{3}\right) = \dfrac{(2)^2}{3} = \dfrac{4}{3}.

Notice that \dfrac{2}{3} = \dfrac{4}{6}, so h\left(\dfrac{4}{6}\right) = \dfrac{(4)^2}{6} = \dfrac{16}{6} = \dfrac{8}{3}.

Notice that:

 h\left(\frac{2}{3}\right) = \frac{4}{3} \neq \frac{8}{3} = h\left(\frac{4}{6}\right) 

Since the function h does not produce the same output given the same input, the function is not well defined.

35. Let U = \{1, 2, 3, 4\}. Student A tries to define a function R: U \to \mathbb{Z} as follows: For each x \in U,

R(x) is the integer y so that (xy) \mod 5 = 1.

Student B claims that R is not well defined. Who is correct: student A or student B? Justify your answer.

Consider R(3) = 2 since (3 \cdot 2) \mod 5 = 1. On the other hand, R(3) = 7 since (3 \cdot 7) \mod 5 = 1.

Since R returns multiple outputs for the same input, it is not well defined, and Student B is correct.

36. Let V = \{1, 2, 3\}. Student C tries to define a function S: V \to V as follows: For each x \in V,

S(x) is the integer y in V so that (xy) \mod 4 = 1.

Student D claims that S is not well defined. Who is right: student C or student D? Justify your answer.

Consider S(1) = 17 since (1 \cdot 17) \mod 4 = 1. On the other hand S(1) = 13 since (1 \cdot 13) \mod 4 = 1.

Since S returns multiple outputs for the same input, it is not well defined, and Student D is correct.

37. On certain computers the integer data type goes from -2,147,483,648 through 2,147,483,647. Let S be the set of all integers from -2,147,483,648 through 2,147,483,647. Try to define a function f: S \to S by the rule f(n) = n^2 for each n in S. Is f well defined? Explain.

No, 2,147,483,247 = 2^{31} - 1, so for values of n greater than 2^{16}, f(n) = n^2 will be greater than 2^{32}, which falls outside of S.

38. Let X = \{a, b, c\} and Y = \{r, s, t, u, v, w\}. Define f: X \to Y as follows: f(a) = v, f(b) = v, and f(c) = t.

a. Draw an arrow diagram for f.

Omitted.

b. Let A = \{a, b\}, C = \{t\}, D = \{u, v\}, and E = \{r, s\}. Find f(A), f(X), f^{-1}(C), f^{-1}(D), f^{-1}(E), and f^{-1}(Y).

 f(A) = \{v\} 

 f(X) = $\{t, v\} 

 f^{-1}(C) = \{c\} 

 f^{-1}(D) = \{a, b\} 

 f^{-1}(E) = \emptyset 

 f^{-1}(Y) = \{a, b, c\} 

39. Let X = \{1, 2, 3, 4\} and Y = \{a, b, c, d, e\}. Define g: X \to Y as follows: g(1) = a, g(2) = a, g(3) = a, and g(4) = d.

a. Draw an arrow diagram for g.

Omitted.

b. Let A = \{2, 3\}, C = \{a\}, and D = \{b, c\}. Find g(A), g(X), g^{-1}(C), g^{-1}(D), and g^{-1}(Y).

 g(A) = \{a\} 

 g(X) = \{a, d\} 

 g^{-1}(C) = \{1, 2, 3\} 

 g^{-1}(D) = \emptyset 

 g^{-1}(Y) = \{1, 2, 3, 4\} 

40. Let X and Y be sets, let A and B be any subsets of X, and let F be a function from X to Y. Fill in the blanks in the following proof that F(A) \cup F(B) \subseteq F(A \cup B).

Proof:

Let y be any element in F(A) \cup F(B). [We must show that y is in F(A \cup B).] By definition of union, __ (i) __.

Case 1 y \in F(A):

In this case, by definition of F(A), y = F(x) for __ (ii) __ x \in A. Since A \subseteq A \cup B, it follows from the definition of union that x \in __ (iii) __. Hence, y = F(x) for some x \in A \cup B, and thus, by definition of F(A \cup B), y \in __ (iv) __.

Case 2, y \in F(B):

In this case, by definition of F(B), __ (v) __ for some x \in B. Since B \subseteq A \cup B it follows from the definition of union that __ (vi) __. Thus y \in F(A \cup B).

Therefore, regardless of whether y \in F(A) or y \in F(B), we have that y \in F(A \cup B) [as was to be shown].

i. y \in F(A) \cup F(B)

ii. some

iii. A \cup B

iv. F(A \cup B)

v. y = F(x)

vi. x \in A \cup B

In 41-49 let X and Y be sets, let A and B be any subsets of X, and let C and D be any subsets of Y. Determine which of the properties are true for every function F from X to Y and which are false for at least one function F from X to Y. Justify your answers.

41. If A \subseteq B then F(A) \subseteq F(B)

Proof:

Let F be a function from X to Y and suppose A \subseteq X, B \subseteq X, and A \subseteq B.

Then, let y be some element such that y \in F(A).

By definition of image of a set, y = F(x) for some x \in A. Thus since A \subseteq B, x \in B, and so y = F(x) for some x \in B. Hence y \in F(B), and therefore F(A) \subseteq F(B).

Q.E.D.

42. F(A \cap B) \subseteq F(A) \cap F(B)

Proof:

Suppose y is some element such that y \in F(A \cap B).

By the supposition and the definition of A \cap B, this means that y = F(x) for some x \in A \cap B.

By the definition of intersection, it follows that x \in A and x \in B.

By the definition of F(A) and F(B), y = F(x) is in F(A) and in F(B).

Hence, by the definition of intersection, y \in F(A) \cap F(B).

Since y \in F(A) \cap F(B), it can be concluded that F(A \cap B) \subseteq F(A) \cap F(B).

Q.E.D.

43. F(A) \cap F(B) \subseteq F(A \cap B)

Disproof (by counterexample):

Let X = \{1, 2, 3\} and Y = \{a, b\}. Then, define a function F: X \to Y such that F(1) = a, F(2) = b, F(3) = b.

Let A = \{1, 2\} and B = \{1, 3\}. Then F(A) = \{a, b\} and F(B) = \{a, b\}.

So F(A) \cap F(B) = \{a, b\}, and F(A \cap B) = F(\{1\}) = \{a\}.

Since \{a\} \neq \{a, b\}, the given statement is false.

Q.E.D.

44. For all subsets A and B of X, F(A - B) = F(A) - F(B).

Disproof (by counterexample):

Let X = \{1, 2\} and Y = \{a\}. Then, define a function F: X \to Y such that F(1) = a and F(2) = a.

Let A = \{1\} and B = \{2\}. Then F(A - B) = F(\{1\}) = \{a\}.

Then F(A) - F(B) = F(\{1\}) - F(\{2\}) = \{a\} - \{a\} = \emptyset.

Since \{a\} \neq \emptyset, the given statement is false.

Q.E.D.

45. For all subsets C and D of Y, if C \subseteq D, then F^{-1}(C) \subseteq F^{-1}(D).

Proof:

Let F be a function from a set X to a set Y, and suppose C \subseteq Y, D \subseteq Y, and C \subseteq D.

Suppose x \in F^{-1}(C). Then F(x) \in C. Since C \subseteq D, F(x) \in D also. Hence, by definition of inverse image, x \in F^{-1}(D). Therefore F^{-1}(C) \subseteq F^{-1}(D).

Q.E.D.

46. For all subsets C and D of Y,

 F^{-1}(C \cup D) = F^{-1}(C) \cup F^{-1}(D) 

Proof:

In order to prove:

 F^{-1}(C \cup D) = F^{-1}(C) \cup F^{-1}(D) 

We must prove:

 F^{-1}(C \cup D) \subseteq F^{-1}(C) \cup F^{-1}(D) 

and:

 F^{-1}(C) \cup F^{-1}(D) \subseteq F^{-1}(C \cup D) 

Proof F^{-1}(C \cup D) \subseteq F^{-1}(C) \cup F^{-1}(D):

Suppose x \in F^{-1}(C \cup D). Then F(x) \in C \cup D. By the definition of union, this means that F(x) \in C or F(x) \in D.

Case F(x) \in C:

Since F(x) \in C, this means that x \in F^{-1}(C). By the definition of union, this means that x \in F^{-1}(C) \cup F^{-1}(D).

Case F(x) \in D:

Since F(x) \in D, this means that x \in F^{-1}(D). By the definition of union, this means that x \in F^{-1}(C) \cup F^{-1}(D).

In both cases, x \in F^{-1}(C) \cup F^{-1}(D). Therefore, any element in F^{-1}(C \cup D) is also in F^{-1}(C), and F^{-1}(C \cup D) \subseteq F^{-1}(C) \cup F^{-1}(D), as was to be shown.

Proof F^{-1}(C) \cup F^{-1}(D) \subseteq F^{-1}(C \cup D):

Suppose x \in F^{-1}(C) \cup F^{-1}(D). By definition of union, this means that x \in F^{-1}(C) or x \in F^{-1}(D).

Case x \in F^{-1}(C):

Since x \in F^{-1}(C), this means that F(x) \in C. It follows by definition of union that F(x) \in C \cup D, or x \in F^{-1}(C \cup D).

Case x \in F^{-1}(D):

Since x \in F^{-1}(D), this means that F(x) \in D. It follows by definition of union that F(x) \in C \cup D, or x \in F^{-1}(C \cup D).

In both cases, x \in F^{-1}(C \cup D). Therefore any element in F^{-1}(C) \cup F^{-1}(D) is in F^{-1}(C \cup D), and so F^{-1}(C) \cup F^{-1}(D) \subseteq F^{-1}(C \cup D). This is what was to be shown.

Conclusion:

Since both subset relations have been proved, it can be concluded that F^{-1}(C \cup D) = F^{-1}(C) \cup F^{-1}(D). This is what was to be shown.

Q.E.D.

47. For all subsets C and D of Y,

 F^{-1}(C \cap D) = F^{-1}(C) \cap F^{-1}(D) 

Proof:

In order to prove:

 F^{-1}(C \cap D) = F^{-1}(C) \cap F^{-1}(D) 

it must be shown that:

 F^{-1}(C \cap D) \subseteq F^{-1}(C) \cap F^{-1}(D) 

and also that:

 F^{-1}(C) \cap F^{-1}(D) \subseteq F^{-1}(C \cap D) 

Proof F^{-1}(C \cap D) \subseteq F^{-1}(C) \cap F^{-1}(D):

Suppose x \in F^{-1}(C \cap D), or F(x) \in C \cap D. By definition of intersection, this means that F(x) \in C and F(x) \in D, or x \in F^{-1}(C) \cap F^{-1}(D). This is what was to be shown.

Proof F^{-1}(C) \cap F^{-1}(D) \subseteq F^{-1}(C \cap D):

Suppose x \in F^{-1}(C) \cap F^{-1}(D), or F(x) \in C and F(x) \in D. By definition of intersection, F(x) \in C \cap D, or x \in F^{-1}(C \cap D). This is what was to be shown.

Conclusion:

Since both subset relations have been proved, it can be concluded that F^{-1}(C \cap D) = F^{-1}(C) \cap F^{-1}(D). This is what was to be shown.

Q.E.D.

48. For all subsets C and D of Y,

 F^{-1}(C - D) = F^{-1}(C) - F^{-1}(D) 

Proof:

In order to prove:

 F^{-1}(C - D) = F^{-1}(C) - F^{-1}(D) 

it must be shown that:

 F^{-1}(C - D) \subseteq F^{-1}(C) - F^{-1}(D) 

and also that:

 F^{-1}(C) - F^{-1}(D) \subseteq F^{-1}(C - D) 

Proof F^{-1}(C - D) \subseteq F^{-1}(C) - F^{-1}(D):

Suppose x \in F^{-1}(C - D), or F(x) \in C - D. By definition of difference of sets, this means that F(x) \in C and F(x) \notin D. By the definition of inverse image, this means x \in F^{-1}(C) and x \notin F^{-1}(D). By the definition of difference, this is x \in F^{-1}(C) - F^{-1}(D). Thus F^{-1}(C - D) \subseteq F^{-1}(C) - F^{-1}(D), which is what was to be shown.

Proof F^{-1}(C) - F^{-1}(D) \subseteq F^{-1}(C - D):

Suppose x \in F^{-1}(C) - F^{-1}(D), or F(x) \in C and F(x) \notin D. By the definition of inverse image, this means that F(x) \in C - D, or x \in F^{-1}(C - D). Thus F^{-1}(C) - F^{-1}(D) \subseteq F^{-1}(C - D), which is what was to be shown.

Conclusion:

Since both subset relations have been proved, it can be concluded that F^{-1}(C - D) = F^{-1}(C) - F^{-1}(D), which is what was to be shown.

Q.E.D.

49. F(F^{-1}(C)) \subseteq C

Proof:

Suppose x \in F(F^{-1}(C)). By definition of image, there exists some a \in F^{-1}(C) such that F(a) = x. By definition of inverse image, a \in F^{-1}(C) means F(a) \in C. Since F(a) = x, we have x \in C. Therefore F(F^{-1}(C)) \subseteq C.

Q.E.D.

50. Given a set S and a subset A, the characteristic function of $A$, denoted \chi_A, is the function defined from S to \mathbb{Z} with the property that for each u \in S,

\chi_{A}(u) = \begin{cases} 1 & \text{if } u \in A \ 0 & \text{if } u \notin A \end{cases}

Show that each of the following holds for all subsets A and B of S and every u \in S.

a. \chi_{A \cap B}(u) = \chi_{A}(u) \cdot \chi_{B}(u)

Omitted.

b. \chi_{A \cup B}(u) = \chi_{A}(u) + \chi_{B}(u) - \chi_{A}(u) \cdot \chi_{B}(u)

Omitted.

Each of exercises 51-53 refers to the Euler phi function, denoted \phi, which is defined as follows: For each integer n \geq 1, \phi(n) is the number of positive integers less than or equal to n that have no common factors with n except \pm 1. For example \phi(10) = 4 because there are four positive integers less than or equal to 10 that have no common factors with 10 except \pm 1 - namely, 1, 3, 7, and 9.

51. Find each of the following:

a. \phi(15)

Omitted.

b. \phi(2)

Omitted.

c. \phi(5)

Omitted.

d. \phi(12)

Omitted.

e. \phi(11)

Omitted.

f. \phi(1)

Omitted.

52. Prove that if p is a prime number and n is an integer with n \geq 1, then \phi(p^n) = p^n - p^{n - 1}.

Omitted.

53. Prove that there are infinitely many integers n for which \phi(n) is a perfect square.

Omitted.