ДискретнаяМатематика / Student Solutions Manual / chapter 4

17.Find two functions F; G : R ! R where F 6= G but F j[0;1) = G j[0;1) .

17. F j[0;1) = Id = G j[0;1)

F jR [0;1) = 3

GjR [0;1) = 2

19. Let A; B; and C be sets, and let F : A ! C be a function. If B A, prove that F jB = F \ (B C):

19. Let x 2 B and (x; y) 2 F jB : F (x) = y 2 C: Therefore, (x; y) 2 B C: Therefore, F jB F \ (B C):

Conversely, let (x; y) 2 F \ (B C): Then (x; y) 2 F so F (x) = y with x 2 A and y 2 C: Since (x; y) 2 B C, x 2 B and y 2 C. Therefore, (x; y) 2 F jB \ (B C):

21.For each of the following functions, prove that the function is 1-1 or nd an appropriate pair of points to show the function is not 1-1.

(a) F : Z ! Z

(b) F : R ! R

F (x) =

x + 1 for x 2 Q 2x for x 62Q

(c) F : R ! R

(d) F : Z ! Z

21. (a) There are four cases: n1 > 0; n2 > 0; n1 = 0; n2 > 0; n1 = 0; n2 < 0; and n1 < 0; n2 < 0: A simple analysis of the value of the function for each kind of argument shows that the function is 1-1.

21. (b) Let x; y 2 R: Note that if x 2 Q and y 62Q, then F (x) = x + 1 2 Q but F (y) = 2y 62Q, so F (x) 6= F (y): If x 62Q and y 2 Q, then F (x) = 2x 62Q and F (y) = y + 1 2 Q, so again, F (x) 6= F (y):

Now, suppose that F (x) = F (y): By the previous argument, there are just two possibilities.

Case 1: x and y are both in Q. Then F (x) = F (y) implies that x+1 = y+1 and so x = y.

Case 2: Neither x nor y is in Q. Then F (x) = F (y) implies that 2x = 2y from which it follows that x = y.

Therefore, F is 1{1.

21.(c) Let x = 0 and y = 21=3: This shows the function is not 1{1.

21.(d) F (2) = F (7) so the function is not 1{1.

23.Construct functions with the following properties.

(a)F : N ! N such that i range(F ) = N and, for each n 2 N, there exist exactly two solutions for the equation F (x) = n:

(b)F : N ! N such that, for each n 2 N, there are exactly n solutions for the equation F (x) = n:

23. (a)

(b)F (j) = bknn+m c where j = kn + m with m 2 f0; 1; 2; : : : ; n 1g:

25.Using the numbering scheme for the letters of the alphabet as given in Section 4.1.8, encrypt the message DISCRETE MATH IS GREAT using the function F (letter) = 17(letter value) + 9 (mod 26). List the letters of the encrypted message. Find the inverse function and decrypt the message. (Hint: 23 17 = 1 (mod 26).)

25.I P D R M Z U Z F J M Y A P D H M Z J M

G(letter) = 23(letter) + 1

27.For the American history fan: Consider the list of U.S. presidents up through Harry Truman. De ne the following \function" on all presidents before Harry Truman: The successor of X is the person who followed X as president. Why is successor not a function?

27. Not a function since one person was elected twice and had two successors.

29. Let A; B; and C be sets, and let F : A ! C and G : B ! C be functions.

(a) What condition must F and G satisfy for F [ G to be a function from

A [ B to C?

(b) Give conditions on A and B such that F [ G is a function for every

F : A ! C and G : B ! C:

29. (a) x 2 A \ B ) F (x) = G(x){that is, F ja\B = GjA\B .

29.(b) A \ B = ;

31.If looked at appropriately, the de nition of a function as a set of ordered pairs and the intuitive notion that a function is something given by a rule are equivalent. Develop that equivalence here. Assume that F has a nite domain f0; 1; 2; : : : ; n 1g and a nite codomain f0; 1; 2; : : : ; m 1g.

(a)Suppose F is a function given as a set of ordered pairs. For an input x1, give a rule for calculating F (x1): Use F (or its graph) in your rule.

(b)Suppose the function F is given by a rule. Express F as a set of ordered pairs.

31. (a) F = f(x; y) : F (x) = yg:

Given x1

Draw the graph of F Now draw the line x = x1

Let y1 be the y-coordinate of intersection of the line x = x1 with the graph of F

31. (b) F is a rule. For every x 2 Dom(F ) apply the rule to x giving y. Now form the ordered pair (x; y).

4.5 Exercises

1.Let X = f1; 2; 3; 4g and Y = f5; 6; 7; 8; 9g: Let F = f(1, 5), (2, 7), (4, 9), (3, 8)g. Show that F is a function from X to Y . Find F 1 and list its elements. Is F 1 a function? Why, or why not.

1. Each element of X is the rst element of one and only one ordered pair

3.Let X = f1; 2; 3; 4g. Let F : X ! R be a function de ned as the set of ordered pairs f(1, 2), (2, 3), (3, 4), (4, 5)g. Let G : R ! R be the function de ned as G(x) = x2. What is G F ?

3. G F (x) = f(1, 4), (2, 9), (3, 16), (4, 25)g

5. De ne the functions F; G; and H as indicated in the following diagrams:

Find the following:

(a)G F

(b)H (G F )

(c)(H G) F

5.(a)

G F : 1 ! e

2g

3 g

4g

5. (b)

H (G F ) : 1 ! r

2t

3 t

4t

5. (c)

(H G) F : 1 ! r

2t

3t

4t

7.Let A = f1; 2; 3; 4g: Let the functions F , G and H be given with domain and codomain A de ned as

F (1) = 3; F (2) = 2; F (3) = 2; F (4) = 4

G(1) = 1; G(2) = 3; G(3) = 4; G(4) = 2

H(1) = 2; H(2) = 4; H(3) = 1; H(4) = 3

Find the following:

(a)F G

(b)H F

(c)G H

(d)F G H

7.

9.For sets X; Y; and Z let F : X ! Y and G : Y ! Z be 1-1 correspondences. Prove that (G F ) 1 = F 1 G 1:

9. By Theorem 2(c) G F is a 1-1 correspondence. By Theorem 3(c) (G F ) 1 is a 1-1 correspondence.

(G F ) (F 1 G 1)(z) = (G F )(F 1(G 1(z)) = G(F (F 1(G 1(z)))

= G(F F 1(G 1(z)) = G(G 1(z))

= (G G 1)(z) = IdZ

(F 1 G 1) (G F ) = IdX is proved similarly.

11.Find the rst six terms of the sequences de ned as:

(a)H(0) = 0 and H(n) = H(n 1) + n3 for n 1

(b)G(0) = 0 and G(n) = 2 G(n 1) + 1 for n 1

(c)F (0) = 2 and F (n) = 3 F (n 1) n + 3 for n 1

11. (a) 0, 1, 9, 36, 100, 225

11. (b) 0, 1, 3, 7, 15, 31

11. (c) 2, 8, 25, 75, 224, 670

13.The formal de nition of a sequence was in terms of a function F , with domain either N or f0; 1; : : : ; n 1g. (If n = 0, then f0; 1; : : : ; n 1g = ;.) The formal de nition of a subsequence involves a sequence F and a strictly increasing sequence S of elements of the domain of F . Since S is a sequence, S is, formally, another function as above. In parts (a) through

(e) of Example 7, identify the functions S and F S as sets of ordered pairs.

13. (a) S : (0; 1); F S : (0; ;)

13. (b) S : (0, 1), (1, 3), (2, 5), (3, 7), (4, 9), . . . ; F S : (0, 1), (1, 6), (2, 120), (3, 5040), . . .

13. (c) S : (0, 0), (1, 1), (2, 2), (3, 3), (4, 4), . . . F S : (0, 1), (1, 1), (2, 2), (3, 6), (4, 24), (5, 120), (6, 720), . . .

13. (d) S : (0, 0); F S : (0, 1)

13.(e) S : (0, 2), (1, 3), (2, 8); F S : (0, 2), (1, 6), (2, 40320)

15.Prove the following:

(a)Theorem 3(a)

(b)Theorem 3(c).

15. (a) ()) Suppose F 1 is a function. Show F is 1-1 and onto.

Let x1; x2 2 X and suppose F (x1) = F (x2). Put y = F (x1) = F (x2). Then, (x; y1); (x; y2) 2 F . This implies that (y; x1); (y; x2) 2 F 1> But

now, x1 = x2, because F 1 is a function. Therefore, F is a 1{1 function.

(() Assume that F is 1{1 and onto. Let y 2 Y . Since F is onto, there is at least one x 2 X such that F (x) = y{that is, such that (x; y) 2 F . Then, (y; x) 2 F 1. Now suppose that also (y; x0) 2 F 1 for some x0 2 X. The (x0; y) 2 F . Since F is 1{1, x0 = x. So, F 1(y) is uniquely de ned.

15. (b) Assume that F is a 1{1 correspondence. Then, F 1 is a function by part (a).

Let y1; y2 2 Y and suppose that F 1(y1) = F 1(y2). Put x = F (y1) = F (y2). Then

(ya; x); (y2; x) 2 F 1 ) (x; y1); (x; y2) 2 F

Therefore, y1 = y2 since F is a function. It follows that F 1 is 1{1

17.Let A and B be nonempty sets, and let F : A ! B be a function. Prove that the following are equivalent:

(a)F is 1-1.

(b)There is a function G : B ! A such that G F = IdA:

(c)For any set C and for functions H1 : C ! A and H2 : C ! A, if

F H1 = F H2; then H1 = H2:

17. We will complete this proof by means of the following series of implications: (a) ) (b) ) (c) ) (a).

(a) ) (b) Let F be an injection. We must nd a function G : B ! A such that G F = IdA: Thus we must produce a function G : B ! A such that G(F (a) = a for every a 2 A: Observe that this condition tells us the action of G on F (A); but gives us complete freedom in de ning the action of G on B F (A): Since F is injective, for each y 2 F (A) there is a unique preimage ay 2 A and in this case we must let G(y) = ay : For y 2 B F (A) we can let G(y) be any element of A; for simplicity let a0 be a xed element of the nonempty set A and let G(y) = a0 for y 2 B F (A): We have the function G de ned by the rule

Since each element of B has a uniquely determined image in A, G is a function. Clearly, G F (a) = G(F (a)) = a: Therefore, G has the required property.

(b) ) (c) Let G be a function such that G F = IdA: Let C be a set and H1 : C ! A and H2 : C ! A be functions such that F H1 = F H2.

Then,

G (F H1) = G (F H2)

The proof that IdA H = H for all functions H : C ! A is left as an exercise.

(c) ) (a) We will prove the contrapositive. Assume that (a) is false. Then, F : A ! B is not an injection, so there are x1; x2 2 A such that x1 6= x2 and F (x1) = F (x2): Let C = fx1; x2g: We de ne

Clearly, H1 6= H2: Put y = F (x1) = F (x2). Then, (F H1)(x1) = y1 and also (F H2)(x1) = (F H2)(x2) = y. Thus,F H1 = F H2, so (c) fails.

19.Let A be any nonempty set, and let FA be the set of all functions from A to R:

(a)Why is F + G 2 FA for all F; G 2 FA:

(b)Prove (F + G) + H = F + (G + H) for all F; G; H 2 FA.

(c)Let Zero 2 FA be de ned by Zero(a) = 0 for all a 2 A: Prove

Zero + F = F for all F 2 FA:

F + F = Zero = F + F for all F 2 FA:

19. The operations take place in R so the operations inherit properties from the real numbers. Speci cally:

19. (a) We must show that F + G is a function from A to R. Let a 2 A. By de nition. (F + G)(a) = F (a) + G(a) and F (A) + G(a) 2 R since the sum of two real numbers is a real number. Therefore, F + G 2 F.

19. (b) We must show that for every a 2 A, ((F + G) + H)(a) = (F + (G + H))(a): Now,

((F + G) + H)(a) = (F + G)(a) + H(a)

=(F (a) + G(a)) + H(a)

=F (a) + (G(a) + H(a))

=(F + (G + H))(a)

So, (F + G) + H = F + (G + H):

19. (c)

(Zero + F )(a) = Zero(a) + F (a) = 0 + F (a) = F (a)

for every a 2 A. So, Zero + F = F .

19. (d) Let a 2 A: Then,

(F + F )(a) = F (a) + F (a) = F (a) + ( F (a)) = 0 = Zero(a)

So F + F = Zero: Similarly, F + F = Zero.

21.(a) Let F : A ! B be a function. Prove that F is onto if and only if F 1(B1) 6= ; for each nonempty subset B1 of B.

(b). Let F : A ! B be a function. Prove that F is onto if and only if F (F 1(B1)) = B1 for all B1 B:

21.(a) Let F be onto. Let B1 B such that ; 6= B1: Let x 2 B1: F onto implies there is a y 2 A such that F (y) = x: Therefore, y 2 F 1(B1): Conversely, let x 2 B: Consider fxg B: Then, fxg 6= ; so F 1(fxg) 6= ;:

Pick y 2 F 1(fxg). Then, F (y) = x. Since x was an arbitrary element of B, F is onto.