Discrete Mathematics with Graph Theory (3rd Edition)
MATH discrete m

Summer 2010
376
Solutions to Review Exercises
15. [BB] Deleting vertex 2 (and the three edges incident with 2) leaves lC 5 Alternatively, cfw_ I, 4, 6, cfw_2, 3, 5 are bipartition sets for lC 3 ,3 after deleting edges 14, 1 6,46 and 35. 16. [BB] Since the graph is no
Discrete Mathematics with Graph Theory (3rd Edition)
MATH discrete m

Summer 2010
Section 0.2
7
23. By Exercise 22, there exists an integer k such that n = 4 k + 1 o r n = 4 k + 3. C ase 1: n
= 4 k + 1.
I f k is even, there exists an integer m such that k = 2 m, so n = 4 (2m) + 1 = 8 m + 1, and the desired conclusion is true. I f k is
Discrete Mathematics with Graph Theory (3rd Edition)
MATH discrete m

Summer 2010
Section 7.7
2 07
hypothesis). This sum is (ktl), b y (5). The third number is the sum o f the numbers to the left and right immediately above it; that is, (~) + This sum is ( ktl) b y (5). I n general, for any r, 1 < r :; k + 1, the r th number is the sum
Discrete Mathematics with Graph Theory (3rd Edition)
MATH discrete m

Summer 2010
306
1 (C,S) D
Solutions to Exercises
( G,2)
F
2
26. (a) Applying the first version o f Dijkstra's algorithm to the digraph produces the labels shown. Only the labels at A and B are correct. There is a negative weight cycle H GJCIH from which every vertex
Discrete Mathematics with Graph Theory (3rd Edition)
MATH discrete m

Summer 2010
302
Solutions to Exercises
(c) p =
[H ~ H
o 100 0 00010
(d) The digraphs are not strongly connected; in 91. for instance, there is no path from Vs to V 4 which respects directions and 92, being isomorphic to 91 cannot be strongly connected either. (e) The
Discrete Mathematics with Graph Theory (3rd Edition)
MATH discrete m

Summer 2010
8
32. (a) [BB] False: x = y = 0 i s a counterexample.
(b) False: a = 6 is a counterexample.
Solutions to Exercises
(c) [BB] False: x = 0 i s a counterexample. (d) False: a = (e)
Vi, b =  Vi is a counterexample. [BB] False: a = b = V i is a counterexample
Discrete Mathematics with Graph Theory (3rd Edition)
MATH discrete m

Summer 2010
208
Solutions to Exercises
(b) The formula is true i f r = n (since then both sides equal 1), so we assume r < n. Then using identity (5)
=
1 + [ (r + 2) _ ( r + 1 ) + ( r + 3 ) _ ( r + 2)
r +1 r +1 r+1 r+1
+
(rr+14) _ (rr+13) + " . + +
as desired. This f
Discrete Mathematics with Graph Theory (3rd Edition)
MATH discrete m

Summer 2010
108
Solutions to Exercises
(b) I f n = 1, n 3 + (n + 1)3 + (n + 2)3 = 13 + 23 + 33 = 36 and 9 I 36. Suppose the result is true for n = k ~ 1; that is, 9 I [k 3 + ( k + 1)3 + (k + 2)3]. Then ( k + 1)3 + ( k + 2)3 + ( k + 3)3 = ( k+ 1)3 + (k + 2)3 + k 3 + 9
Discrete Mathematics with Graph Theory (3rd Edition)
MATH discrete m

Summer 2010
Section 11.3
307
30. The answer is no, unless there are negative weight cycles. Suppose there is a walk u . . ~ . . w with a repeated vertex v. We may assume that the part o f the walk from v b ack to v (which we have denoted C) is a cycle. I f this walk
Discrete Mathematics with Graph Theory (3rd Edition)
MATH discrete m

Summer 2010
Chapter 0
9
38. We have proven in the text that y'2 is irrational. Thus, i f y'2./2 i s rational, w e are done (with a =
b = y'2). O n t he other hand, i f y'2./2 is irrational, then let a a b = y'22 = 2 i s rational.
= y'2./2 a nd b = y'2 i n which case
Discrete Mathematics with Graph Theory (3rd Edition)
MATH discrete m

Summer 2010
106
(c)
Solutions to Review Exercises
I
cfw_x E Z I X 2
 12 is divisible by 3 cfw_ x E Z I x 2  1 = 3 k for some integer k cfw_ x E Z I x I = 3 Uor some t' E Z or x + 1 (3Z + 1) U (3Z  1)
= 3 Uor some t' E
Z
(Since x 2

1
=
(x
+ l )(x 
1), i f 31 (
Discrete Mathematics with Graph Theory (3rd Edition)
MATH discrete m

Summer 2010
C hapter 4
103
(c) [BB] AIR corresponds to the number 10918. E = M 5 ( mod 391) = 259. (d) BYE corresponds to the number 22505. E
= M 5 ( mod 391) = 97. (e) N OW corresponds to the number 141523. E = M 5 ( mod 391) = 104. = 1(5)  1(4) and 1 = 5(5)  4(6)
Discrete Mathematics with Graph Theory (3rd Edition)
MATH discrete m

Summer 2010
4
Solutions to Exercises
Exercises 0.2
1. (a) [BB] Hypothesis: a and b are positive numbers.
Conclusion: a + b is positive.
(b) Hypothesis: T is a right angled triangle with hypotenuse o f length c and the other sides o f lengths a and b.
Conclusion: a2
+
Discrete Mathematics with Graph Theory (3rd Edition)
MATH discrete m

Summer 2010
Section 11.2
303
(c) The given matrices are the adjacency matrices o f the digraphs 91 and 92 shown.
The map 'P defined by
is an isomorphism 91 ~ 92. Thus, A2
= P A 1 PT where
0 0 1 0 0 1 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 1 0
P=
17. (a) [BB] Each o f thes
Discrete Mathematics with Graph Theory (3rd Edition)
MATH discrete m

Summer 2010
Section 0.2
5
(b) Converse: I f n~l is not an integer, then n is an integer. This is false: n
(n
=
1) n +1  3"
_
! is a counterexample
Contrapositive: I f n~l is an integer, then n is not an integer. This is false: n = 0 is a counterexample. Negation: T
Discrete Mathematics with Graph Theory (3rd Edition)
MATH discrete m

Summer 2010
Section 7.7
7. (a) 18 terms (b) There are two middle terms (since 17 is odd): C ;)x9(3y)8 and C ;)x 8(3y)9.
205
8. [ BB] T he required term is C~) ( 4x)3(5yf = 120(64x3)(78125y 7). T he coefficient is 6 00,000,000.
9. [ BB] T he t erm in question is (1J)x
Discrete Mathematics with Graph Theory (3rd Edition)
MATH discrete m

Summer 2010
304
Solutions to Exercises
(c) The possible outdegree sequences listed for the vertices o f the bipartition sets o f sizes 2 and 3 respectively are 3 ,3,0,0,0 2 ,2,1,1,0 1 ,1,2,1,1 3,2,0,0,1 2,1,2,1,0 1,0,2,2,1 3 ,1,0,1,1 2 ,1,1,1,1 0 ,0,2,2,2 3 ,0,1,1,1
Discrete Mathematics with Graph Theory (3rd Edition)
MATH discrete m

Summer 2010
6
17. n 2  n
Solutions to Exercises
+ 5 = n (n  1) + 5. Now either n  1 or n is even, since these integers are consecutive. So n ( n  1) is even. Since the sum o f an even integer and the odd integer 5 is odd, the result follows. + 3 = 2 (x 1)2
18. [
Discrete Mathematics with Graph Theory (3rd Edition)
MATH discrete m

Summer 2010
Section 1.1
11
16. L et the rational numbers be ~ and~. We may al!sume that a, b, e, d are positive integers and that a C a b < d ' Thus a d < be. The h int suggests that g+d IS b a d C and tho S IS the case. b < g b ' etween b and' I. b +d is equivalent
Discrete Mathematics with Graph Theory (3rd Edition)
MATH discrete m

Summer 2010
Section 1.1
(b) [ BB]
13
p
T T
F F
T he final statement is a contradiction.
,p (,p) A q ,q p V (,q) (,p) A q) A (p V (,q) T F F T F F F FF F T T F F TT T F FT F T T F column shows that ( ,p) A q) A (p V (,q) i s false for all values o f p a nd q,
Discrete Mathematics with Graph Theory (3rd Edition)
MATH discrete m

Summer 2010
Chapter 7
213
C50)
m 2520 groups of 2, 3 and 5;
=
~ (~O) (~) = 1575 groups o f 2, 4 and 4; and
~ (~) (~) = 2100 groups o f 3, 3 and 4.
B y the addition rule, 630 + 2520 + 1575 + 2100 = 6825 groups can be formed. 11. She can purchase the books in ways. F
Discrete Mathematics with Graph Theory (3rd Edition)
MATH discrete m

Summer 2010
312
Solutions to Exercises
2. (a) [BB] Since the times required for the tasks depend on which other tasks have already been completed and since tasks cannot take place at the same time, this is a Type I scheduling problem.
s
2
E (C,17)
T (E,19)
C cfw_A,14
Discrete Mathematics with Graph Theory (3rd Edition)
MATH discrete m

Summer 2010
Section 5.1
(h) When n = 1, the left side is
k 2': 1 and that
/5
113
/5 and the right side is i, so the formula holds.
Now assume that
+ 5~9 + 9.~3 + . + (4k3)\4k+1) = 4k~1' We have
1 + . + [4(k + 1)  3)][4(k + 1) + 1)] 1 1 1 + 5 9 + 9 13 + . + ( 4k + 1