Solution Manual for Connecting Discrete Mathematics and Computer Science 1st Edition By David Liben-

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Connecting Discrete Mathematics and Computer Science Solution Manual

This version: January 21, 2022

David Liben-Nowell

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2

Basic Data Types

2.2

Booleans, Numbers, and Arithmetic 2.1 112 and 201 2.2 111 and 201 2.3 18 and 39 2.4 17 and 38 2.5 Yes: there’s no fractional part in either x or y, so when they’re summed then there’s still no fractional part.

2.6 Yes: if we can write x = ba and y = dc with b ̸= 0 and d ̸= 0, then we can write x + y as ad+cb (and bd ̸= 0 because bd neither b = 0 nor d = 0).

2.7 No, and here’s a counterexample: π and 1 − π are both irrational numbers, but π + (1 − π) = 1, which is rational.

2.8 6, because ⌊2.5⌋ = 2 and ⌈3.75⌉ = 4 (and 2 + 4 = 6).

2.9 3, because ⌊3.14159⌋ = 3 and ⌈0.87853⌉ = 1 (and 3 · 1 = 3).

2.10 34 = 3 · 3 · 3 · 3 = 81, because ⌊3.14159⌋ = 3 and ⌈3.14159⌉ = 4.

2.11 The functions floor and truncate differ on negative numbers. For any real number x, we have ⌊x⌋ ≤ x—even if x is negative. That’s not true for truncation. For example, ⌊−3.14159⌋ = −4 but trunc(−3.14159) = −3 .14159 = −3.

2.12 ⌊x + 0.5⌋ 2.13 0.1 · ⌊10x + 0.5⌋

2.14 10−k · 10k x + 0.5

2.15 10−k · 10k x 2.16 If x is an integer, then ⌊x⌋ + ⌈x⌉ = 2x; if x isn’t an integer, then ⌊x⌋ + ⌈x⌉ = 2 + 3 for any x in [2, 3]. Thus the expression x − ⌊x⌋+⌈x⌉ is equivalent to x − x = 0 for x = 2 or 3 (yielding values 0 and 0), and x − 2.5 for noninteger x.

2 So the largest possible value for this expression is 0.4999999 · · · , when x = 2.9999999 · · · .

2.17 By the same logic as in Exercise 2.16, the smallest value of the expression x − ⌊x⌋+⌈x⌉ is −0.5 + ϵ, which occurs for 2 x = 2 + ϵ, for arbitrarily small values of ϵ. For example, the value is −0.4999999 when x = 2.00000001.

2.18 ⌊x⌋ 2.19 ⌈x⌉

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4

Basic Data Types

2.20 ⌈x⌉ 2.21 ⌊x⌋ 2.22 No: for example, |⌊−3.5⌋| = |−4| = 4, but ⌊|−3.5|⌋ = ⌊3.5⌋ = 3.

2.23 Yes: we have that x − ⌊x⌋ = (x + 1) − ⌊x + 1⌋ because both x and x + 1 have the same fractional part (so both x − ⌊x⌋ and (x + 1) − ⌊x + 1⌋ denote the fractional part of x).

2.24 No: if x = y = 0.5, then ⌊x⌋ + ⌊y⌋ = ⌊0.5⌋ + ⌊0.5⌋ = 0 + 0 = 0, but ⌊x + y⌋ = ⌊0.5 + 0.5⌋ = ⌊1⌋ = 1.

2.25 The key observation is that ⌈x⌉ = ⌊x⌋ when x is an integer (and otherwise ⌈x⌉ = ⌊x⌋ + 1). Thus 1 + ⌊x⌋ − ⌈x⌉ is 1 when x is an integer, and it’s 0 when x is not an integer.

2.26 Observe that ⌊ n+1 ⌋ = ⌈ n2 ⌉: it’s true when n is even, and it’s true when n is odd. Thus there are ⌈ n2 ⌉ − 1 elements less 2 than the specified entry, and n − (⌈ 2n ⌉ − 1 + 1) = n − ⌈ 2n ⌉ = ⌊ n2 ⌋ elements larger.

2.27 310 is bigger: 310 > 39 = (33 )3 = 273 , which is definitely (much) larger than 103 .

2.28 48 = (22 )8 = 216 = 65,536 1 , which is (by calculator) 0.00001525878.

2.29 1/216 = 65,536

2.30 65,536 2.31 −4 · 65,536 = −262,144 2.32 4, because 44 = 256.

2.33 2.34

√ 4 √ 4

√ 8 ≈ 1.6818, because 1.68184 ≈ 8.0001. (Alternatively, 4 8 = (23 )1/4 = 23/4 .)

512 ≈ 4.7568, because 4.75684 ≈ 511.9877. (Alternatively, 83/4 =

√ 4

83 = (29 )1/4 = 29/4 .)

2.35 Undefined (because we don’t include imaginary numbers in this book): there’s no real number x for which x4 = −9.

2.36 log2 8 = 3, because 23 = 8.

2.37 log2 (1/8) = −3, because 2−3 = 1/23 = 1/8.

2.38 log8 2 = 1/3, because 81/3 = 2 (which is true because 23 = 8).

2.39 log1/8 2 = −1/3, because (1/8)−1/3 = 81/3 = 2.

2.40 log10 17 is larger than 1 (because 101 < 17), but log17 10 is smaller than 1 (because 171 > 10). So log10 17 is larger.

2.41 By the definition of logarithms, logb 1 is the real number y such that by = 1. By Theorem 2.8.1, for any real number b we have b0 = 1, so logb 1 = 0.

2.42 By the definition of logarithms, logb b is the real number y such that by = b. By Theorem 2.8.2, for any real number b we have b1 = b, so logb b = 1.

2.43 Let q = logb x and r = logb y. By the definition of logarithms, then, bq = x and br = y. Using Theorem 2.8.3, we have x · y = bq · br = bq+r . Because xy = bq+r , we have by the definition of logs that logb xy = q + r.

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2.2 Booleans, Numbers, and Arithmetic

5

2.44 Let q = logb x. By the definition of logarithms, we have bq = x. Thus, raising both sides to the yth power, we have (bq )y = xy . We can rewrite (bq )y = bqy using Theorem 2.8.4, so xy = bqy . Again using the definition of logarithms, therefore logb xy = qy. By the definition of q, this value is y · logb x.

2.45 Let q = logc b and r = logb x, so that cq = b and br = x. Then x = (cq )r = cqr by the definition of logs and Theorem 2.8.4, and therefore logc x = qr = (logc b) · (logb x); rearranging yields the claim.

2.46 We show that b[logb x] = x by showing that logb of the two sides are equal: i h logb b[logb x] = [logb x] · logb b = [logb x] · 1 = logb x.

by Theorem 2.10.5

by Theorem 2.10.2

2.47 We show that n[logb a] = a[logb n] by showing that logb of the two sides are equal: h i logb n[logb a] = [logb a] · logb n = [logb n] · logb a h i = logb a[logb n] .

by Theorem 2.10.5 x · y = y · x for any x and y

by Theorem 2.10.5, applied “backward”

2.48 The property logb xy = logb x − logb y follows directly from the rule for the log of a product: logb xy = logb (x · y−1 )

x = x · y−1 for any x and y y

−1

= logb x + logb y = logb x + (−1) · logb y.

by Theorem 2.10.3

by Theorem 2.10.5

2.49 The hyperceiling of n can be defined as ⌈n⌉ = 2⌈log2 n⌉ 2.50 The number of columns needed to write down n in standard decimal notation is   if n > 0 ⌈log10 (n + 1)⌉ 1 if n = 0  1 + ⌈log (−n + 1)⌉ if n < 0, 10 where the “1 + ” in the last case accounts for the negative sign, which takes up a column.

2.51 202 mod 2 = 0, because 202 = 0 + 101 · 2.

2.52 202 mod 3 = 1, because 202 = 1 + (67 · 3).

2.53 202 mod 10 = 2, because 202 = 2 + (20 · 10).

2.54 −202 mod 10 = 8, because −202 = 8 + (−21 · 10).

2.55 17 mod 42 = 17, because 17 = 17 + (0 · 42).

2.56 42 mod 17 = 8, because 42 = 8 + (2 · 17).

2.57 17 mod 17 = 0, because 17 = 0 + (1 · 17).

2.58 −42 mod 17 = 9, because −42 = 9 + (−3 · 17).

2.59 −42 mod 42 = 0, because −42 = 0 + (−1 · 42).

2.60 Here is a definition of the behavior of % in Python that’s consistent with the reported behavior:

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6

Basic Data Types

• If k > 0, then n mod k is n − k · ⌊ nk ⌋.

• If k < 0, then n mod k is −(−n mod −k) = n − k · ⌊ nk ⌋.

(And Python generates a ZeroDivisionError if k = 0.)

2.61 30 2.62 1 2.63 10 2.64 68 2.65 53 2.66 A solution in Python is shown in Figure S.2.1 on p. 7.

2.67 A solution in Python is shown in Figure S.2.2 on p. 7.

2.68 A solution in Python is shown in Figure S.2.3 on p. 7. The numbers produced as output are: 4, 9, 25, 49, 121, 169, 289, 361, 529, 841, and 961. Note that these output values are 22 , 32 , 52 , 72 , 112 , 132 , . . . , 312 —the squares of all sufficiently small prime numbers. For a prime number p, the three factors of p2 are 1, p, and p2 (and p2 has no other factor).

2.69 6 + 6 + 6 + 6 + 6 + 6 = 36 2.70 1 + 4 + 9 + 16 + 25 + 36 = 91 2.71 4 + 16 + 64 + 256 + 1024 + 4096 = 5460 2.72 1(2) + 2(4) + 3(8) + 4(16) + 5(32) + 6(64) = 642 2.73 (1 + 2) + (2 + 4) + (3 + 8) + (4 + 16) + (5 + 32) + (6 + 64) = 147 2.74 6 · 6 · 6 · 6 · 6 · 6 = 46,656 2.75 1 · 4 · 9 · 16 · 25 · 36 = 518,400 2.76 22 · 24 · 26 · 28 · 210 · 212 = 242 = 4,398,046,511,104 2.77 1(2) · 2(4) · 3(8) · 4(16) · 5(32) · 6(64) = 1,509,949,440 2.78 (1 + 2) · (2 + 4) · (3 + 8) · (4 + 16) · (5 + 32) · (6 + 64) = 10,256,400 2.79 21 + 42 + 63 + 84 + 105 + 126 = 441 2.80 21 + 40 + 54 + 60 + 55 + 36 = 266 2.81 1 + 6 + 18 + 40 + 75 + 126 = 266 2.82 8 + 14 + 18 + 20 + 20 + 18 + 14 + 8 = 120 2.83 36 + 35 + 33 + 30 + 26 + 21 + 15 + 8 = 204 2.84 44 + 49 + 51 + 50 + 46 + 39 + 29 + 16 = 324

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