Exercise 1 - Positional Number Systems - Solutions PDF

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MATH 1310 – Technical Math for IT

EXERCISE 1 SOLUTIONS Positional Number Systems TERMINOLOGY 1. Exactly how many bits are in each of the following? a. 2 kilobits b. 64 KB

2048 bits

d.

524 288 bits

c. 400 MB

1.2 gigabits

e. 0.5 TB

1 288 490 188.8 bits

4 398 046 511 104 bits

3 355 443 200 bits

BASE CONVERSION 2. Convert each of the following numbers from the given base to an equivalent decimal representation. a. (1011001)2

= (89)10

= (11851)10

b. (2E4B)16 c. (372)8

= (250)10 = (16965)10

d. (212021100)3

= (0.328125)10

e. (0.010101)2 f. (31.72)8

= (25.90625)10

g. (102.0201102)3

h. (2C.3D)16

524

= 11 2187 ≈ 11.2395976223

= (44.23828125)10

3. Convert each of the following decimal numbers to an equivalent representation in the indicated base. a. 45 to binary

b. 1463 to binary

= (101101)2

= (10110110111)2

MATH 1310 – Technical Math for IT

c. 54 to base 3

= (2000)3

d. 676 to octal

= (1244)8 = (1417)16

e. 5143 to hexadecimal

f. 98.25 to binary

= (1100010.01)2

g. 3.40625 to binary

= (11.01101)2

h. 0.474609375 to octal

= (0.363)8

i. 972.815185546875 to hexadecimal

j. 0.1 to binary

= (3CC.D0B)16

= (0.00011)2

k. 0.2 to binary Let x = 0.2 10x = 2.2 Subtract the first line from the second: 9x = 2 so x = 29 Perform the steps of the conversion : algorithm using this fraction: 2 × 2= 49 integer part 0 9 4 × 2= 89 integer part 0 9 8 × 2= 169 integer part 1 9 Reset integer part to 0 (subtract 1): 7 × 2= 149 integer part 1 9 Reset integer part to 0 (subtract 1): 5 × 2= 109 integer part 1 9 Reset integer part to 0 (subtract 1): 1 × 2= 92 inte ger part 0 9 The steps in the algorithm will : now start repeating infinitely: 0.2 = 0.001110

( ) ( 10

)

2

MATH 1310 – Technical Math for IT

4. Convert each of the following numbers from the base in which they are given to the indicated base. a. (39)16 to base 7 = (57)10 = (111)7 b. (1745372.25137)8 to binary = (1111100101011111010.010101001011111)2 c. (22101022.2011202)3 to base 9 = (8338.6466)9 Note that 9 = 32 so we can use a shortcut conversion. Starting at the radix point, group the digits of the base-3 number into groups of two and convert each group to a single base-9 digit. Pad the base-3 number with leading and trailing zeros as necessary. d. (101110100010.1010011)2 to hexadecimal = (BA2.A6)16 e. (B2.4C)16 to octal = (10110010.010011)2 = (262.23)8 f. (143.203)5 to binary (CHALLENGE QUESTION) = (48.424)10 = (110000. 0.01101100100010110100001110010101100000010000011000100100110111 010010111100011010100111111011111001110 )2 Fractional part repeats after 100 digits. 5. Determine the loss of accuracy (express your answer in each case as a decimal number) in the following problems: a. Determine the loss of accuracy if the number (0.4 )10 must be represented in binary on only six bits. Assume that the binary representation is rounded off to six bits (rather than being truncated).

(0.4 )10 = (0.0110 )2 To round off to six bits we must consider the first seven bits:

(0.0110011... )2

rounds to ( 0.011010) 2 = ( 0.40625)10

The loss of accuracy is thus (0.00625 )10 .

MATH 1310 – Technical Math for IT

b. Determine the loss of accuracy if the number (0.55 )10 must be truncated in binary to only eight bits.

(0.55 )10 = ( 0.100011) 2 Truncating to eight bits we get:

(0.10001100 )2 = ( 0.546875)10 The loss of accuracy is thus (0.003125 )10 .

( 13 )10 must be truncated in

c. Determine the loss of accuracy if the number binary to only four bits.

Use the conversion algorithm on the fraction: 1 × 2= 32 integer part 0 3 2 × 2= 43 integer part 1 3 Reset integer part to 0 (subtract 1): 1 × 2= 32 integer part 0 3 The steps now start repeating inf initely: 1 = 0.01

() ( 3 10

)

2

Truncating to four bits we get:

(0.0101 )2 = (0.3125 )10 = ( 165 )10 The loss of accuracy is thus

( ) −( ) =( ) 1 3 10

5 16 10

1 48 10

(

or 0.02083

)

10

....