Higher Engineering Mathematics, Sixth Edition

90 Higher Engineering Mathematics

Problem 8. Perform the binary addition:

11111 + 10101

11111

+ 10101

sum 110100

carry 11111

Problem 9. Perform the binary addition:

1101001 + 1110101

1101001

+ 1110101

sum 11011110

carry 11 1

Problem 10. Perform the binary addition:

1011101 + 1100001 + 110101

1011101

1100001

+ 110101

sum 11110011

carry 11111 1

Now try the following exercise

Exercise 40 Further problems on binary

addition

Perform the following binary additions:

1. 10+ 11 [101]


2. 101+ 110 [1011]


3. 1101+ 111 [10100]


4. 1111+ 11101 [101100]


5. 110111+ 10001 [1001000]


6. 10000101+ 10000101 [100001010]


7. 11101100+ 111001011 [1010110111]


8. 110011010+ 11100011 [1001111101]


9. 10110+ 1011 + 11011 [111100]


10. 111+ 10101 + 11011 [110111]


11. 1101+ 1001 + 11101 [110011]


12. 100011+ 11101 + 101110 [1101110]

10.3 Octal numbers

For decimal integers containing several digits, repeat-

edly dividingby 2 can be a lengthyprocess. In this case,

it is usually easier to convert a decimal number to a

binary number via the octal system of numbers. This

system has a radix of 8, using the digits 0, 1, 2, 3, 4,

5, 6 and 7. The decimal number equivalent to the octal

number 4317 8 is:

4 × 83 + 3 × 82 + 1 × 81 + 7 × 80

i.e. 4 × 512 + 3 × 64 + 1 × 8 + 7 ×1 or 2255 10

An integer decimal number can be converted to a cor-

responding octal number by repeatedly dividing by 8

and noting the remainder at each stage, as shown below

for 493 10.

Remainder

755

8 493

861

87

5

5

70

Thus 49310 = (^7558)
Thefractional part of adecimal number can beconverted
to an octal number by repeatedly multiplying by 8, as
shown below for the fraction 0.4375 10
4.3

3. 5


4. 0

0.4375 3 8 5

0.5 3 8 5

For fractions, the most significant bit is the top integer

obtained by multiplication of the decimal fraction by

8, thus,

0. 437510 = 0. (^348)
   The natural binary code for digits 0 to 7 is shown
   in Table 10.1, and an octal number can be converted
   to a binary number by writing down the three bits
   corresponding to the octal digit.
   Thus 437 8 =100 011 111 2
   and 26. 358 =010 110.011 101 2