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Chapter 2.4, Problem 10QE

Explanation of Solution

Determine the addition of two binary numbers if the pattern represents values stored in two’s complement notation:

  • In two’s complement system, the leftmost bit of pattern represents the sign of the value. Negative values are indicated by the binary number 1, while positive values are indicated by the binary number 0.
  • Similar to the binary addition, the addition in two’s complement system is done by adding both the numbers, but the carry is truncated in this case so that the number of bits in all the patterns including the answer is same.
  • The addition of 01101010 and 11001100 is:

    0110101011001100_100110110

  • Since, the number of bits in answer is 9 and number of bits in input values is 8, so the carry part is truncated from the sum and the result is reduced to 00110110.

Therefore, the sum of 01101010 and 11001100 is "00110110".

Determine the addition of two binary numbers if the pattern represents values stored in floating-point notation:

  • Floating-point notation is the method in which numbers are represented as decimal fractions and exponents.
  • In floating point notation, the number consists of only 8 bits. First bit from the left is assigned as sign bit, next three bits are assigned as exponents and the last four bits are assigned as mantissa. Negative values are indicated by the binary number 1, while positive values are indicated by the binary number 0.
  • The procedure to perform addition in floating-point notation is:
    • Convert both the binary numbers in floating point numbers.
    • Add both the numbers.
    • Convert the result into binary notation.
  • Decode the bit pattern 01101010 in floating point notation. The exponent field is 110 and it represents 2 according to the excess notation system for 3 bits.
    • Extract the mantissa to obtain .1010
    • Move the radix 2 bit to the right, since the exponent represents +2. So, the value is 10.10.
    • Convert 10.10 to base 10 representations.
      • First consider the integer part 10,

        1×ばつ21+0×ばつ20=2+0=2

      • Now consider the fractional part .10,

        1×ばつ12+0×ばつ122=12+0=12

    • Combine integer and fractional part results to obtain the base 10 value of 10.10. The value is 212.
  • Decode the bit pattern 11001100 in floating point notation. The exponent field is 100 and it represents 0 according to the excess notation system for 3 bits.
    • Extract the mantissa to obtain .1100
    • The radix will not move in any direction since the exponent represents 0. So, the value is .1100.
    • Convert .1100 to base 10 representations.

      1×ばつ12+1×ばつ122+0×ばつ123+0×ばつ124=12+14+0+0=34

    • The base 10 value is 34

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Chapter 2 Solutions

Computer Science: An Overview (13th Edition) (What's New in Computer Science)

Chapter 2.3, Problem 1QE Chapter 2.3, Problem 2QE Chapter 2.3, Problem 3QE Chapter 2.3, Problem 4QE Chapter 2.4, Problem 1QE Chapter 2.4, Problem 2QE Chapter 2.4, Problem 3QE Chapter 2.4, Problem 4QE Chapter 2.4, Problem 5QE Chapter 2.4, Problem 6QE Chapter 2.4, Problem 7QE Chapter 2.4, Problem 8QE Chapter 2.4, Problem 9QE Chapter 2.4, Problem 10QE Chapter 2.4, Problem 11QE Chapter 2.4, Problem 12QE Chapter 2.5, Problem 1QE Chapter 2.5, Problem 2QE Chapter 2.5, Problem 3QE Chapter 2.6, Problem 1QE Chapter 2.6, Problem 2QE Chapter 2.6, Problem 3QE Chapter 2.6, Problem 4QE Chapter 2.6, Problem 6QE Chapter 2.7, Problem 1QE Chapter 2.7, Problem 2QE Chapter 2.7, Problem 3QE Chapter 2, Problem 1CRP Chapter 2, Problem 2CRP Chapter 2, Problem 3CRP Chapter 2, Problem 4CRP Chapter 2, Problem 5CRP Chapter 2, Problem 6CRP Chapter 2, Problem 7CRP Chapter 2, Problem 8CRP Chapter 2, Problem 9CRP Chapter 2, Problem 10CRP Chapter 2, Problem 11CRP Chapter 2, Problem 12CRP Chapter 2, Problem 13CRP Chapter 2, Problem 14CRP Chapter 2, Problem 15CRP Chapter 2, Problem 16CRP Chapter 2, Problem 17CRP Chapter 2, Problem 18CRP Chapter 2, Problem 19CRP Chapter 2, Problem 20CRP Chapter 2, Problem 21CRP Chapter 2, Problem 22CRP Chapter 2, Problem 23CRP Chapter 2, Problem 25CRP Chapter 2, Problem 26CRP Chapter 2, Problem 27CRP Chapter 2, Problem 28CRP Chapter 2, Problem 29CRP Chapter 2, Problem 30CRP Chapter 2, Problem 31CRP Chapter 2, Problem 32CRP Chapter 2, Problem 33CRP Chapter 2, Problem 34CRP Chapter 2, Problem 35CRP Chapter 2, Problem 36CRP Chapter 2, Problem 37CRP Chapter 2, Problem 38CRP Chapter 2, Problem 39CRP Chapter 2, Problem 40CRP Chapter 2, Problem 41CRP Chapter 2, Problem 42CRP Chapter 2, Problem 43CRP Chapter 2, Problem 44CRP Chapter 2, Problem 45CRP Chapter 2, Problem 46CRP Chapter 2, Problem 47CRP Chapter 2, Problem 48CRP Chapter 2, Problem 49CRP Chapter 2, Problem 50CRP Chapter 2, Problem 51CRP Chapter 2, Problem 52CRP Chapter 2, Problem 53CRP Chapter 2, Problem 54CRP Chapter 2, Problem 55CRP Chapter 2, Problem 56CRP Chapter 2, Problem 57CRP Chapter 2, Problem 58CRP Chapter 2, Problem 59CRP Chapter 2, Problem 1SI Chapter 2, Problem 2SI Chapter 2, Problem 3SI Chapter 2, Problem 4SI Chapter 2, Problem 5SI Chapter 2, Problem 6SI Chapter 2, Problem 7SI Chapter 2, Problem 8SI
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