Arithmetic Quiz - Answers

1. For an 8-bit group, work out the representation for −37₁₀ in

   37₁₀ = 100101₂

   1. Sign & Magnitude

      10100101

   2. One's Complement

      11011010

   3. Two's Complement

      11011011

   4. Excess-256

      −37 = −37 + 256 = 219 = 11011011

      Note: the range of representable numbers is −256 to −1 (!)

   5. Excess-128

      −37 = −37 + 128 = 91 = 01011011

      This is the more usual range. i.e. −128 to +127

2. For a 10-bit group, what range of integers can be represented using
   1. Sign & Magnitude

      −2ⁿ⁻¹−1 to +2ⁿ⁻¹−1, where n = 10, i.e. −511 to +511

   2. One's Complement

      −2ⁿ⁻¹−1 to +2ⁿ⁻¹−1, where n = 10, i.e. −511 to +511

   3. Two's Complement

      −2ⁿ⁻¹ to +2ⁿ⁻¹−1, where n = 10, i.e. −512 to +511

   4. Excess-512

      −2ⁿ⁻¹ to +2ⁿ⁻¹−1, where n = 10, i.e. −512 to +511

3. Express 98765₁₀ in Binary Coded Decimal

9	8	7	6	5
1001	1000	0111	0110	0101

4. Form the negative equivalent of the following 8-bit Two's Complement numbers

   • 00011001 = 16 + 8 + 1 = 25₁₀

     "invert the bits and add 1" 11100110 + 1 = 11100111

     check:

     11100111 = −128 + (64 + 32 + 4 + 2 + 1) = −25₁₀

   • 00011110 = 16 + 8 + 4 + 2 = 30₁₀

     "invert the bits and add 1" 11100001 + 1 = 11100010

     check:

     11100010 = −128 + (64 + 32 + 2) = −30₁₀

   • 01101000 = 64 + 32 + 8 = 104₁₀

     "invert the bits and add 1" 10010111 + 1 = 10011000

     check:

     10011000 = −128 + (16 + 8) = −104₁₀

   • 01110100 = 64 + 32 + 16 + 4 = 116₁₀

     "invert the bits and add 1" 10001011 + 1 = 10001100

     check:

     10001100 = −128 + (8 + 4) = −116₁₀

   by comparing the resulting bitpatterns to the originals, can you spot a "short cut" method for the conversion?

Take another look at the bit patterns:

positive: 00011001   00011110   01101000   01110100
negative: 11100111   11100010   10011000   10001100

"starting from the rightmost bit (lsb), copy each bit unchanged up to and including the first 1 then invert all the remaining bits"

5. Perform the following 12-bit two’s complement subtraction

    1010 1010 1011
   −1011 0000 1101
   

   Two's Complement subtraction: "negate the subtrahend and add"

   Two's Complement negation: "invert the bits and add 1"

   101100001101 = 010011110010 + 1 = 010011110011

    1010 1010 1011
   +0100 1111 0011
    --------------
    1111 1001 1110

   Check your answer by determining the decimal representation of the numbers and the result

    1010 1010 1011 =   −2048 +  683 = −1365
   −1011 0000 1101 = −(−2048 +  781 = −1267)
    --------------                    -----
    1111 1001 1110 =   −2048 + 1950 =   -98
   

last updated: 24-Oct-06 Ian Harries <ih@doc.ic.ac.uk>