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Multiplying two N bit signed numbers together

  1. Sep 3, 2012 #1
    1. The problem statement, all variables and given/known data

    Determine the number of bits needed to represent the range of results obtained when two N bit signed numbers are multiplied together.


    2. Relevant equations

    None.

    3. The attempt at a solution

    I considered the case when N = 3, i.e. a 3 bit signed number. The values that can be expressed with this are +3 - > -4. Hence with two 3 bit signed numbers we can get a maximum positive number of +16 and a maximum negative number of -12. We would therefore need 6 bits to represent this range of results (2N).

    Looking at the case of 4 bits, the values that can be expressed are +7 -> -8. Here our largest positive number would be 64 and largest negative number -56. We would therefore need 8 bits (2N).

    So my answer is 2N, is this correct ?
     
  2. jcsd
  3. Sep 3, 2012 #2

    lewando

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    Gold Member

    For 2's complement, this is true. For sign-magnitude, one less so.
     
  4. Sep 3, 2012 #3
    That's pretty much the same thing that's bothering me... I have written the question as it has been asked. I'm assuming they mean sign magnitude ?
     
  5. Sep 3, 2012 #4

    lewando

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    Gold Member

    I suggest clarify question with instructor or state your assumptions in the answer. If you can't do that and if it is one of these computer-based answer entry systems (no assumptions permitted) I'd go with 2N. It allows you to do both.
     
  6. Sep 3, 2012 #5

    rcgldr

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    Homework Helper

    Or ones complement. For 3 bit sign-magnitude or one's complement numbers the range of products is limited to -9 to +9, which only requires 5 bits.
     
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