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Complement - What is wrong?

  1. May 6, 2009 #1
    1. The problem statement, all variables and given/known data
    Represent (–3042)5 in the 4-digit 4’s complement system.


    2. Relevant equations
    –X = R^n – X – 1


    3. The attempt at a solution
    5^4 – 3042 – 1 = 625 – 3042 – 1 = ???

    Answer should be (1402)4s
     
  2. jcsd
  3. May 6, 2009 #2
    You can obtain the 4's complement of a four-digit number by subtracting from 4, digit by digit. Give it a try.
    For better understand how to do it and why we need to do this, you can look up different articles, such as
    http://en.wikipedia.org/wiki/Ten's_complement
    which explains the nine's complement in detail.
     
  4. May 6, 2009 #3
    Thanks. I'll try again :)
     
  5. May 6, 2009 #4
    Sorry, should I convert (–3042)5 to a decimal or binary before I proceed?
     
  6. May 6, 2009 #5
    For the complement, you are better off doing it in the original base.
    For example, the nines complement of 1423 is 8576. You'd be doing unnessary extra work if you do the conversion, say from base 6 to base 10, and then reverting it to base 6.
    Another example:
    5's complement of 1452 (base 6) would be 4103, namely take each digit from 5 and write down the new number.
    5-1=4, 5-4=1, 5-5=0, 5-2=3 to give 4103 (base 6).
     
  7. May 6, 2009 #6
    How about dealing with negatives? Just treat them as positive?

    In my case, 4's complement of (–3042)5
    4-3=1, 4-0=4, 4-4=0, 4-2=2 giving (1402)4s
    I'll get the answer with the method ignoring the negative.
     
  8. May 6, 2009 #7
    This is the idea, by converting the negative value to the four's complement, you end up with a positive number that you can ADD instead of subtract.
    This way, we can subtract by adding the 4's complement PLUS ONE, and ignore any carry past the maximum number of digits.
    For example, to subtract 2421 (base 5) from 4213 (base 5), we proceed as follows (all calculations use base 5):
    4213-2421
    =4213+(4's complement of 2421) + 1
    =4213+2023+1
    =(1)1241+1 [ ignore the carry of 1]
    =1242
    If you proceed to do the calculations using base 5, or by conversion to base 10, and reconvert to base 5, you would get the same result.

    I suggest you read and understand the following article which explains the method in detail:
    http://en.wikipedia.org/wiki/Ten's_complement
     
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