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unsigned binary math


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Hello all! (and yes this is a hw question, before anyone asks, Im looking for help, not the answer)

I have to subtract two unsigned binary numbers, with the second larger than the first.




I have gotten most of the answer:

X0101000, but I dont know what to do for the last number. I have 0 - 1, and nowhere to borrow from. What am I supposed to do in this situation? I dont know if the fact that this is supposed to be unsigned changes anything? Any help is appreciated.

Edited by cardboardtoast
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Ok, so odd issue here. Your final answer makes sense (94-182 = -88), but the grading program flagged it as wrong, claiming that the second step's answer: 1010 1000 (168) was correct. I only did the complement once in all of the other questions, and it said correct. Is there a reason why one would NOT do the complement twice?

Thanks for the quick response!

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Funny you should bring up assembler, because this is for an assembly language class (the prof is going over a little about how a computer actually does the math) :p

Ok, so I believe I have figured out the answer. Because this was defined as a system that uses "unsigned (ie positive)" numbers only and ignores all negatives, it does not take the possibility of a negative answer into account (as it cannot have a negative at all). Using only 1 two's complement, I can get the correct answer to any subtraction that will result in a positive answer:

just to be easy: 8-3 = 5 (I did a few more on paper, but yeah, that would make this a big post)

8: 0000 1000
3: 0000 0011

Doing the two's comp only once gets:

0000 1000
1111 1101
0000 0101 = 5

So if you dont ever think you will have a negative number, this is all that is needed. Because my problem was for "unsigned numbers" only, I did not have to check if the problem would result in a negative answer (which is impossible), and could stop at that step. As you pointed out, the next step involved putting a negative sign in front of the answer, which I cannot do as it will be rejected (and positive 88 would be wrong anyways). Well, thats all I can think of to explain it, I will see if I can ask in class, but that will be a while.

Once again, thanks for your help and the quick responses Octarine!

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