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Binary Arithmetic
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the sum of the digits now in the eights column is 1 1 1 1 1 1 10 1 11, hence we write down 1 and carry a 1 into the sixteens column, giving the answer 11000, as shown to the right above with decimal equivalents alongside.
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In the above we found the sum of just two binary numbers, but suppose the sum of, say, one thousand such numbers must be found. In the internal circuitry of a digital computer, this is most conveniently done by adding the binary numbers together two at a time until the nal sum is reached. As mentioned before, we must remember that a digital computer can execute millions of such routine operations per second. Next, let us consider the subtraction of one number from another number. Subtraction makes use of the borrow operation, illustrated rst in the following decimal-system example. PROBLEM 9 4 6 7 3 7 7 6 minuend subtrahend difference
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8 15 13 4 4 7 8 7 6
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Discussion. Beginning at the right-hand side in the PROBLEM we rst have 7 6 1, which presents no di culty because 7 is larger than 6. Continuing on, from right to left, we next must subtract 7 from 3 (actually, 70 from 30), which does present a di culty because 3 is smaller than 7. To get around this di culty we now, in the minuend, borrow a 1 from the 6, and transfer the borrowed 1 over to the 3; however, since a 1 in the third column has ten times the value of a 1 in the second column, this e ectively makes the 3 become 13, as shown in the SOLUTION. Since 13 is larger than 7 we now have 13 7 6, as shown. At this point we must not forget that the 6, in the third column, is now changed to 5 (because of the previous borrowing of the 1 from the 6). Therefore, continuing on in the PROBLEM, we must now subtract 7 from 5, which again presents a di culty because 5 is smaller than 7. To get around this di culty we now, in the minuend, borrow a 1 from the 9 and transfer the borrowed 1 to the 5, e ectively making the 5 become 15, as shown in the SOLUTION; thus we have 15 7 8, as shown. Because of the borrowing of the 1, the 9 becomes 8, as shown in the SOLUTION; hence the last step is to subtract 4 from 8, giving the nal answer 4861. We must remember that the same basic arithmetic procedures apply to all positional number systems, whatever the particular radix. Thus the borrow procedure, illustrated above for the decimal system, is used in the same way to subtract one binary number from another binary number, as the following example illustrates. PROBLEM 1 0 0 1 1 1 1 0 0 1 1 1 minuend subtrahend difference
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1 1 0 answer
Discussion. Beginning at the right-hand side in the PROBLEM we have 1 1 0, as shown in the SOLUTION. Continuing on, from right to left in the PROBLEM, the next step is to subtract 1 from 0, which presents a di culty because 0 is smaller than 1. To surmount this di culty we now, in the minuend, borrow a 1 from the third
CHAPTER 12 Binary Arithmetic
column and transfer it over to the second column in the minuend. In the binary system, however, we must remember that the value of a 1 doubles each time we move one position to the left; hence the 1 transferred from the third column e ectively becomes two in the second column, as shown by the 10 in the SOLUTION. We therefore have 10 1 1, as shown in the SOLUTION. Next, referring to the SOLUTION, note that the transferral of the 1 has left 0 in the third column of the minuend; therefore the next two steps in the subtraction present no di culty, because 0 0 0 and 1 1 0, as shown in the SOLUTION. However, in the fth column of the PROBLEM we again run into the di culty of subtracting 1 from 0, a di culty we overcome by borrowing a 1 from the minuend in the sixth column; the borrowed 1 e ectively becomes 10 (two) in the minuend in the fth column, as shown in the SOLUTION. Thus in the fth column we have 10 1 1, giving the nal answer 10010, as shown (in equivalent decimal notation, 45 27 18). In our work so far we ve found that it s generally easier, and less time-consuming, to do addition than it is to do subtraction. This is true in both pencil and paper work and in terms of digital computer circuitry requirements. It would therefore be an advantage if subtraction could somehow be performed in a way that used only the addition operation. Fortunately it is possible to do this by making use of what is called the ONE s COMPLEMENT of a binary number; the basic procedure can most easily be developed by considering the subtraction of one whole binary number from another whole binary number, as follows. Let N be any whole binary number. Since R 2 for the binary system, and since N is to be a whole number, eq. (553) will contain no negative exponents and will thus be of the form N d 2 n d 23 d 22 d 21 d 20 555
in which the digit d is, in any individual term, equal to either 1 or 0. Note that, for convenience, we re expressing the value of the binary number N in terms of the decimal digits 2, 3, and so on. In eq. (555) let n be called the order of the binary number N; therefore, in accordance with the terminology of eq. (555), it should be noted that a binary number of order n contains n 1 digits. With the foregoing in mind, let us now de ne that, if N is a binary number of order n, then the 1 s complement of N 2n 1 N 1 556 *
The reason the above de nition is useful will become clear later on, but rst let s investigate the 1 s complement of a binary number as de ned above. To do this, let N be a binary number of order n, and note that eq. (556) can also be written as N 1 s comp of N 2n 1 1 Now remembering that, in binary notation, 20 1 21 10 22 100 23 1000 24 10000 25 100000 557
* It is also possible to work in terms of the 1 s complement plus 1, which is called the 2 s complement.
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