qr code vb.net library Again, we ve found the difference, Y N, by use of addition only. in .NET framework

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Again, we ve found the difference, Y N, by use of addition only.
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If Y and N are not integers (whole numbers), exactly the same procedure is used, but we must remember to line up the binary points just as we line up the decimal points in decimal addition. For example, suppose we are to nd Y N where, let us say, Y 1 1 0 1 : 0 1 1 0 decimal 13:3750 N 1 0 0 1 : 1 0 1 1 decimal 9:6875 Note that the answer in decimal notation is Y N 13:3750 9:6875 3:6875. To work out the problem let us use the 1 s complement of N in the usual way; thus Y 1 1 0 1 : 0 1 1 0 1 s comp of N 0 1 1 0 : 0 1 0 0 r 0 0 1 1 : 1 0 1 0 j 1 ! ! 0 0 1 1 : 1 0 1 1; answer dec: 3:6875
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Note: The above can also be worked in terms of whole numbers by rst shifting the binary point four places to the right and then multiplying by 0.0001; thus Y 11010110 0:0001 1 s comp of N 01100100 0:0001 00111010 0:0001 1 00111011 0:0001 0011:1011; as before As you may have noticed, in the foregoing examples the magnitude of Y is greater than the magnitude of N, and hence the values of Y N are all positive numbers. If, however, the magnitude of Y is less than the magnitude of N, then Y N is a negative number. A digital computer must, of course, be able to detect, store, and use both positive and negative numbers. One way a computer can sense whether a di erence Y N is positive or negative is to detect the presence or absence of the over ow 1 when computing Y N by use of the 1 s complement of N. This is based upon the fact that if Y is greater than N an over ow 1 will be generated, but if Y is less than N no over ow 1 will be generated. Thus, if Y N is found by use of the 1 s complement of N, then Y N is a POSITIVE number if an over ow 1 is produced, but is a NEGATIVE number if no over ow 1 is produced. To illustrate this, let s return to example 1 above, and this time let Y 10011 and N 11001, so that the problem now becomes Y 1 0 0 1 1 N 1 1 0 0 1 Y 19 dec N 25 dec
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Now using the 1 s complement of N in the usual way we nd that Y 1 0 0 1 1 1 s comp of N 0 0 1 1 0 Y 1 s comp of N 0 1 1 0 0 1 % Note that no over ow 1 is produced, which tells the computer that Y is less than N, and therefore that (a) the answer to Y N is a negative number whose magnitude is therefore found by
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CHAPTER 12 Binary Arithmetic
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(b) changing the minuend number to the 1 s complement form and adding; thus 1 s comp of Y 0 1 1 0 0
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N 1 1 0 0 1 r 0 0 1 0 1 j 1 ! ! 0 0 1 1 0 magnitude of Y N The computer must now have some way of indicating that the answer, 00110, is a negative value, minus six, and one way of doing this is to use a sign digit in the following manner. In the present example, the numbers in the computer are represented in ve-digit binary form, such as 11101, 00010, and so on. In this case an additional binary 1 or 0 could be used in the sixth place as a SIGN DIGIT to indicate plus or minus ; thus, if 0 indicates plus and 1 indicates minus or negative, the answer to the foregoing problem would be registered in the computer as 100110, indicating minus six. In the same way, 000110 would indicate positive six, and so on. The sign digit, 1 or 0, is generated when the computer processes the di erence Y N. Thus (assuming 0 and 1 to denote plus and minus respectively) if the computer senses, for example, that an over ow 1 is not produced, this fact signals the computer to change Y to its 1 s complement form, add N to it, and put a 1 in front of the magnitude value of Y N. On the other hand, if an over ow 1 is produced, this causes the computer to put the digit 0 in front of the result, to show that Y N is a positive quantity. Problem 266 Convert 67 decimal to binary form. Problem 267 Convert 383 decimal to binary form. Problem 268 Convert 118.182 decimal to binary form (to 9 binary places). Problem 269 Convert 1110101 binary to decimal form. Problem 270 Convert 1001.01101 binary to decimal form. Problem 271 Using binary addition, write the sums of the following binary numbers, with answers in binary form. a 0 1 0 1 1 0 1 1 0 1 1 0 0 1 b 1 0 1 1 0 1 0 1 1 1 0 1 c 1 1 : 0 1 1 0 1 : 1 0 1
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