1 INTRODUCTION The previous lesson introduced binary and hexadecimal numbers. In this lesson we look at simple arithmetic operations using these number systems. In particular, we examine the problem of representing negative numbers. YOUR AIMS On completing this lesson you should be able to: add and subtract binary numbers describe two ways in which negative numbers can be represented by a binary code perform arithmetic operations using two s complement notation convert a two s complement number back into a recognizable magnitude. STUDY ADVICE Again, this is a short unit in terms of text but it will require considerable time to complete all the examples and Self-Assessment Questions. Have to hand a pencil and scrap paper!
2 ADDITION OF BINARY NUMBERS In principle, binary numbers are added in exactly the same way as in denary addition but remember we are working to a base of two! This means that (1 + 1) will generate a carry. The rules of binary addition are summarised below: 0 0 1 1 + 0 + 1 + 0 + 1 0 1 1 0 carry 1 You may find it worthwile to write these rules down several times so that you remember them. Then study the following examples. Example 1 Add the binary numbers 1010 and 0011. 1 0 1 0 + 0 0 1 1 1 1 0 1 carry The bits in each column are added together and any carry bit generated is added into the column on the left.
3 Example 2 Add the binary numbers 1011 and 0011. * 1 0 1 1 + 0 0 1 1 1 1 1 0 carry In this example the column marked with an asterisk represents the addition (1 + 1 + 1) = 11. This means that 1 is entered into this column and one is carried over to the next most significant column. Example 3 Perform 11101110 2 + 01101101 2 1 1 1 0 1 1 1 0 0 1 1 0 1 1 0 1 1 0 1 0 1 1 0 1 1 carry A harder example, this, with lots of carries. Note that the most significant column also generates a carry. Microprocessors have a carry flag that can be set when an overflow carry is generated more on this in a later lesson.
4 BINARY SUBTRACTION There is no theoretical reason why binary numbers cannot be subtracted using the same rules as for denary subtraction, i.e. starting with the least significant column we subtract one bit from another. The rules for binary subtraction will then be: 0 1 1 0 0 0 1 1 (borrow 1 from left hand column) 0 1 0 1 In the fourth rule, 1 from 0 won t go and we have to borrow a bit from the adjacent column. This now forms the equivalent of (2 1) = 1. Example 4 Subtract 0111 from 1111. 1 1 1 1 0 1 1 1 1 0 0 0 Note that the answer can be checked by adding the result to the number subtracted. 0 1 1 1 + 1 0 0 0 1 1 1 1
5 Example 5 Subtract 0011 from 1101. 1 1 0 1 (In column 2 we have to borrow 0 0 1 1 from column 3.) 1 0 1 0 borrow This is as far as we shall go in binary subtraction. The reason for this is that computers do not perform subtraction in this way. There are two reasons why they do not: (a) Binary addition is carried out in the arithmetic logic unit (ALU) built into the CPU. The addition is done by special adder hardware. If subtraction were to be done in the conventional way then subtractor hardware would also be required. (b) Subtraction can give rise to negative results. Thus we need a means of representing positive and negative numbers. Conventionally we do this by using the symbols + and. We must remember, however, that a binary machine will only recognise two symbols (namely 0 and 1 ). We cannot therefore use extra symbols to denote a number s sign.
6 REPRESENTATION OF NEGATIVE NUMBERS Two methods will be considered. THE SIGN-MAGNITUDE METHOD This method is widely used in data communications systems. The magnitude of a number is its value, ignoring its sign. Thus the magnitude of +3 is 3 and -8 is 8. The magnitude of minus one hundred is equal to the magnitude of plus one hundred. In the sign magnitude method, the sign of a binary number is given by the MSB. The remaining bits give the magnitude of the number. If positive numbers are represented by the MSB being set to 0, then negative numbers are represented by the MSB being set to 1. To illustrate the idea, the binary equivalents of -5 to +5 are listed below (using 1 byte numbers): SIGN BIT MAGNITUDE -5 1 0000101-4 1 0000100-3 1 0000011-2 1 0000010
7-1 1 0000001-0 1 0000000 +0 0 0000000 +1 0 0000001 +2 0 0000010 +3 0 0000011 +4 0 0000100 +5 0 0000101 As a matter of curiosity, note that this method gives two zeros, 10000000 (minus zero) and 00000000 (plus zero)! This, however, does not cause any problems. Also, there is, as a matter of fact, no hard and fast rule as to whether a 1 or a 0 is used to represent a positive number. Which convention is adopted will depend upon the particular system. The advantage of the method is the simplicity with which it allows the magnitude of the number to be quickly established. A serious disadvantage though, is that the method cannot easily be used for binary arithmetic. Consider the following examples.
8 Example 6 Add (+3) to ( 3) using binary signed numbers +3 = 00000011 and 3 = 10000011 0 0 0 0 0 0 1 1 + 1 0 0 0 0 0 1 1 1 0 0 0 0 1 1 0 Clearly the result should have been zero ((+3) + ( 3) = 0) but the addition has come up with 6. Example 7 Add ( 5) to (+2). 5 = 10000101 and +2 = 00000010 1 0 0 0 0 1 0 1 0 0 0 0 0 0 1 0 1 0 0 0 0 1 1 1 This time the answer is 7 and not 3 as it should be. You may care to try a few examples yourself and I am sure you will soon be convinced that this method is quite hopeless for arithmetic. Nevertheless it does have the value of simplicity and is used in systems where the data does not have to be processed arithmetically. The sign-magnitude method is used, for example, in the PCM (plus code modulation) system of digital audio applications.
9 THE TWO S COMPLEMENT METHOD The one s complement of a binary number is obtained by changing all the 1 s to 0 s and vice versa. (This process is known as INVERTING the number.) A few examples of binary numbers and their one s complements are given below. NUMBER ONE S COMPLEMENT 0 0 0 0 1 1 1 1 1 0 1 0 0 1 0 1 0 0 1 1 1 1 0 0 1 1 1 1 0 0 0 0 Note that in each of the above examples adding the binary number to its complement always yields 1111. Moreover we note that in performing the operation of adding a binary number to its complement +1 will always yield 0000 or zero. The two s complement of a binary number is formed by taking the one s complement and adding 1 to it. Example 8 Find the two s complement of 14. Express the answer as an 8 bit number. 14 10 = 00001110 One s complement = 11110001 Two s complement = ( one s complement + 1)
10 = (11110001 + 1) = 11110010 Example 9 Find the two s complement of 25. 25 10 = 00011001 2 Two s complement = (11100110 + 1) = 11100111 We have shown above that adding a binary number to its one s complement +1 will always yield 0000 or zero. As the one s complement +1 is in fact the two s complement of the binary number then we can write: N + N = 0 where N is the binary number and N is its two s complement. Thus N = N Thus two s complement form can be used to represent the negative of a binary number that can be used in arithmetic operations. It may be a bit tricky for us to use and recognise the two s complement form, but relatively simple arithmetic circuits in the CPU are quite at home with them. To convince yourself that the two s complement representation of binary numbers works, study the following examples.
11 Example 10 Using the two s complement method perform the operation (25 14). (25 14) may be rewritten as (25 + ( 14)). The procedure is to represent ( 14) in two s complement form and then add this to 25. 25 10 = 0 0 0 1 1 0 0 1 add ( 14 10 ) = 1 1 1 1 0 0 1 0 [1] 0 0 0 0 1 0 1 1 carry As we are working in 8 bit numbers, the carry one is ignored. The result is, therefore, 00001011 2 = 11 10. Example 11 Using the two s complement representation show that 25 25 = 0. 25 = 0 0 0 1 1 0 0 1 add ( 25) = 1 1 1 0 0 1 1 1 [1] 0 0 0 0 0 0 0 0 carry Again, the carry bit is ignored and a zero result is obtained.
12 Example 12 Using the two s complement representation evaluate (14 25). 14 = 0 0 0 0 1 1 1 0 add ( 25) = 1 1 1 0 0 1 1 1 1 1 1 1 0 1 0 1 This answer highlights one of the difficulties of the two s complement form. In the complement system all negative numbers have a MSB of 1, so we know our result is negative, but the magnitude of the number is by no means obvious! Before attempting to evaluate, in denary, the solution to example 12, let's investigate our new systems a little more closely. Initially, for simplicity, we will use 4 bit signed numbers. This gives a numbering range of ( 8) to (+7) as indicated on the number line shown below. DENARY 8 7 6 5 4 3 2 1 0 +1 +2 +3 +4 +5 +6 +7 1000 1001 1010 1011 1100 1101 1110 1111 ( 8+0) ( 8+1) ( 8+2) ( 8+3) ( 8+4) ( 8+5) ( 8+6) ( 8+7) 0000 0001 0010 0011 0100 0101 0110 0111 BINARY The clue to evaluating the magnitude of a negative number represented in two s complement form is to note that, for example, ( 3) = ( 8 + 5). Looking at the number here we see that ( 3) = 1101. Thus the two s complement form can be regarded as consisting of a mixture of two numbers: (a) a negative number represented by the MSB and equal to ( 2 n 1 ) where n is the bit length of the number
13 (b) a positive number evaluated from the remaining bits. NEGATIVE NUMBER POSITIVE NUMBER 1 1 0 1 ( 2n 1 ) = ( 2 3 ) = 8 (2 2 + 2 0 ) = 5 The result is, of course, obtained by adding the two numbers together. We are now in a position to return to Example 12. The result is an 8-bit signed number and thus the number range will be from ( 128) to (+127). This range was evaluated using the formulae: (a) lower limit = ( 2 n 1 ) giving 128 for n = 8 (b) upper limit = (2 n 1 1) giving +127 for n = 8 Splitting (11110101) into two parts yields: 11110101 ( 2 8 1 ) = 128 (2 6 + 2 5 + 2 4 + 2 2 + 2 0 ) = 117 The answer, in denary is, therefore, ( 128 + 117) = 11 The system works!
14 A SIMPLE TRICK! Before concluding this lesson we introduce you to a simple rule for converting binary numbers into a two s complement form, which can save a lot of work. The rule is: (a) Write down the binary number, starting with the LSB, up to and including the first 1. (b) Invert the remaining bits. Consider example 13. Example 13 Write down the two s complement of 01110010. 01110010 invert these bits to give 'up to and including the first '1" gives '1 0 0 0 1 1' '1 0' Thus two s complement of 01110010 is 10001110. Example 14 Perform the arithmetic operation (17 97) in binary. Express your answer in two s complement form and then convert it to denary. Work in 8 bit numbers.
15 17 = 00010001 97 = 01100001 2 ( 97) = 10011111 (in two s complement) Adding ( 97) to (+17): 0 0 0 1 0 0 0 1 + 1 0 0 1 1 1 1 1 1 0 1 1 0 0 0 0 Thus in two s complement the result is 10110000 (which is clearly negative as the MSB is set to 1 ). Converting this to denary: 10110000 = ( 2 8 1 ) + (2 5 + 2 4 ) = ( 128 + 48) = 80 The trick can also be used in reverse to convert a negative number expressed in two complement form back into a magnitude. Consider the result to example 14, namely 10110000. In reversing the trick we: (a) write down the number up to and including the first 1 (yielding 10000) (b) invert all the other bits (giving 01010000 ). The magnitude is thus 01010000 = 80 As we know we are dealing with a negative number the answer is 80.
16 Which method you use (the more formal [ 2 n -1 + (positive number)] or the trick ) is entirely up to you there is little to choose between them in terms of speed of conversion. This completes our work on computer arithmetic, but to further develop and consolidate your understanding try the following Self-Assessment Questions.
17 SELF-ASSESSMENT QUESTIONS 1. Add the following binary numbers. In each case express the answer as an 8 bit number. In each case state if a carry is generated beyond the MSB. (a) 0 0 0 0 1 1 1 1 + 0 0 0 0 1 0 1 0 (b) 1 0 1 0 1 0 1 0 + 1 1 0 1 0 1 0 1 (c) 0 1 1 1 1 0 0 1 + 1 0 1 1 0 0 0 1 (d) 0 0 0 0 0 0 1 0 + 1 1 1 1 1 1 1 1 2. Add the following HEX numbers by converting to binary. Express the answer in HEX. (a) 08 + 03 (b) 0F + 02 (c) 0ABC + DA0E 3. Convert the following numbers to two s complement form: (a) 0 0 0 0 0 0 0 1 (b) 0 1 1 0 1 0 0 0 (c) 0 0 1 1 1 1 0 0 (d) 0 1 1 1 1 1 1 1
18 4. In the table below, negative numbers are represented by the two s complement form. Identify the odd negative numbers: (a) 0 0 1 1 0 1 0 1 (b) 0 0 0 0 0 0 0 1 (c) 1 0 0 0 0 0 0 1 (d) 1 0 1 0 1 0 1 0 (e) 0 1 0 1 0 1 0 1 (f) 1 0 0 1 0 1 1 0 (g) 1 0 1 0 1 1 1 1 (h) 0 0 1 1 0 0 0 1 (i) 1 0 0 0 0 0 1 0 (j) F8 (k) 8A (l) CB 5. A sixteen bit microprocessor is to be used to perform arithmetic operations using signed numbers in two s complement form. Calculate the numbering range that can be represented using 16 bits. 6. Convert the denary numbers in (a) (d) to 8 bit binary numbers and evaluate using the two s complement method. Leave your answers in binary form. (a) 17 8 (b) 126 33 (c) 4 48 (d) 97 125 7. When performing an arithmetic operation using signed numbers a microprocessor gives the result F6. Express this in denary. 8. Subtract C006 from A0F7 using two s complement arithmetic. Express the answer in Hex.
19 9. Perform the operations: (a) 7A + 85 (b) 7A 7B Comment upon the result. 10. State the effect upon the magnitude of an 8 bit binary number by: (a) shifting each bit one position to the left (b) shifting each bit one position to the right. The answers are on page 20.
20 ANSWERS TO SELF-ASSESSMENT QUESTIONS 1. (a) 0 0 0 0 1 1 1 1 + 0 0 0 0 1 0 1 0 0 0 0 1 1 0 0 1 (no carry) (b) 1 0 1 0 1 0 1 0 + 1 1 0 1 0 1 0 1 1 0 1 1 1 1 1 1 1 (carry generated) (c) 0 1 1 1 1 0 0 1 + 1 0 1 1 0 0 0 1 1 0 0 1 0 1 0 1 0 (carry generated) (d) 0 0 0 0 0 0 1 0 + 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 1 (carry generated) 2. (a) 0 8 0 0 0 0 1 0 0 0 + 0 3 0 0 0 0 0 0 1 1 0 B 0 0 0 0 1 0 1 1 (b) 0 F 0 0 0 0 1 1 1 1 + 0 2 0 0 0 0 0 0 1 0 1 1 0 0 0 1 0 0 0 1
21 (c) 0ABC 0 0 0 0 1 0 1 0 1 0 1 1 1 1 0 0 + DA0E 1 1 0 1 1 0 1 0 0 0 0 0 1 1 1 0 E4CA 1 1 1 0 0 1 0 0 1 1 0 0 1 0 1 0 3. (a) Using the trick the complements may be written straight down: (a) 1 1 1 1 1 1 1 1 (b) 1 0 0 1 1 0 0 0 (c) 1 1 0 0 0 1 0 0 (d) 1 0 0 0 0 0 0 1 4. Negative odd numbers will have MSB and LSB at 1. Therefore (c), (g) and (l) are odd negative numbers. 5. The numbering range will extend from ( 2 n 1 ) to (2 n 1 1) where n = 16. Thus the range is ( 2 15 ) to (2 15 1) = 32768 to 32767 6. (a) 17 = 0 0 0 1 0 0 0 1 8 = 0 0 0 0 1 0 0 0 ( 8) = 1 1 1 1 1 0 0 0 (17) 0 0 0 1 0 0 0 1 + ( 8) 1 1 1 1 1 0 0 0 1 0 0 0 0 1 0 0 1 The carry one is ignored and thus the answer is 00001001.
22 (b) 126 = 0 1 1 1 1 1 1 0 33 = 0 0 1 0 0 0 0 1 ( 33) = 1 1 0 1 1 1 1 1 (126) 0 1 1 1 1 1 1 0 + ( 33) 1 1 0 1 1 1 1 1 1 0 1 0 1 1 1 0 1 Again, the carry is ignored, leaving the result 01011101. (c) 4 = 0 0 0 0 0 1 0 0 48 = 0 0 1 1 0 0 0 0 ( 48) = 1 1 0 1 0 0 0 0 (4) 0 0 0 0 0 1 0 0 + ( 48) 1 1 0 1 0 0 0 0 1 1 0 1 0 1 0 0 (d) 97 = 0 1 1 0 0 0 0 1 125 = 0 1 1 1 1 1 0 1 ( 125) = 1 0 0 0 0 0 1 1 (97) = 0 1 1 0 0 0 0 1 + ( 125) = 1 0 0 0 0 0 1 1 1 1 1 0 0 1 0 0
23 7. F6 = 11110110 which is obviously negative. Thus 11110110 = ( 2 7 ) + (2 6 + 2 5 + 2 4 + 2 2 + 2 1 ) = 128 + 118 = 10 (N.B. You will see in later work that this kind of conversion needs to be applied when using certain types of computer instructions.) 8. A0F7 = 1 0 1 0 0 0 0 0 1 1 1 1 0 1 1 1 C006 = 1 1 0 0 0 0 0 0 0 0 0 0 0 1 1 0 ( C006) = 0 0 1 1 1 1 1 1 1 1 1 1 1 0 1 0 A0F7 1 0 1 0 0 0 0 0 1 1 1 1 0 1 1 1 + ( C006) 0 0 1 1 1 1 1 1 1 1 1 1 1 0 1 0 E0F1 1 1 1 0 0 0 0 0 1 1 1 1 0 0 0 1 9. (a) 7A = 0 1 1 1 1 0 1 0 85 = 1 0 0 0 0 1 0 1 1 1 1 1 1 1 1 1 (b) 7B = 0 1 1 1 1 0 1 1 ( 7B) = 1 0 0 0 0 1 0 1 (7A) = 0 1 1 1 1 0 1 0 + ( 7B) = 1 0 0 0 0 1 0 1 1 1 1 1 1 1 1 1
24 Comment The answers are identical. Note, however, that in (a) we are using unsigned, eight bit numbers, whilst in (b) signed numbers are being used. Hence in converting (a) back to denary we would read it as +255 whereas (b) represents 1. 10. The table below shows the effect of shifting an 8 bit binary number to the left. The effect is to double the number. A right shift will halve the number. This is an obvious consequence of positional notation. 0 0 0 0 0 0 0 1 (1) 0 0 0 0 0 0 1 0 (2) 0 0 0 0 0 1 0 0 (4) 0 0 0 0 1 0 0 0 (8) 0 0 0 1 0 0 0 0 (16) 0 0 1 0 0 0 0 0 (32) 0 1 0 0 0 0 0 0 (64) 1 0 0 0 0 0 0 0 (128) A microprocessor will have instructions that permit us to shift left or shift right which are useful in, for example, binary multiplication. Note, however, that if we were using signed numbers then we would have to be careful. For instance, the last result of the above table would then be negative!
25 SUMMARY (a) The rules of binary addition are: 0 0 1 1 +0 +1 +0 1 0 1 1 0 carry 1 to left hand column (b) The rules of binary subtraction are: 0 1 1 0 0 0 1 1 0 1 0 1 borrow 1 from left hand column (c) Two common ways of representing negative numbers are: (i) Sign-magnitude method when the MSB gives the sign of the number and the remaining bits the magnitude. (ii) Two s complement method obtained by inverting the number and adding 1 to the result. In the next lesson we consider some simple hardware which can be used to perform simple operations on binary data.