3.0 DIGITAL ARITHMETIC
Data presentation and representation is for a
purpose, haven discussed different ways of representing digital data, and the
next step is to study how this data are manipulated in specific operations.
The types of manipulative operations data are
performed on data include arithmetic
operations20 and logic operations21.
Basic arithmetic operations include addition, subtraction, multiplication and
division. While basic logic operations are; AND, OR and NOT. In this chapter,
we will discuss the rules that govern arithmetic operations on digital data. In
later chapters we will study logic operations on digital data.
3.1 Binary
Addition and Subtraction Rules
Arithmetic rules for binary
numbers are quite straightforward, and similar to those used in decimal
arithmetic. The rules for addition of binary numbers are:
Table 3.1: Rules of binary addition
Addition
|
Sum
|
Carry
|
0 + 0
|
0
|
0
|
0 + 1
|
1
|
0
|
1 + 0
|
1
|
0
|
1 + 1
|
0
|
1
|
1 + 1 + 1
|
1
|
1
|
Binary addition22 is carried out just like
the decimal addition that we are used to, in decimal addition, when the result
of the addition in a particular digit is equal or greater than the radix/base
value of the system (i.e. ten for decimal), it produces a carry to the next
higher bit and the sum will be the ‘result’ of addition minus the radix/base
value.
For example:
For 7 + 3,
the result is ‘10’ which is equal to the radix/base of decimal system a carry
is produce to the next higher digit and radix value (‘10’) is deducted from the
result ‘10’ remain ‘0’ as the sum. For 8 + 7, result is ‘15’ obviously greater
than the radix value ‘10’ carry ‘1’ is produced to the next higher digit, radix
value ‘10’ is deducted from ‘15’, thus remaining sum of
‘5’. In the case of 2 + 8, the result is 8 which is less than the radix
value, no carry is produce and the sum is ‘8’.
This same process applies
for the binary arithmetic rule described in table 3.1 above, 0 + 0 result to
‘0’, this result is less than the binary radix value which is ‘2’, thus leaving
the sum as ‘0’ and no carry is produced. Same for 0 + 1 and 1 + 0 which
produces a sum of ‘1’ and no carry i.e. ‘0’.
For 1 + 1, the result of addition is 2 which is
equal/greater than the radix value ‘2’, thus produces a carry ‘1’ and by
deducting the radix value ‘2’ from the result of addition ‘2’ it leaves the sum
as ‘0’. For 1 + 1 +1, the result is ‘3’
obviously will produce a carry ‘1’, deducting radix value ‘2’ from addition
result ‘3’, we are left with sum of ‘1’.
Binary subtraction rules
Again likened to the decimal
rules of subtraction, when we deduct a lesser number from a larger number the
subtraction is straight forward e.g. 8 –
2 = ‘difference’ 6 (in
decimal). But when a large number is being subtracted from a lesser number, a
borrow is required from the next higher digit e.g. 5 – 8 = ‘Difference’ 7 ‘borrow’ 1, in this case, the borrow from
the higher digit takes up the radix value ‘10’ in the next lower digit. The rules
for binary subtraction23 are summarized in table 3.2 bellow.
Table 3.2: Rules of binary subtraction
Subtraction
|
Difference
|
Borrow
|
0 - 0
|
0
|
0
|
0 - 1
|
1
|
1
|
1 - 0
|
1
|
0
|
1 - 1
|
0
|
1
|
1 – 1 -1
|
1
|
1
|
3.2
Binary Addition and Subtraction with larger bits
Our illustration so far has been on one
bit binary and in a practical situation; addition and subtraction in binary are
usually in larger bits. We discuss with examples the process of adding and
subtracting binary numbers of larger bits.
In the examples above we have
illustrated how binary numbers of more bits can be added or subtracted from one
another. But there is a clause; what if we have a situation where we are
required to subtract 1110 (1011)
from 910 (1001) i.e.
1011 – 1001.
for example;
9 1001
- 11 - 1011This
operation is not possible with unsigned binary representation
- 2
So far we have dealt with binary
additions of only POSITIVE numbers or subtractions that yield only POSITIVE
results. The reason for this is that it is not possible in PURE (or unsigned) binary
to signify whether a number is positive or negative. This we lead us to the
next topic on Signed binary numbers.
3.3 Signed Binary24
To represent a binary number signifying
its sign, we have to device a means of identifying the sign within the range of
symbols available to the binary number system, i.e. {0 and 1}. In signed binary
number, we represent the sign of a particular number, using the MSB of the
number as the sign bit and remaining bits as the magnitude bits. ‘0’ denotes Positive (+) and ‘1’ mean Negative (-), for example;
In the eight bit binary representation
of decimal +27 and -27 above, the MSB denotes the sign of the number, while the
remaining seven bits represents the magnitude of the number.
The use of signed binary with sign bit
to perform arithmetic operation in digital system involves a lot more work,
because certain bit in the stream must be treated as special, this means it
will require a more complex and thus expensive circuits. Aim of digital
electronics designers is to optimize design by making the circuit simpler, more
functional and less expensive.
A better means of performing digital arithmetic
with adequate representation and interpretation of signed numbers is the use of
complements of binary numbers. These
are One’s complement and Two’s complement representation.
3.4 Complements of binary numbers
Complements are used in digital
computers to simplify the subtraction operation and for logical manipulation.
Simplifying operations leads to simpler, less expensive circuits to implement
the operations. For binary numbers, we have 1’s complement and 2’s complement.
In mathematics, subtraction can be implemented
in a variety of different ways as A – B, is the same as saying A + (-B) or –B +
A. Using complements, subtraction of two binary numbers can be performed by
simply using addition, i.e. A – B = A + (-B), -B is the complement of B.
3.4.1 One’s Complement25
The complement (or opposite)
of +4 is −4. One’s Complement (1’s
complement) is a way of representing negative binary numbers in a signed binary
number system. In 1’s complement, positive numbers (also called
non-complements) remain unchanged as before with the sign-magnitude numbers.
Negative numbers however,
are represented by taking the 1’s complement of the positive number. To obtain
1’s compliment of a binary number, we subtract each digit of the number from 1.
For example;
We will observe in the
examples above that, when subtracting binary digit from ‘1’, we can have either
1 – 0 = 1 or 1 – 1 =
0, which causes the bit to change from 0 to 1 or from 1 to 0m
respectively. Therefore, 1’s complement of a binary number is formed by
changing 1’s to 0’s and 0’s to 1’s.
For example:
+510 is 000001012 and −510 is 111110102 (1’s
Complement)
In the example above, the most significant
bit (MSB) in the negative number –510 is 1, just as in signed
binary. The remaining 7 bits of the negative number however are not the same as
in signed binary notation. They are just the complement of the remaining 7
bits, and these give the value or magnitude of the number.
Binary
Addition26 and Subtraction with 1’s Complement
Both addition and subtraction in 1’s
complement are addition; subtraction is just the addition of the 1’s complement
of the Subtrahend i.e. 8 – 3 = 8 + (-3).
The steps are as follows;
1. Convert
negative numbers to its equivalent 1’s complements
2. Leave
all positive numbers unchanged
3. Perform
binary addition on the numbers
4. Add
any carry out of the sign bit to the result.
5. It
the result is negative (i.e. MSB = 1) find its 1’s complement.
6. If
the result is positive (i.e. MSB = 0) leave the result as it is.
Let’s illustrate these steps with
examples.
Example 3.4 Add -4 and + 6 using ones
compliment
The result in this case is a negative
number because the MSB = 1, therefore we
have to determine the complement of the number i.e. the positive value of the
number by finding its 1’compliment;
1’s complement of 11111000 ≡ 00000111
= +7, therefore the answer 11111000 = -7 (correct)
Note that, n-bit notation can be used to represent numbers in the range from
–(2n-1 – 1) to +(2n-1-1) using the 1’s complement format.
Using 1’s complement in binary addition
and subtraction has limitation in that we have to consider the overflow into a
non-existing bit to obtain a correct value of our operation. This took us to
the next better option which is 2’s complement.
3.4.2 Two’s
Complement27
Two’s complement (2’s Complement)
notation solves the problem of the relationship between positive and negative numbers,
and achieves accurate results in subtractions without the need to consider the
carry into a non-existing binary bit.
To perform binary subtraction28 the twos complement system uses the technique
of complementing the number to be subtracted.
To produce a negative number in
two complement system, we simply find the 1’s complement of the number and add
1.
For example;
2’s complement of +6 ≡ 000000110
= 11111001 (1’s complement)
+
1
11111010 (2’s complement)
Addition
and Subtraction using 2’s complement
1. Convert
negative numbers to its equivalent 2’s complements
2. Leave
all positive numbers unchanged
3. Perform
binary addition on the numbers
4. Discard any carry out of the sign bit.
5. It
the result is negative (i.e. MSB = 1) find its 2’s complement.
6. If
the result is positive (i.e. MSB = 0) leave the result as it is.
Example 3.6: Add +5 and -5 in
binary using 2’s complement.
Again, n-bit notation can be used to represent numbers in the range from
–(2n-1 ) to +(2n-1-1) using the 1’s complement format.
Overflow29
In order to obtain a correct
answer, we must ensure that the result had a sufficient number of bits to
accommodate the sum. If we start with two n-bit numbers and the sum occupies n
+ 1 bits, we say that an overflow occurs.
When one performs the addition with paper and pencil, an overflow is not
a problem, because we are not limited by the width of the page. We just add
another 0 to a positive number or another 1 to a negative number in the most
significant position to extend the number to n+1 bits and then perform the
addition. Overflow is a problem in computer because the number of bits that
hold a number is finite, and a result that exceeds the finite value by 1 cannot
be accommodated.
For example;
Example 3.8:
With
four bits perform -6 -3 using 2’s complement..
+
6 ≡ 0110
+
3 ≡ 0011
- 6 ≡ 1010 (2’s complement)
- 3 ≡
1101 (2’s complement)
The result obtain in example 3.8
is obviously wrong, -6-3 = -9 not +7, discrepancy in the answer is due to the
fact that the expected result can not be accommodate in four bit binary, for a
sign binary representation as mentioned earlier, N bit number will represent
from – 2N-1-1 to +2N-1-1
decimal number. In this example, 4 bit
can represent -24-1to +24-1-1 i.e. -8 to +7. The expected
result of -9 is not within this range and thus overflow occurs. To detect
overflow in signed binary addition; if the carry-in to the sign bit is not
equal to carry-out of the sign bit, overflow has occur. That, is to obtain
correct result with addition in 2’s complement, the carry-in to the sign bit must
be the same with the carry out of the sign bit. For example 3.8, carry-in to
the sign bit is ‘0’ while carry-out of the sign bit is ‘1’.
Earlier in the this chapter, we stated that the basic binary arithmetic operations are addition, subtraction, multiplication and division. We have been able to discuss how digital circuit perform addition and subtraction as addition of negative numbers.We will not cover binary multiplication and division in this text, but it is worth mentioning that multiplication is performed as repeated addition while subtraction is performed as repeated subtraction in digital circuit. In this chapter we have
extensively discussed the process of binary addition and subtraction, earlier
we mentioned that the basic binary arithmetic operations are addition,
subtraction, multiplication and division. We will not cover binary
multiplication and binary division in this text, but it is helpful for exploration reader to mention here that, binary multiplication is simply repeated addition30 while binary division is simply repeated subtraction31.
Exercise
3
a.
Obtain
the 1’s and 2’s Complements of the following binary numbers:
i. 00010000 ii.
00000000 iii. 11011010 iv. 10101010
v.
10000101 vi. 11111111
b.
Write
the eight-bit 2’s complement representation of (-23)10
c.
What
is the first thing you must do in order to solve the subtraction problem 1110
– 410 using 2’s compliment.
d.
What
is the general technique for subtracting binary numbers using 2’s compliment.
e.
Solve
each of the following 5-bit subtraction problems using 2’s complement
representation.
i.
001102
- 001012
ii.
011002
- 010102
iii.
001002
- 001012
iv.
010012
-010112
v.
000112
- 011002
f.
Solve
each of the following 8-bit subtraction problems using 2’s complement
representation:
i.
011111112
- 7810
ii.
001100102
- 12310
iii.
010010012
- 11110
iv.
000001112
- 3510
g.
Perform
operation (3E91)16 – (1DEF)16 using 2’s complement
arithmetic.
h.
Convert
your answers from question (3) to decimal.
i.
Prove
that 16-bit 2’s complement arithmetic cannot be used to add +18150 and +14618,
while it can be sued to add -18150 and -14618
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