Review Questions Of Number Systems

Review Questions
1. What is meant by the radix or base of a number system? Briefly describe why hex representation is
used for the addresses and the contents of the memory locations in the main memory of a computer.
2. What do you understand by the l’s and 2’s complements of a binary number? What will be the
range of decimal numbers that can be represented using a 16-bit 2’s complement format?
3. Briefly describe the salient features of the IEEE-754 standard for representing floating-point
numbers.
4. Why was it considered necessary to carry out a revision of the IEEE-754 standard? What are the
main features of IEEE-754r (the notation for IEEE-754 under revision)?
5. In a number system, what decides (a) the place value or weight of a given digit and (b) the maximum
numbers representable with a given number of digits?
6. In a floating-point representation, what represents (a) the range of representable numbers and (b)
the precision with which a given number can be represented?
7. Why is there a need to have floating-point standards that can take care of decimal data and decimal
arithmetic in addition to binary data and arithmetic?
Problems
1. Do the following conversions:
(a) eight-bit 2’s complement representation of (−23)10;
(b) The decimal equivalent of (00010111)2 represented in 2’s complement form.
(a) 11101001; (b) +23
2. Two possible binary representations of (−1)10 are (10000001)2 and (11111111)2. One of them
belongs to the sign-bit magnitude format and the other to the 2’s complement format. Identify.
(10000001)2
= sign-bit magnitude and (11111111)2
= 2’s complement form
3. Represent the following in the IEEE-754 floating-point standard using the single-precision format:
(a) 32-bit binary number 11110000 11001100 10101010 00001111;
(b) (−118.625)10.
(a) 01001111 01110000 11001100 10101010;
(b) 11000010 11101101 01000000 00000000
4. Give the next three numbers in each of the following hex sequences:
(a) 4A5, 4A6, 4A7, 4A8, ;
(b) B998, B999,
(a) 4A9, 4AA, 4AB; (b) B99A, B99B, B99C
5. Show that:
(a) (13A7)16
= (5031)10;
(b) (3F2)16
= (1111110010)2.
6. Assume a radix-32 arbitrary number system with 0–9 and A–V as its basic digits. Express the mixed
binary number (110101.001)2 in this arbitrary number system.

Range of Numbers and Precision

Range of Numbers and Precision :

          The range of numbers that can be represented in any machine depends upon the number of bits in the
exponent, while the fractional accuracy or precision is ultimately determined by the number of bits
in the mantissa. The higher the number of bits in the exponent, the larger is the range of numbers
that can be represented. For example, the range of numbers possible in a floating-point binary number
format using six bits to represent the magnitude of the exponent would be from 2−64 to 2+64, which
is equivalent to a range of 10−19to 10+19. The precision is determined by the number of bits used to
represent the mantissa. It is usually represented as decimal digits of precision. The concept of precision
as defined with respect to floating-point notation can be explained in simple terms as follows. If the
mantissa is stored in n number of bits, it can represent a decimal number between 0 and 2n−1 as the
mantissa is stored as an unsigned integer. If M is the largest number such that 10M −1 is less than or
equal to 2n−1, then M is the precision expressed as decimal digits of precision. For example, if the
mantissa is expressed in 20 bits, then decimal digits of precision can be found to be about 6, as 220 −1
equals 1 048 575, which is a little over 106−1. We will briefly describe the commonly used formats
for binary floating-point number representation.

Floating Point Number Formats :

           The most commonly used format for representing floating-point numbers is the IEEE-754 standard.
The full title of the standard is IEEE Standard for Binary Floating-point Arithmetic (ANSI/IEEE STD
754-1985). It is also known as Binary Floating-point Arithmetic for Microprocessor Systems, IEC
60559:1989. An ongoing revision to IEEE-754 is IEEE-754r. Another related standard IEEE 854-
1987 generalizes IEEE-754 to cover both binary and decimal arithmetic. A brief description of salient
features of the IEEE-754 standard, along with an introduction to other related standards, is given below.
ANSI/IEEE-754 Format

The IEEE-754 floating point is the most commonly used representation for real numbers on
computers including Intel-based personal computers, Macintoshes and most of the UNIX platforms.
It specifies four formats for representing floating-point numbers. These include single-precision,
double-precision, single-extended precision and double-extended precision formats. Table 1.1 lists
characteristic parameters of the four formats contained in the IEEE-754 standard. Of the four formats
mentioned, the single-precision and double-precision formats are the most commonly used ones. The
single-extended and double-extended precision formats are not common.
Figure 1.1 shows the basic constituent parts of the single- and double-precision formats. As shown in
the figure, the floating-point numbers, as represented using these formats, have three basic components
including the sign, the exponent and the mantissa. A ‘0’ denotes a positive number and a ‘1’ denotes
a negative number. The n-bit exponent field needs to represent both positive and negative exponent
values. To achieve this, a bias equal to 2n−1− 1 is added to the actual exponent in order to obtain the
stored exponent. This equals 127 for an eight-bit exponent of the single-precision format and 1023 for
an 11-bit exponent of the double-precision format. The addition of bias allows the use of an exponent
in the range from −127 to +128, corresponding to a range of 0–255 in the first case, and in the range
from −1023 to +1024, corresponding to a range of 0–2047 in the second case. A negative exponent
is always represented in 2’s complement form. The single-precision format offers a range from 2−127
to 2+127, which is equivalent to 10−38 to 10+38. The figures are 2−1023 to 2+1023, which is equivalent to 10−308 to 10+308 in the case of the double-precision format.

           The extreme exponent values are reserved for representing special values. For example, in the case
of the single-precision format, for an exponent value of −127, the biased exponent value is zero,
represented by an all 0s exponent field. In the case of a biased exponent of zero, if the mantissa is zero
as well, the value of the floating-point number is exactly zero. If the mantissa is nonzero, it represents
a denormalized number that does not have an assumed leading bit of ‘1’. A biased exponent of +255,
corresponding to an actual exponent of +128, is represented by an all 1s exponent field. If the mantissa
is zero, the number represents infinity. The sign bit is used to distinguish between positive and negative
infinity. If the mantissa is nonzero, the number represents a ‘NaN’ (Not a Number). The value NaN is
used to represent a value that does not represent a real number. This means that an eight-bit exponent
can represent exponent values between −126 and +127. Referring to Fig. 1.1(a), the MSB of byte 1
indicates the sign of the mantissa. The remaining seven bits of byte 1 and the MSB of byte 2 represent
an eight-bit exponent. The remaining seven bits of byte 2 and the 16 bits of byte 3 and byte 4 give a
23-bit mantissa. The mantissa m is normalized. The left-hand bit of the normalized mantissa is always


‘1’. This ‘1’ is not included but is always implied. A similar explanation can be given in the case of
the double-precision format shown in Fig. 1.1(b).
Step-by-step transformation of (23)10 into an equivalent floating-point number in single-precision
IEEE format is as follows:
• (23)10
= (10111)2
= 1.0111e + 0100.
• The mantissa = 0111000 00000000 00000000.
• The exponent = 00000100.
• The biased exponent = 00000100 + 01111111 = 10000011.
• The sign of the mantissa = 0.
• (+23)10
= 01000001 10111000 00000000 00000000.
• Also, (–23)10
= 11000001 10111000 00000000 00000000.
IEEE-754r Format
As mentioned earlier, IEEE-754r is an ongoing revision to the IEEE-754 standard. The main objective of
the revision is to extend the standard wherever it has become necessary, the most obvious enhancement
to the standard being the addition of the 128-bit format and decimal format. Extension of the standard
to include decimal floating-point representation has become necessary as most commercial data are
held in decimal form and the binary floating point cannot represent decimal fractions exactly. If the
binary floating point is used to represent decimal data, it is likely that the results will not be the same as
those obtained by using decimal arithmetic.
In the revision process, many of the definitions have been rewritten for clarification and consistency.
In terms of the addition of new formats, a new addition to the existing binary formats is the 128-bit
‘quad-precision’ format. Also, three new decimal formats, matching the lengths of binary formats,
have been described. These include decimal formats with a seven-, 16- and 34-digit mantissa, which
may be normalized or denormalized. In order to achieve maximum range (decided by the number of
exponent bits) and precision (decided by the number of mantissa bits), the formats merge part of the
exponent and mantissa into a combination field and compress the remainder of the mantissa using
densely packed decimal encoding. Detailed description of the revision, however, is beyond the scope
of this book.
IEEE-854 Standard
The main objective of the IEEE-854 standard was to define a standard for floating-point arithmetic
without the radix and word length dependencies of the better-known IEEE-754 standard. That is why
IEEE-854 is called the IEEE standard for radix-independent floating-point arithmetic. Although the
standard specifies only the binary and decimal floating-point arithmetic, it provides sufficient guidelines
for those contemplating the implementation of the floating point using any other radix value such
as 16 of the hexadecimal number system. This standard, too, specifies four formats including single,
single-extended, double and double-extended precision formats.
Example 1.11
Determine the floating-point representation of −14210 using the IEEE single-precision format.
Solution
• As a first step, we will determine the binary equivalent of (142)10. Following the procedure outlined
in an earlier part of the chapter, the binary equivalent can be written as (142)10
= (10001110)2.
• (10001110)2
= 1.000 1110 × 27 = 1.0001110e + 0111.
• The mantissa = 0001110 00000000 00000000.
• The exponent = 00000111.
• The biased exponent = 00000111 + 01111111 = 10000110.
• The sign of the mantissa = 1.
• Therefore, −14210
= 11000011 00001110 00000000 00000000.
Example 1.12
Determine the equivalent decimal numbers for the following floating-point numbers:
(a) 00111111 01000000 00000000 00000000 (IEEE-754 single-precision format);
(b) 11000000 00101001 01100 45 0s (IEEE-754 double-precision format).
Solution
(a) From an examination of the given number:
The sign of the mantissa is positive, as indicated by the ‘0’ bit in the designated position.
The biased exponent = 01111110.
The unbiased exponent = 01111110−01111111 = 11111111.
It is clear from the eight bits of unbiased exponent that the exponent is negative, as the 2’s
complement representation of a number gives ‘1’ in place of MSB.
The magnitude of the exponent is given by the 2’s complement of (11111111)2, which is
(00000001)2= 1.
Number Systems 17
Therefore, the exponent =−1.
The mantissa bits = 11000000 00000000 00000000 (‘1’ in MSB is implied).
The normalized mantissa = 1.1000000 00000000 00000000.
The magnitude of the mantissa can be determined by shifting the mantissa bits one position to the left.
That is, the mantissa = (.11)2
= (0.75)10.
(b) The sign of the mantissa is negative, indicated by the ‘1’ bit in the designated position.
The biased exponent = 10000000010.
The unbiased exponent = 10000000010−01111111111 = 00000000011.
It is clear from the 11 bits of unbiased exponent that the exponent is positive owing to the ‘0’ in
place of MSB. The magnitude of the exponent is 3. Therefore, the exponent = +3.
The mantissa bits = 1100101100 45 0s (‘1’ in MSB is implied).
The normalized mantissa = 1.100101100 45 0s.
The magnitude of the mantissa can be determined by shifting the mantissa bits three positions to
the right.
That is, the mantissa = (1100.101)2
= (12.625)10.
Therefore, the equivalent decimal number =−12625.

Floating Point Numbers

          Floating-point notation can be used conveniently to represent both large as well as small fractional
or mixed numbers. This makes the process of arithmetic operations on these numbers relatively much
easier. Floating-point representation greatly increases the range of numbers, from the smallest to the
largest, that can be represented using a given number of digits. Floating-point numbers are in general
expressed in the form
N = m×be (1.1)
where m is the fractional part, called the significant or mantissa, e is the integer part, called the
exponent, and b is the base of the number system or numeration. Fractional part m is a p-digit number
of the form (±d.dddd dd), with each digit d being an integer between 0 and b – 1 inclusive. If the
leading digit of m is nonzero, then the number is said to be normalized.
Equation (1.1) in the case of decimal, hexadecimal and binary number systems will be written as
follows:
Decimal system
N = m×10e (1.2)
Number Systems 13
Hexadecimal system
N = m×16e (1.3)
Binary system
N = m×2e (1.4)
For example, decimal numbers 0.0003754 and 3754 will be represented in floating-point notation
as 3.754 × 10−4 and 3.754 × 103 respectively. A hex number 257.ABF will be represented as
2.57ABF × 162. In the case of normalized binary numbers, the leading digit, which is the most
significant bit, is always ‘1’ and thus does not need to be stored explicitly.
Also, while expressing a given mixed binary number as a floating-point number, the radix point is
so shifted as to have the most significant bit immediately to the right of the radix point as a ‘1’. Both
the mantissa and the exponent can have a positive or a negative value.
The mixed binary number (110.1011)2 will be represented in floating-point notation as .1101011
× 23 = .1101011e+0011. Here, .1101011 is the mantissa and e+0011 implies that the exponent is
+3. As another example, (0.000111)2 will be written as .111e−0011, with .111 being the mantissa
and e−0011 implying an exponent of −3. Also, (−0.00000101)2 may be written as −.101 × 2−5 =
−.101e−0101, where −.101 is the mantissa and e−0101 indicates an exponent of −5. If we wanted
to represent the mantissas using eight bits, then .1101011 and .111 would be represented as .11010110
and .11100000.

The Four Axioms

                 Conversion of a given number in one number system to its equivalent in another system has been discussed at length in the preceding sections. The methodology has been illustrated with solved examples. The complete methodology can be summarized as four axioms or principles, which, if understood properly, would make it possible to solve any problem related to conversion of a given number in one number system to its equivalent in another number system.
These principles are as follows:
1. Whenever it is desired to find the decimal equivalent of a given number in another number system,
it is given by the sum of all the digits multiplied by their weights or place values. The integer and
fractional parts should be handled separately. Starting from the radix point, the weights of different
digits are r0, r1, r2 for the integer part and r−1, r−2, r−3 for the fractional part, where r is the radix
of the number system whose decimal equivalent needs to be determined.
2. To convert a given mixed decimal number into an equivalent in another number system, the integer
part is progressively divided by r and the remainders noted until the result of division yields a
zero quotient. The remainders written in reverse order constitute the equivalent. r is the radix of
the transformed number system. The fractional part is progressively multiplied by r and the carry
recorded until the result of multiplication yields a zero or when the desired number of bits has been
obtained. The carrys written in forward order constitute the equivalent of the fractional part.
3. The octal–binary conversion and the reverse process are straightforward. For octal–binary
conversion, replace each digit in the octal number with its three-bit binary equivalent. For
hexadecimal–binary conversion, replace each hex digit with its four-bit binary equivalent. For
binary–octal conversion, split the binary number into groups of three bits, starting from the binary
point, and, if needed, complete the outside groups by adding 0s, and then write the octal equivalent
of these three-bit groups. For binary–hex conversion, split the binary number into groups of four
bits, starting from the binary point, and, if needed, complete the outside groups by adding 0s, and
then write the hex equivalent of the four-bit groups.
4. For octal–hexadecimal conversion, we can go from the given octal number to its binary equivalent
and then from the binary equivalent to its hex counterpart. For hexadecimal–octal conversion, we
can go from the hex to its binary equivalent and then from the binary number to its octal equivalent. 
Example 1.9
Assume an arbitrary number system having a radix of 5 and 0, 1, 2, L and M as its independent digits.
Determine:
(a) the decimal equivalent of (12LM.L1);
(b) the total number of possible four-digit combinations in this arbitrary number system.
Solution
(a) The decimal equivalent of (12LM) is given by
M×50 +L×51+2×52 +1×53 = 4×50 +3×51+2×52 +1×53L = 3M= 4
= 4+15+50+125 = 194
The decimal equivalent of (L1) is given by
L×5−1+1×5−2 = 3×5−1+5−2 = 064
Combining the results, (12LM.L1)5
= (194.64)10.
(b) The total number of possible four-digit combinations = 54 = 625.
Example 1.10
The 7’s complement of a certain octal number is 5264. Determine the binary and hexadecimal
equivalents of that octal number.
Solution
• The 7’s complement = 5264.
• Therefore, the octal number = (2513)8.
• The binary equivalent = (010 101 001 011)2
= (10101001011)2.
• Also, (10101001011)2
= (101 0100 1011)2
= (0101 0100 1011)2
= (54B)16.
• Therefore, the hex equivalent of (2513)8
= (54B)16 and the binary equivalent of (2513)8
=(10101001011)2.

Hex–Octal and Octal–Hex Conversions

       For hexadecimal–octal conversion, the given hex number is firstly converted into its binary equivalent
which is further converted into its octal equivalent. An alternative approach is firstly to convert the
given hexadecimal number into its decimal equivalent and then convert the decimal number into an
equivalent octal number. The former method is definitely more convenient and straightforward. For
octal–hexadecimal conversion, the octal number may first be converted into an equivalent binary
number and then the binary number transformed into its hex equivalent. The other option is firstly to
convert the given octal number into its decimal equivalent and then convert the decimal number into
its hex equivalent. The former approach is definitely the preferred one. Two types of conversion are
illustrated in the following example.
 
Example 1.8
Let us find the octal equivalent of (2F.C4)16 and the hex equivalent of (762.013)8
Number Systems 11
Solution
• The given hex number = (2F.C4)16.
• The binary equivalent = (0010 1111.1100 0100)2
= (00101111.11000100)2
= (101111.110001)2
= (101 111.110 001)2
= (57.61)8.
• The given octal number = (762.013)8.
• The octal number = (762.013)8
= (111 110 010.000 001 011)2
= (111110010.000001011)2
= (0001 1111 0010.0000 0101 1000)2
= (1F2.058)16.

Binary Octal and Octal Binary Conversions

           An octal number can be converted into its binary equivalent by replacing each octal digit with its
three-bit binary equivalent. We take the three-bit equivalent because the base of the octal number
system is 8 and it is the third power of the base of the binary number system, i.e. 2. All we have then
to remember is the three-bit binary equivalents of the basic digits of the octal number system. A binary
number can be converted into an equivalent octal number by splitting the integer and fractional parts
into groups of three bits, starting from the binary point on both sides. The 0s can be added to complete
the outside groups if needed.
 
Example 1.6
Let us find the binary equivalent of (374.26)8 and the octal equivalent of (1110100.0100111)2
Solution
• The given octal number = (374.26)8
• The binary equivalent = (011 111 100.010 110)2
= (011111100.010110)2
• Any 0s on the extreme left of the integer part and extreme right of the fractional part of the equivalent
binary number should be omitted. Therefore, (011111100.010110)2
= (11111100.01011)2
• The given binary number = (1110100.0100111)2
• (1110100.0100111)2
= (1 110 100.010 011 1)2
= (001 110 100.010 011 100)2
= (164.234)8

Hex–Binary and Binary–Hex Conversions

        A hexadecimal number can be converted into its binary equivalent by replacing each hex digit with its
four-bit binary equivalent. We take the four-bit equivalent because the base of the hexadecimal number
system is 16 and it is the fourth power of the base of the binary number system. All we have then to
remember is the four-bit binary equivalents of the basic digits of the hexadecimal number system. A
given binary number can be converted into an equivalent hexadecimal number by splitting the integer
and fractional parts into groups of four bits, starting from the binary point on both sides. The 0s can
be added to complete the outside groups if needed.
Example 1.7
Let us find the binary equivalent of (17E.F6)16 and the hex equivalent of (1011001110.011011101)2.
Solution
• The given hex number = (17E.F6)16
• The binary equivalent = (0001 0111 1110.1111 0110)2
= (000101111110.11110110)2
= (101111110.1111011)2
• The 0s on the extreme left of the integer part and on the extreme right of the fractional part have
been omitted.
• The given binary number = (1011001110.011011101)2
= (10 1100 1110.0110 1110 1)2
• The hex equivalent = (0010 1100 1110.0110 1110 1000)2
= (2CE.6E8)16

Decimal to Hexadecimal Conversion

           The process of decimal-to-hexadecimal conversion is also similar. Since the hexadecimal number
system has a base of 16, the progressive division and multiplication factor in this case is 16. The
process is illustrated further with the help of an example.
Example 1.5
Let us determine the hexadecimal equivalent of (82.25)10
Solution
• The integer part = 82
Divisor Dividend Remainder
16 82 —
16 5 2
— 0 5
• The hexadecimal equivalent of (82)10
= (52)16
• The fractional part = 0.25
• 0.25 × 16 = 0 with a carry of 4
• Therefore, the hexadecimal equivalent of (82.25)10
= (52.4)16

Decimal to Octal Conversion

         The process of decimal-to-octal conversion is similar to that of decimal-to-binary conversion. The
progressive division in the case of the integer part and the progressive multiplication while working
on the fractional part here are by ‘8’ which is the radix of the octal number system. Again, the integer
and fractional parts of the decimal number are treated separately. The process can be best illustrated
with the help of an example.
Example 1.4
We will find the octal equivalent of (73.75)10
Solution
• The integer part = 73
Divisor Dividend Remainder
8 73 —
8 9 1
8 1 1
— 0 1
Number Systems 9
• The octal equivalent of (73)10
= (111)8
• The fractional part = 0.75
• 0.75 × 8 = 0 with a carry of 6
• The octal equivalent of (0.75)10
= (.6)8
• Therefore, the octal equivalent of (73.75)10
= (111.6)8

Decimal to Binary Conversion

       As outlined earlier, the integer and fractional parts are worked on separately. For the integer part,
the binary equivalent can be found by successively dividing the integer part of the number by 2
and recording the remainders until the quotient becomes ‘0’. The remainders written in reverse order
constitute the binary equivalent. For the fractional part, it is found by successively multiplying the
fractional part of the decimal number by 2 and recording the carry until the result of multiplication
is ‘0’. The carry sequence written in forward order constitutes the binary equivalent of the fractional
part of the decimal number. If the result of multiplication does not seem to be heading towards zero in the
case of the fractional part, the process may be continued only until the requisite number of equivalent bits
has been obtained. This method of decimal–binary conversion is popularly known as the double-dabble
method. The process can be best illustrated with the help of an example.
 
Example 1.3
We will find the binary equivalent of (13.375)10.
Solution
• The integer part = 13
Divisor Dividend Remainder
2 13 —
2 6 1
2 3 0
2 1 1
— 0 1
• The binary equivalent of (13)10 is therefore (1101)2
• The fractional part = .375
• 0.375 × 2 = 0.75 with a carry of 0
• 0.75 × 2 = 0.5 with a carry of 1
• 0.5 × 2 = 0 with a carry of 1
• The binary equivalent of (0.375)10
= (.011)2
• Therefore, the binary equivalent of (13.375)10
= (1101.011)2

Finding the Decimal Equivalent

       The decimal equivalent of a given number in another number system is given by the sum of all
the digits multiplied by their respective place values. The integer and fractional parts of the given
number should be treated separately. Binary-to-decimal, octal-to-decimal and hexadecimal-to-decimal
conversions are illustrated below with the help of examples.

Binary-to-Decimal Conversion :
The decimal equivalent of the binary number (1001.0101)2 is determined as follows:
• The integer part = 1001
• The decimal equivalent = 1 × 20 + 0 × 21 + 0 × 22 + 1 × 23 = 1 + 0 + 0 + 8 = 9
• The fractional part = .0101
• Therefore, the decimal equivalent = 0 × 2−1 + 1 × 2−2 + 0 × 2−3 + 1 × 2−4 = 0 + 0.25 + 0
+ 0.0625 = 0.3125
• Therefore, the decimal equivalent of (1001.0101)2
= 9.3125

Octal-to-Decimal Conversion :

The decimal equivalent of the octal number (137.21)8 is determined as follows:
• The integer part = 137
• The decimal equivalent = 7 × 80 + 3 × 81 + 1 × 82 = 7 + 24 + 64 = 95
Number Systems 7
• The fractional part = .21
• The decimal equivalent = 2 × 8−1 + 1 × 8−2 = 0.265
• Therefore, the decimal equivalent of (137.21)8
= (95.265)10

Hexadecimal-to-Decimal Conversion :
 The decimal equivalent of the hexadecimal number (1E0.2A)16 is determined as follows:
• The integer part = 1E0
• The decimal equivalent = 0 × 160 + 14 × 161 + 1 × 162 = 0 + 224 + 256 = 480
• The fractional part = 2A
• The decimal equivalent = 2 × 16−1 + 10 × 16−2 = 0.164
• Therefore, the decimal equivalent of (1E0.2A)16
= (480.164)10
 
Example 1.2
Find the decimal equivalent of the following binary numbers expressed in the 2’s complement format:
(a) 00001110;
(b) 10001110.
Solution
(a) The MSB bit is ‘0’, which indicates a plus sign.
The magnitude bits are 0001110.
The decimal equivalent = 0×20 +1×21+1×22 +1×23+0×24 +0×25+0×26
= 0+2+4+8+0+0+0 = 14
Therefore, 00001110 represents +14
(b) The MSB bit is ‘1’, which indicates a minus sign
The magnitude bits are therefore given by the 2’s complement of 0001110, i.e. 1110010
The decimal equivalent = 0×20 +1×21+0×22 +0×23+1×24 +1×25
+1×26
= 0+2+0+0+16+32+64 = 114
Therefore, 10001110 represents −114

Introduction to Number Systems

1.2 Introduction to Number Systems :

                  We will begin our discussion on various number systems by briefly describing the parameters that are common to all number systems. An understanding of these parameters and their relevance to number systems is fundamental to the understanding of how various systems operate. Different characteristics that define a number system include the number of independent digits used in the number system, the place values of the different digits constituting the number and the maximum numbers that can
be written with the given number of digits. Among the three characteristic parameters, the most
fundamental is the number of independent digits or symbols used in the number system. It is known as
the radix or base of the number system. The decimal number system with which we are all so familiar
can be said to have a radix of 10 as it has 10 independent digits, i.e. 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9.
Similarly, the binary number system with only two independent digits, 0 and 1, is a radix-2 number
system. The octal and hexadecimal number systems have a radix (or base) of 8 and 16 respectively.
We will see in the following sections that the radix of the number system also determines the other
two characteristics. The place values of different digits in the integer part of the number are given by
r0, r1, r2, r3 and so on, starting with the digit adjacent to the radix point. For the fractional part, these
are r−1, r−2, r−3 and so on, again starting with the digit next to the radix point. Here, r is the radix
of the number system. Also, maximum numbers that can be written with n digits in a given number
system are equal to rn.
 
1.3 Decimal Number System :

                The decimal number system is a radix-10 number system and therefore has 10 different digits or
symbols. These are 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9. All higher numbers after ‘9’ are represented in terms
of these 10 digits only. The process of writing higher-order numbers after ‘9’ consists in writing the
second digit (i.e. ‘1’) first, followed by the other digits, one by one, to obtain the next 10 numbers
from ‘10’ to ‘19’. The next 10 numbers from ‘20’ to ‘29’ are obtained by writing the third digit (i.e.
‘2’) first, followed by digits ‘0’ to ‘9’, one by one. The process continues until we have exhausted all
possible two-digit combinations and reached ‘99’. Then we begin with three-digit combinations. The
first three-digit number consists of the lowest two-digit number followed by ‘0’ (i.e. 100), and the
process goes on endlessly.
The place values of different digits in a mixed decimal number, starting from the decimal point, are
100, 101, 102 and so on (for the integer part) and 10−1, 10−2, 10−3 and so on (for the fractional part).
Number Systems 3
The value or magnitude of a given decimal number can be expressed as the sum of the various digits
multiplied by their place values or weights.
As an illustration, in the case of the decimal number 3586.265, the integer part (i.e. 3586) can be
expressed as
3586 = 6×100 +8×101+5×102 +3×103 = 6+80+500+3000 = 3586
and the fractional part can be expressed as
265 = 2×10−1+6×10−2 +5×10−3 = 02+006+0005 = 0265
We have seen that the place values are a function of the radix of the concerned number system and
the position of the digits. We will also discover in subsequent sections that the concept of each digit
having a place value depending upon the position of the digit and the radix of the number system is
equally valid for the other more relevant number systems.

1.4 Binary Number System :
                 The binary number system is a radix-2 number system with ‘0’ and ‘1’ as the two independent digits.
All larger binary numbers are represented in terms of ‘0’ and ‘1’. The procedure for writing higherorder
binary numbers after ‘1’ is similar to the one explained in the case of the decimal number system.
For example, the first 16 numbers in the binary number system would be 0, 1, 10, 11, 100, 101, 110,
111, 1000, 1001, 1010, 1011, 1100, 1101, 1110 and 1111. The next number after 1111 is 10000, which
is the lowest binary number with five digits. This also proves the point made earlier that a maximum
of only 16 (= 24 numbers could be written with four digits. Starting from the binary point, the place
values of different digits in a mixed binary number are 20, 21, 22 and so on (for the integer part) and
2−1, 2−2, 2−3 and so on (for the fractional part).
 
Example 1.1
Consider an arbitrary number system with the independent digits as 0, 1 and X. What is the radix of
this number system? List the first 10 numbers in this number system.
Solution
• The radix of the proposed number system is 3.
• The first 10 numbers in this number system would be 0, 1, X, 10, 11, 1X, X0, X1, XX and 100.
1.4.1 Advantages
Logic operations are the backbone of any digital computer, although solving a problem on computer
could involve an arithmetic operation too. The introduction of the mathematics of logic by George
Boole laid the foundation for the modern digital computer. He reduced the mathematics of logic to a
binary notation of ‘0’ and ‘1’. As the mathematics of logic was well established and had proved itself
to be quite useful in solving all kinds of logical problem, and also as the mathematics of logic (also
known as Boolean algebra) had been reduced to a binary notation, the binary number system had a
clear edge over other number systems for use in computer systems. Yet another significant advantage of this number system was that all kinds of data could be conveniently represented in terms of 0s and 1s. Also, basic electronic devices used for hardware implementation could be conveniently and efficiently operated in two distinctly different modes. For example, a bipolar transistor could be operated either in cut-off or in saturation very efficiently. Lastly, the circuits required for performing arithmetic operations such as addition, subtraction, multiplication, division, etc., become a simple affair when the data involved are represented in the form of 0s and 1s.
1.5 Octal Number System :
 
         The octal number system has a radix of 8 and therefore has eight distinct digits. All higher-order
numbers are expressed as a combination of these on the same pattern as the one followed in the case
of the binary and decimal number systems described in Sections 1.3 and 1.4. The independent digits
are 0, 1, 2, 3, 4, 5, 6 and 7. The next 10 numbers that follow ‘7’, for example, would be 10, 11, 12,
13, 14, 15, 16, 17, 20 and 21. In fact, if we omit all the numbers containing the digits 8 or 9, or both,
from the decimal number system, we end up with an octal number system. The place values for the
different digits in the octal number system are 80, 81, 82 and so on (for the integer part) and 8−1, 8−2,
8−3 and so on (for the fractional part).
 
1.6 Hexadecimal Number System :
 
          The hexadecimal number system is a radix-16 number system and its 16 basic digits are 0, 1, 2, 3,
4, 5, 6, 7, 8, 9, A, B, C, D, E and F. The place values or weights of different digits in a mixed
hexadecimal number are 160, 161, 162 and so on (for the integer part) and 16−1, 16−2, 16−3 and so on
(for the fractional part). The decimal equivalent of A, B, C, D, E and F are 10, 11, 12, 13, 14 and 15
respectively, for obvious reasons.
The hexadecimal number system provides a condensed way of representing large binary numbers
stored and processed inside the computer. One such example is in representing addresses of different
memory locations. Let us assume that a machine has 64K of memory. Such a memory has 64K (= 216
= 65 536) memory locations and needs 65 536 different addresses. These addresses can be designated
as 0 to 65 535 in the decimal number system and 00000000 00000000 to 11111111 11111111 in the
binary number system. The decimal number system is not used in computers and the binary notation
here appears too cumbersome and inconvenient to handle. In the hexadecimal number system, 65 536
different addresses can be expressed with four digits from 0000 to FFFF. Similarly, the contents of the
memory when represented in hexadecimal form are very convenient to handle.
 
1.7 Number Systems – Some Common Terms
In this section we will describe some commonly used terms with reference to different number systems.
 
1.7.1 Binary Number System
Bit is an abbreviation of the term ‘binary digit’ and is the smallest unit of information. It is either ‘0’
or ‘1’. A byte is a string of eight bits. The byte is the basic unit of data operated upon as a single unit
in computers. A computer word is again a string of bits whose size, called the ‘word length’ or ‘word
size’, is fixed for a specified computer, although it may vary from computer to computer. The word
length may equal one byte, two bytes, four bytes or be even larger.
Number Systems 5
The 1’s complement of a binary number is obtained by complementing all its bits, i.e. by replacing
0s with 1s and 1s with 0s. For example, the 1’s complement of (10010110)2 is (01101001)2. The 2’s
complement of a binary number is obtained by adding ‘1’ to its 1’s complement. The 2’s complement
of (10010110)2 is (01101010)2.
 
1.7.2 Decimal Number System
Corresponding to the 1’s and 2’s complements in the binary system, in the decimal number system we
have the 9’s and 10’s complements. The 9’s complement of a given decimal number is obtained by
subtracting each digit from 9. For example, the 9’s complement of (2496)10 would be (7503)10. The
10’s complement is obtained by adding ‘1’ to the 9’s complement. The 10’s complement of (2496)10
is (7504)10.
 
1.7.3 Octal Number System
In the octal number system, we have the 7’s and 8’s complements. The 7’s complement of a given
octal number is obtained by subtracting each octal digit from 7. For example, the 7’s complement of
(562)8 would be (215)8. The 8’s complement is obtained by adding ‘1’ to the 7’s complement. The 8’s
complement of (562)8 would be (216)8.
 
1.7.4 Hexadecimal Number System
The 15’s and 16’s complements are defined with respect to the hexadecimal number system. The 15’s
complement is obtained by subtracting each hex digit from 15. For example, the 15’s complement of
(3BF)16 would be (C40)16. The 16’s complement is obtained by adding ‘1’ to the 15’s complement.
The 16’s complement of (2AE)16 would be (D52)16.