Number Systems
Introduction
Binary Number System
The goal of this handout is to make you comfortable with the binary number system. We
will correlate your previous knowledge of the decimal number system to the binary
number system. That will lay the foundations on which our discussion of various
representation schemes for numbers (both integer and real numbers) will be based.
Binary means base 2 (the prefix bi). Based on our earlier discussion of the decimal
number system, the digits that can be used to count in this number system are 0 and 1.
The 0,1 used in the binary system are called binary digits (bits)
The bit is the smallest piece of information that can be stored in a computer. It can have
one of two values 0 or 1. Think of a bit as a switch that can be either on or off. For
example,
Decimal Number System
Counting as we have been taught since kindergarten is based on the decimal number
system. Decimal means base 10 (the prefix dec). In any number system, given the base
(often referred to as radix), the number of digits that can be used to count is fixed. For
example in the base 10 number system, the digits that can be used to count are
0,1,2,3,4,5,6,7,8,9.
Generalizing that for any base b, the first b digits (starting with 0) represent the digits that
are used to count. When a number ≥ b has to be represented, the place values are used.
Example 1. Consider the number 1234. It can be represented as
1*103+ 2*102 + 3*101 + 4*100
From the hardware perspective, ON and OFF can be represented as voltage levels
(typically 0V for logic 0 and +3.3 to +5V for logic 1). Since only two values can be
stored in a bit, we combine a series of bits to represent more information. Again the
concept of place values is applicable here as well.
Example 3. Consider the binary number 1101. It can be represented as
(1)
Where:
 1 is in the thousand’s place
 2 is in the hundred’s place
 3 is in the ten’s place
 4 is in the one’s place.
1*23+ 1*22 + 0*21 + 1*20
(3)
This expanded notation also gives you the means of converting binary numbers directly
into the equivalent decimal number.
8 + 4 + 0 + 1 = 13
The equation (1) is an expanded representation of 1234. The expanded representation has
the advantage of making the base of the number system explicit.
Example 2. Consider the number 1234.567. It is represented as
1*103+ 2*102 + 3*101 + 4*100+ 5*101+ 6*102+ 7*103
Bit
Value
0 OFF / FALSE
1
ON/ TRUE
Table 1. Interpreting Bit Values
(2)
Where:
 5 is in the tenth’s place
 6 is in the hundredth’s place
 7 is in the thousandth’s place
In equation (2), the representation includes digits both to the left and to the right of the
decimal point.
Example 4. Consider the binary number 1101.101. It can be represented as:
1*23+ 1*22 + 0*21 + 1*20+ 1*21+ 0*22+ 1*23
(4)
The same notation is applicable to real numbers represented in binary notation. The
equivalent decimal number is
13 + 0.5 + 0 + 0.125 = 13.625
To represent larger numbers, we have to group series of bits. Two of these groupings are
of importance:
 Nibble
A nibble is a group of four bits
 Byte
A byte is a group of eight bits The byte is the smallest addressable unit in most computers. The key components of a
byte are shown below in figure 1.
msb
lsb
a7 a6 a5 A4 a3 a2 a1 a0
Figure 1. Byte
Byte BigEndian LittleEndian
0
00000100
00000001
1
00000001
00000100
Table 2. BigEndian versus LittleEndian Representation of 00000100 00000001
The endian system may sometimes be used to represent the bit order within a byte. In this
context, equation (5) refers to the littleendian ordering of bits within the byte.
The most significant bit (msb) is the bit with the highest place value, while the least
significant bit (lsb) denotes the bit position that has the lowest place value. To convert the
byte to the equivalent integer number, the formula in (5) is used.
a7 * 27 + a6 * 26 + a5 * 25 + a4 * 24 + a3 * 23 + a2 * 22 + a1 * 21 + a0 * 20
Operations on Numbers
(5)
Self Exercise: What is the value of the bit pattern 11111111 ?
Answer: 255
Self Exercise: What is the range of values represented by an 8bit binary number?
Answer: 0 to 255.
Note: When we look at negative number representation using the same 8bit number, the
range will change.
When we want to represent a value larger than 255, we have to group bytes together to
form words. The size of words is dependent on the underlying processor, but is usually an
even number of bytes (typically 4 bytes).
In the bigendian system, the byte with the largest significance is stored first (bigendfirst). Conversely, in the littleendian system, the byte with the least significance is stored
first (littleendfirst). The number 1025 in binary is 00000100 00000001.
The sequencing of bytes to form larger numbers leads to the issue of which is the first
byte in the sequence. Many mainframe computers, particularly IBM mainframes, use a
bigendian architecture. Most modern computers, including PCs, use the littleendian
system. The PowerPC system is biendian because it can understand both systems
The term endian comes from Swift's "Gulliver's Travels" via the famous paper "On Holy
Wars and a Plea for Peace" by Danny Cohen, USC/ISI IEN 137, 19800401. The
Lilliputians, being very small, had correspondingly small political problems. The BigEndian and LittleEndian parties debated over whether softboiled eggs should be opened
at the big end or the little end.
For any given number system, the operations of addition and subtraction are fundamental.
Operations such as multiplication and division can be implemented using addition and
subtraction. Again, we will correlate the addition and subtraction operations in the
decimal number system to the binary number system.
Example 5. Consider the operation 145 + 256
Carry 1
Carry 1
+
145
256
401
The algorithm works by starting at the least significant digit and working from right to
left. At each place position, the digits are added and if the resulting number is a single
digit, it is entered in the same place position in the sum. If on the other hand, the
summation operation results in 2 digits, the digit of lower significance is entered into the
place position of the sum and the digit of higher significance is added along with the
other digits in the next most significant place (or is carried over to the next most
significant place).
The same principle applies to binary addition.
Rule
1
2
3
4
Step Result Carry
0+0
0
0
0+1
1
0
1+0
1
0
1+1
0
1
Table 3.Rules for Binary Addition Signed Magnitude Representation
Example 6. Consider the operation:
Carry 1
+
11100100
00000101
11101001
Note that at the 2nd bit position, there is a carry of 1 into the 3rd bit position (counting
from the right with 0 being the first position).
The signed magnitude representation uses the most significant bit to determine if the
number is positive or negative. The advantage of this notation is that by examining the
msb alone, it is possible to determine if the number is positive or negative. The
disadvantage however is that one bit pattern is wasted (there are two possible
representations for zero) and subtraction cannot be performed using addition alone. Table
4. shows the signed magnitude representation of numbers using 4 bits.
SelfExercise: Convert the numbers above into decimal to verify that the answer is
correct.
Bit Pattern Number
0000
0
0001
1
0010
2
0011
3
0100
4
0101
5
0110
6
0111
7
1000
0
1001
1
1010
2
1011
3
1100
4
1101
5
1110
6
1111
7
Answer: 1 1 1 0 0 1 0 0 = 228, 0 0 0 0 0 1 0 1 = 5, 1 1 1 0 1 0 0 1 = 233 = 228+ 5
Decimal subtraction works very similar to decimal addition, the numbers are aligned to
the same place values and the algorithm proceeds from right to left. The bottom digit is
subtracted from the top digit, and the result written in the place value position in the
result. If the top digit is less than the bottom digit, then we must 'borrow' from the next
place value position. That means decrementing the top digit in the next significant
position and adding the base to the top digit of this position before performing the
subtraction. This operation gets even more complicated when there is a ‘0’ in the next
significant position.
Example 7. Consider the decimal operation 445  256:
445
256
189
The decimal subtraction algorithm can get very complicated and the time taken to
perform the subtraction can vary greatly. Since the binary addition algorithm is already
understood and already implemented in hardware, it can be reused to also perform
subtraction.
Table 4. Numbers using 4bit signed magnitude representation
Note: +0 and –0 have different bit patterns.
Example 8. Consider the following operation 7 – 2. Substituting the bit patterns from the
table:
The operation x – y can be rewritten as x+ (y). That brings up the question – How are
negative numbers represented in binary notation?
+
0111
1010
1 0001
Negative Numbers
The bit pattern 0001 is 1 but the result should by 5 0101.
Negative binary numbers are represented by the ‘’ sign followed by the magnitude of the
number. The computer however does not have a means of representing signs. The sign
has to be captured in the bit pattern itself.
One’s Complement The one’s complement notation represents a negative number by inverting the bits in
each place. The one’s complement notation for a 4bit number is shown in Table 5. Again
the limitations of the sign magnitude representation are not overcome (there are two bit
patterns used to represent 0 and the addition operation cannot be used to perform
subtraction). The one’s complement is important because it is very easy to perform the
inversion operation in hardware and it forms the basis of computing the two’s
complement.
Bit Pattern Number
0000
0
0001
1
0010
2
0011
3
0100
4
0101
5
0110
6
0111
7
1111
0
1110
1
1101
2
1100
3
1011
4
1010
5
1001
6
1000
7
Table 5. Numbers using 1’s complement representation
The two’s complement notation has the advantages that the sign of the number can be
computed by looking at the msb. The addition operation can be used to perform
subtraction. Also, there is only one bitpattern to represent ‘0’ so an extra number can be
represented. Table 6 summarizes the 2’s complement notation for a 4bit number.
Bit Pattern Number
0000
0
1
0001
0010
2
0011
3
0100
4
5
0101
0110
6
0111
7
1
1111
1110
2
1101
3
4
1100
1011
5
1010
6
7
1001
1000
8
Table 6. Numbers using 2’s complement representation
Example 10. Consider the following operation 7 – 2. Substituting the bit patterns from the
table:
Example 9. Consider the following operation 7 – 2. Substituting the bit patterns from the
table:
+
+
0111
1101
1 0100
The bit pattern 0100 is 4 but the result should by 5 0101.
0111
1110
1 0101
The bit pattern 0101 is 5, which is the expected result.
The limitation with the 2’s complement notation is that the bit patterns are not in order
i.e. comparing the bit patterns alone does not provide any information as to which
number is larger.
Two’s Complement
The two’s complement notation builds on the one’s complement notation. The algorithm
goes as follows:
1. Compute the 1’s complement
2. Add 1 to the result to get the 2’s complement.
Excess Notation
The excess notation is a means of representing both negative and positive numbers in a
manner in which the order of the bit patterns is maintained. The algorithm for computing
the excess notation bit pattern is as follows: 1. Add the excess value (2N1, where N is the number of bits used to represent the
number) to the number.
SelfExercise: What is the range of numbers that can be represented using the Excess2N1
notation?
2. Convert the resulting number into binary format.
Answer: 2N1 1 to 2N1
The 2N1 is often referred to as the Magic Number for computing the excess
representation of the number (except that there is no magic in it). Table 7 presents all the
numbers that can be represented using the excess8 notation.
Number Excess Number Bit Pattern
7
15
1111
6
14
1110
5
13
1101
4
12
1100
3
11
1011
2
10
1010
1
9
1001
0
8
1000
1
7
0111
2
6
0110
3
5
0101
4
4
0100
5
3
0011
6
2
0010
7
1
0001
8
0
0000
Table 7. Numbers using the Excess8 representation
The number of bits used to represent a code in excess8 is 4 bits (241 = 8). Also, the bit
patterns are in sequence (the largest number that can be represented has the bit pattern
1111).
Example 11. Consider the following operation 7 – 2. Substituting the bit patterns from the
table:
+
1111
0110
1 0101
The result of the addition operation is the bitpattern used for 5 in binary. The excess
notation representation however takes longer to compute than the 2’s complement
notation. The excess notation will however play an important part in computing floatingpoint representations.
Bias Notation
The excess notation is a special case of the biased notation. For instance, excess8 is
biased around 8 (i.e.0 has the bit pattern associated with decimal 8). Instead of using the
magic number, any number (bias) can be used.
Note: This concept becomes important when we address the IEEE Single Precision
FloatingPoint standard.
FloatingPoint Notation
The floatingpoint notation is used:
a. To represent integers that are larger than the maximum value that can be held by a
bitpattern (the maximum value that can be held by 8 bits is 255).
b. To represent real numbers.
Large Integers
Consider a really large number 1,234,567. The number requires seven places to represent
the value. If the number of places available to represent the number is limited to say four
places, certain digits have to be dropped. The selection of digits to be dropped is based on
the value associated with the digit. In this case, we will drop the last three digits ‘567’.
The resulting number is:
1,234,000
The loss of ‘567’ is a loss of precision but if the most significant digits were to be
eliminated, say ‘123’, then the resulting number is 4,567, which presents an even greater
loss of precision.
Rules for determining significance (integers):
1. A nonzero digit is always significant
2. The digit '0' is significant if it lies between other significant digits
3. The digit '0' is never significant if it precedes all the nonzero digits
SelfExercise: What are the significant digits in 0012340?
Answer: 0012340 1,234,000 can be represented using only four digits as 1234 * 103. This representation is
called the exponential form.
In the binary fixedpoint notation, the radix position is fixed at a certain point within the
bit pattern. To illustrate the notation, we will consider an 8bit bitpattern (byte) as shown
below.
The exponential form consists of two parts:
mantissa – (the significand) the significant digits (1234)
exponent – the power to which the base is raised before being multiplied by the
mantissa. (3).
a7 a6 a5 a4 a3 a2 a1 a0
We can define the fixedpoint notation to be:
Two special forms of representation are:
6
1.234 * 10 – the scientific notation, in which the decimal point is to the right of
the most significant digit.
0.1234 * 107 – the normalized notation, in which the decimal point is to the left
of the most significant digit.
Real Numbers
Mathematically, real numbers are set of rational and irrational numbers. A rational
number is any number that can be represented as a ratio of two integers (a/b, b ≠ 0).
Irrational numbers are real numbers that are not rational i.e. they cannot be represented as
a ratio of two integers (typically numbers whose decimal expansion never ends and never
enters a periodic pattern).
There are a number of ways of representing a real number in a computer.
1. One notation is the fixed point, wherein the decimal point (radix) is fixed at some
position between the digits. The digits to the left of the radix are integer values
and those to the right of the radix are fractions of some fixed unit. For example:
10.82 = 1 x 101 + 0 x 100 + 8 x 101 + 2 x 102, the radix is between 0 and 8. This
notation is limited both in the range of numbers that can be represented as well as
in the precision of the numbers that can be represented.
2. Another notation is to use rational notation (represent the real number as a ratio of
two integers). This representation is not always possible because all real numbers
are not necessarily rational!
3. The floatingpoint notation is the most common of the representation schemes. It
is based on either the scientific or normalized representation of the number.

a7, a6, a5, a4 – contain the integer part in two’s complement form
a3, a2, a1, a0 – contain the fractional part in normal binary form
a7a6a5a4 Value
0000
0
0001
1
0010
2
3
0011
0100
4
0101
5
6
0110
0111
7
1111
1
1110
2
1101
3
1100
4
1011
5
1010
6
1001
7
1000
8
Table 8. Integer part values
a3 a2 a1 a0
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
Value
0
0.0625
0.125
0.1875
0.250
0.3125
0.375
0.4375
0.5
0.5625
0.625
0.6875
0.75
0.8125
0.875
0.9375
Table 9. Fractional Part Values
Tables 8,9 show the possible values that the different segments of the fixedpoint notation
can take.
SelfExercise: What is the range of values that the fixedpoint notation can take?
Answer: 8.9375 to 7.93725
SelfExercise: What is the precision of the notation?
Answer: The precision is 0.0625.
Binary FixedPoint Notation
The notation makes an assumption about the representation (primarily the location of the
radix and the format used for the integer part). For two computers (or two programs for that matter), they must understand the representation that is being used. This is captured
by means of standards. At the end of this handout, we will discuss the IEEE standard
used for representing floating point numbers.
Example12. What is the value represented by the bit pattern 11011110? Assume the radix
point is between bit positions 3 and 4.
The bit pattern can be split into the integer part 1101 and the fractional part 1110.
1101 =
1110 =
11011110 =
3
0.875
3.875
1. Determine the value of the exponent
2. If the exponent x is negative,
a. Add x leading 0’s to the fractional part
b. Insert a radix point before the 0’s
3. If the exponent is positive,
a. Move the radix point to the right x places.
4. Convert the binary number into decimal
5. Add the sign based on the a7 bit.
SelfExercise: Why does step 3 in the algorithm above not address adding zeroes?
Answer: From the excess table above, it is clear that the maximum positive exponent
is 3. Moving the radix point right does not move to the right of the fraction part,
hence, padding (adding leading 0’s or trailing 0’s) to the right is not needed.
Modified Significance Rules
Rules for determining significance:
1. A ‘1’ bit is always significant
2. The bit '0' is significant if it lies between other significant bits
3. The bit '0' is never significant if it precedes all 0’s, even if it follows an embedded
radix (in the example above, the radix is embedded between bit positions 3 and 4).
Excess Number Actual Value BitPattern
7
3
111
6
2
110
5
1
101
4
0
100
3
1
011
2
010
2
1
3
001
0
4
000
4. The bit '0' is significant if it follows an embedded radix point and other significant
bits
Table 10. Excess4 Conversion Table
SelfExercise: What are the significant bits in the bit pattern00.010000?
Example 13. What is the decimal value associated with 01111001(assume that the bit
pattern is in 8bit floating point format)?
Answer: 00.010000
Answer: Split the byte into the respective components,
8bit Floating Point Notation
0
111
1001
Step 1. Convert the excessfour exponent into its decimal equivalent.
The 8bit floating point notation can be describe based on the byte shown below:
111 = 3, hence the exponent = 3.
a7 a6 a5 a4 a3 a2 a1 a0
Where
 a7
 a6 a5 a4
 a3 a2 a1 a0
Step 2. Since the exponent is positive, move the radix point three place to the
right.
– sign bit
– exponent in excess4 notation
– fractional part in normal binary
The algorithm for converting from 8bit floating point to decimal is detailed below:
The binary number
= 0.1001 * 23
= 100.1
Step 3. Convert the binary number to decimal, 100.1 = 4.5 Step 4. Add the sign, hence 01111001 = 4.5
IEEE 754 Single Precision Floating Point Notation
The algorithm for converting from decimal to 8bit floating point is detailed below:
1. Set the sign bit (a7) to 1 if the number is negative, 0 otherwise
2. Convert the number into binary representation.
3. Normalize the binary number (move the radix point to the left of the most
significant bit).
4. Convert the exponent into excess4 notation
5. Set a6 a5 a4 to the exponent value.
6. Select the 4 most significant bits and enter them into the fraction part (a3 a2 a1 a0).
Given that the normalization procedure always has a 1 in the first significant bit position,
it is possible to use scientific notation instead and always implicitly assume that the first
bit is 1.This allows us to gain an additional bit of precision in the representation.
Sign Bit
There are some limitations of the 8bit floatingpoint notation. Some of them are listed
below:
 The maximum positive exponent is 3
 The lowest negative exponent is –4
 The precision is determined by the exponent
Example 14: Convert 11/4 into 8bit floatingpoint notation.
Answer:
8 bit exponent in
bias 127 notation
23 bits to represent a 24 bit
mantissa
Figure 2. IEEE 32 bit floating point notation
The single precision representation uses 32 bits as shown in Figure 2 above. The standard
also defines a double precision standard that uses 64 bits. We will only be discussing the
single precision standard.
The method for converting decimal numbers into the 754 standard representation is as
follows:
Step 1.
0 a6 a5 a4 a3 a2 a1 a0
Step 2. Convert the number into binary form,11/4 = 2 ¾ = 0010.11
Step 3. Normalize 0010.11 = 0.1011 * 22
Step 4. Convert the exponent to excess4, 2 = 110
Step 5. Fill in the exponent
0 1 1 0 a3 a2 a1 a0
1. If the number is negative
a. set the sign bit to ‘1’
2. If the number is positive
a. Set the sign bit to ‘0’
3. Convert the decimal number into binary form.
4. Convert the binary number into scientific notation (move the radix immediately
right of the most significant bit).
5. Convert the exponent into bias127 notation
6. Fill in the 8 bits demarcated for the exponent
7. Fill in the bits (except the most significant bit) into the mantissa from left to right.
Pad any remaining spaces with 0’s.
Step 6. Fill in the fraction
0 1 1 0 1 0 1 1
Hence 2 ¾ = 01101011
The exponent is computed as bias127. The algorithm for computing the biased exponent
is as follows:
1. Add 127 to the decimal value of the exponent (The exponent value is negative if
the radix is moved to the left).
2. Convert the decimal value into binary Example 15. Convert 194.375 into IEEE754 single precision representation.
Answer:
Step1. Fill in the sign bit
0
Step 2. Convert the decimal number into binary.
194 =
0.375 =
194.375 =
11000010
0.011
11000010.011
Step 3. Convert to scientific notation
1.1000010011 * 27
Step 4.Convert the exponent into bias127 notation
7 + 127 = 134 = 10000110
Step 5.Fill in the exponent
010000110
Step6. Fill in the mantissa without the most significant bit.
01000011010000100110000000000000
Hence 194.375 = 0 1 0 0 0 0 1 1 0 1 0 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0
Endnote
This handout was developed for Unified Engineering, Department of Aeronautics and
Astronautics, MIT. Any errors of commission or omission are mine
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