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Computing Number Systems(Binary, Decimal, Octal and Hexadecimal)
(1’s and 2’s complement)
Binary
Digit
|
Octal Digit |
Decimal
Digit
|
Hexadecimal Digit |
|
0 |
0 |
0 |
0 |
|
1 |
1 |
1 |
1 |
|
|
2 |
2 |
2 |
|
|
3 |
3 |
3 |
|
|
4 |
4 |
4 |
|
|
5 |
5 |
5 |
|
|
6 |
6 |
6 |
|
|
7 |
7 |
7 |
|
|
|
8 |
8 |
|
|
|
9 |
9 |
|
|
|
|
A (decimal value of 10) |
|
|
|
|
B (decimal value of 11) |
|
|
|
|
C (decimal value of 12) |
|
|
|
|
D (decimal value of 13) |
|
|
|
|
E (decimal value of 14) |
|
|
|
|
F (decimal value of 15) |
Comparing the binary, octal, decimal and hexadecimal
number systems.
|
Attribute |
Binary |
Octal
|
Decimal |
Hexadecimal |
|
Base |
2 |
8 |
10 |
16 |
|
Lowest digit |
0 |
0 |
0 |
0 |
|
Highest digit |
1 |
7 |
9 |
F |
Positional values in the decimal number system
|
Decimal digit |
9 |
3 |
7 |
|
Positional name |
Hundreds |
Tens |
Ones |
|
Positional value |
100 |
10 |
1 |
|
Positional value as a Power of the base (10) |
102 |
101 |
100 |
For longer decimal numbers, the next
positions to the left would be the thousands position (10 to the 3rd power),
the ten-thousands position (10 to the 4th power), the hundred-thousands
position (10 to the 5th power), the millions position (10 to the 6th power)_,
the ten-millions position (10 to the 7th power) and so on.
Positional values in the binary number system
|
Binary digit |
1 |
0 |
1 |
|
Position name |
Fours |
Twos |
Ones |
|
Positional value |
4 |
2 |
1 |
|
Positional value as a Power of the bade (2) |
22 |
21 |
20 |
For longer binary numbers, the next positions to the left
would be the eights position (2 to the 3rd power), the sixteen’s position (2 to
the 4th power), the thirty-twos position (2 to the 5th power), the sixty–fours
position (2 to the 6th power) and so on.
In the octal number 425, we say that the 5 is written in
the ones position, the 2 is written in the eights position, and the 4 is written
in the sixty-fours position. Notice that each of these positions is a power of
the base (base 8), and that these powers begin at 0 and increase by 1 as we
move left in the number, given in below figure:
Positional values in the octal number system
|
Decimal Digit |
4 |
2 |
5 |
|
Position name |
Sixty-fours |
Eights |
Ones |
|
Positional value |
64 |
8 |
1 |
|
Positional value as a Power of the bade (8) |
82 |
81 |
80 |
For longer octal numbers, the next positions to the left would
be the five-hundred-and-twelve position (8 to the 3rd power), the
four-thousand-and-ninety-sixes position (8 to the 4th power), the
thirty-two-thousand-seven-hundred-and-sixty eights position (8 to the 5th
power) and so on.
IN the Hexadecimal number #3DA, we say that the A is written in the ones position, the
D is written in the sixteen’s position is a power of the base (base 16), and
that these powers begin at 0 and increase by 1 as we move left in the number,
given below:
Positional values in the hexadecimal number system
|
Decimal Digit |
3 |
D |
A |
|
|
Position name |
Two-hundred-and-fifty-sixes |
Eights |
Ones |
|
|
Positional value |
64 |
8 |
1 |
|
|
Positional value as a Power of the bade (16) |
82 |
81 |
80 |
|
For longer hexadecimal numbers, the nest positions to the
left would be the four-thousand-and-ninety-sixes position (16 to the 3rd
power), the sixty-five-thousand-five-hundred-and-thirty-six position (16 to the
4th power) and so on.
Abbreviating binary numbers as octal and hexadecimal numbers
The main use for octal and hexadecimal numbers in
computing is for abbreviating lengthy binary representations. The figure below
highlights the fact that lengthy binary numbers can be expressed concisely in
number systems with higher bases than the binary number system.
Decimal, Binary, Octal and Hexadecimal equivalents
|
Decimal Number |
Binary Representation |
Octal Representation |
Hexadecimal Representation |
|
0 |
0 |
0 |
0 |
|
1 |
1 |
1 |
1 |
|
2 |
10 |
2 |
2 |
|
3 |
11 |
3 |
3 |
|
4 |
100 |
4 |
4 |
|
5 |
101 |
5 |
5 |
|
6 |
110 |
6 |
6 |
|
7 |
111 |
7 |
7 |
|
8 |
1000 |
10 |
8 |
|
9 |
1001 |
11 |
9 |
|
10 |
1010 |
12 |
A |
|
11 |
1011 |
13 |
B |
|
12 |
1100 |
14 |
C |
|
13 |
1101 |
15 |
D |
|
14 |
1110 |
16 |
E |
|
15 |
1111 |
17 |
F |
|
16 |
10000 |
20 |
10 |
Binary Number Octal
Equivalent Hexadecimal Equivalent
100011010001 4321 8D1
To see how the binary number converts easily to octal, simply
break the 12-digit binary number into groups of three consecutive bits each,
and write those groups over the corresponding digits of the octal number as
follow:
100 011 010 001
4 3 2 1
Notice that the octal digit you have written under each
group of the bits corresponds precisely to the octal equivalent of that 3-digit
binary number as shown in the above figure.
The same kind of relationship may be observed in
converting numbers from binary to hexadecimal. In particular, break the
12-digit binary number into groups of four consecutive bits each and write
those groups over the corresponding digits of the hexadecimal number as follow:
1000 1101 0001
8 D 1
Converting Octal and Hexadecimal numbers to Binary
numbers
In the section mentioned above , we saw how to convert
binary numbers to their octal and hexadecimal equivalents by forming groups of
binary digits and simply rewriting these groups as their equivalent octal digit
values or hexadecimal digit values. This process may be used in reverse to
produce the binary equivalent of a given octal or hexadecimal number.
For example, the octal number 653 is converted to binary
simply by writing the 6 as its 3-digit binary equivalent 110, the 5 as its
3-digit binary equivalent 101 and the 3 as its 3-digit binary equivalent 011 to
form the 9-digit binary number 110101011.
The hexadecimal number FAD5 is converted to binary simply
by writing the F as its 4-digit binary equivalent 1111, the A as its 4-digit
binary equivalent 1010, the D as its 4-digit binary equivalent 1101 and the 5
as its 4-digit binary equivalent 0101 to form the 16 digit 1111101011010101.
CONVERTING FROM BINARY, OCTAL OR HEXADECIMAL TO DECIMAL
Because we are accustomed to working in decimal, it is often
convenient to convert a binary, octal or hexadecimal number to decimal to get
sense of what the number is “really” worth. To convert a number to decimal from
another base, multiply the decimal equivalent of each digit by its positional
value, and sum these products. For example, the binary number 110101 is
converted to decimal 53 in figure below:
Converting a binary number to decimal
|
Positional Values |
32 |
16 |
8 |
4 |
2 |
1 |
|
Symbol Values |
1 |
1 |
0 |
1 |
0 |
1 |
|
Products |
1*32=32 |
1*16=16 |
0*8=0 |
1*4=4 |
0*2=0 |
1*1=1 |
|
Sum |
=32+16+0+4+0+1 = 53 |
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To convert octal number 7614 to decimal 3980, we use the
same technique, the time using appropriate octal positional value as shown in
figure below:
Converting a octal number to decimal
|
Positional Values |
512 |
64 |
8 |
1 |
|
Symbol Values |
7 |
6 |
1 |
4 |
|
Products |
7*512=3584 |
6*64=384 |
1*8=8 |
4*1=4 |
|
Sum |
=584+384+8+4=3980 |
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To convert hexadecimal AD3B to decimal 44347, we use the same
technique, this time using appropriate hexadecimal positional values as shown
in fig below:
Converting a hexadecimal number to decimal
|
Positional Values |
4096 |
256 |
16 |
1 |
|
Symbol Values |
A |
D |
3 |
B |
|
Products |
A*4096=40960 |
D*256=3328 |
3*16=48 |
B*1=11 |
|
Sum |
=40960+3328+48+11=44347 |
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CONVERTING FROM DECIMAL TO BINARY,OCTAL OR HEXADECIMAL
The conversions of the previous section follow naturally from
the positional notation conventions. Converting from decimal to binary, octal
or hexadecimal also follows these conventions.
Suppose we wish to convert decimal 57 to binary. We begin
by writing the positional values of the columns right to left until we reach a
column whose positional values is greater than the decimal number. We do not
need that column, so we discard it. Thus, we first write;
Positional Values: 64 32 16 8 4 2 1
then we discard the column with positional value 64
leaving:
Positional Values: 32 16 8 4 2 1
NEXT we work from the leftmost column to the right. We
divide 32 into 57 and observe that there is one 32 in 57 with a remainder of
25, so we write 1 in the 32 column. We divide 16 into 25 and observe that there
is one 16 in 25 with a remainder of 9 and write 1 in the 16 column. We divide 8
into 9 and observe that there is 8 in 9 with a remainder of 1. The next two
columns each produce quotients of zero when their positional values are divided
into 1 so we write 0s in the 4 and 2 column. Finally, 1 into 1 is 1 so we write
1 in the 1 column. This yields:
Positional Values: 32 16 8 4 2 1
Symbol Values: 1 1 1 0 0 1
and thus decimal 57 is equivalent to binary 111001
To convert decimal 103 to octal, we begin by writing the positional
values of the columns until we reach a column whose positional value is greater
than the decimal number:
We do not need that column, so we discard it. Thus, we
first write:
Positional Values: 512 64 8 1
Then we discard the column with positional value 512,
yielding:
Positional Values: 64 8 1
NEXT we work from the leftmost column to the right. We
divide 64 into 103 and observe that there is one 64 in 103 with a remainder of
39, so we write 1 in the 64 column. We divide 8 into 39 and observe that there
are four 8s In 39 with a reminder of 7 and write 4 in the 8 column. Finally, we
divide 1 into 7 and observe that there are seven 1s in 7 with no remainder so
we write 7 in the 1 column. This yields:
Positional Values: 64 8 1
Symbol Values: 1 4 7
and thus decimal 103 is equivalent to octal 147.
To convert decimal 375 to hexadecimal, we begin by
writing the positional values of the columns until we reach a column whose
positional value is greater than the decimal number. We do not need that
column, so we discard it. Thus, we first write:
Positional Values: 4096 256 16 1
Then we discard the column with positional value 4096,
yielding
Positional Values: 256 16 1
NEXT we work from the leftmost column to the right. We
divide 256 into 375 and observe that there is one 256 in 375 and observe that
there is one 256 in 375 with a remainder of 119, so we write 1 in the 256
column. We divide 16 into 119 and observer that there are seven 16s in 119 with
a remainder of 7 and write 7 in the 16 columns. Finally, We divide 1 into 7 and
observe that there are seven 1s in 7 with no remainder so we write 7 in the 1
column. This yield:
Positional Values: 256 16 1
Symbol Values: 1 7 7
and thus decimal 375 is equivalent to hexadecimal 177.
NEGATIVE BINARY NUMBERS: TWO’S COMPLEMENTS NOTATION
Here we will see how computers represent negative numbers
using two’s complement notation.
Consider a machine with 32-bit integers. Suppose
VAR VALUE = 13;
The 32-bit representation of value is
00000000 00000000 000000000 00001101
To form the negative of value all the ones becomes ZERO
and all ZEROs becomes 1’s as follows:
11111111 11111111 11111111 11110010
To form the two’s complement of value we simply add one
to value’s one’s complement,
Thus, Two’s
complement of the above is
11111111 11111111 11111111 11110011
Now if this is in fact equal to –13, we should be able to
add it to binary 13 and obtain a result of 0. Let us try this:
00000000 00000000 000000000 00001101
11111111 11111111 11111111 11110011
00000000 00000000 000000000 00000000
The carry bit coming out of the leftmost column is
discarded and we indeed get zero as a
result. If we add the one’s complement of a number to the number, the result
would be all 1s. The key to getting a result of all zeros is that the twos
complement is 1 more than the one’s complement. The addition of 1 causes each
column to add to 0 with a carry of 1. The carry keeps moving leftward until it
is discarded from the leftmost bit, and the resulting number is all zeros.
Computers actually perform a subtraction such as
x = a – value
by adding the two’s complement of value to a as follow:
x = a + (~value + 1);
Suppose A is 27 and value is 13 as before. If the two’s
complement of value is actually the negative of value, then adding the two’s
complement of value to a should produce the result 14. Let us try this:
a(i.e. 27) 00000000 00000000 000000000 00011011
+(~value + 1) 11111111 11111111 11111111 11110011
----------------------------------------------------------------------------------
00000000 00000000 000000000
00001110
which is indeed equal to 14.
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