From time immemorial, we have used the decimal system for counting.  The reason is that we have a total of ten digits on our hands which are valuable in addition.  There are ten unique digits in the system, namely 0 1 2 3 4 5 6 7 8 and 9.  All other numbers are constructed by combining two or more of these digits in any order.  But a computer does not have fingers and thumbs.  Instead it has switches.  Switches have only two positions, Off and On.  So we use a system based on two digits known as Base two or Binary.  Off is represented by 0 and On by 1.  The two binary digits 0 and 1 are also called bits.   Since 8 and 16 are powers of 2, it is easy to convert a binary system to a Base 8 or Octal and a Base 16 or Hexadecimal system.  The eight digits in the Octal system are 0 1 2 3 4 5 6 and 7, while the 16 digits in the Hexadecimal system are 0 1 2 3 4 5 6 7 8 9 A B C D E and F.  The base is also called Radix.

The table below gives the decimal numbers with their binary, octal, and hexadecimal equivalents.
 
 
 

Decimal	Binary	Octal	Hexadecimal
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
20	10100	24	14
30	11110	36	1E
40	101000	50	28
50	110010	62	32
60	111100	74	3C
70	1000110	106	46
80	1010000	120	50
90	1011010	132	5A
100	1100100	144	64
1000	1111101000	1750	3E8
2989	101110101101	5655	BAD

To convert a decimal number into any other divide by the radix.   Divide by the radix, that is by 2 to convert to binary, by 8 for octal and by 16 for hexadecimal, etc.  Write the quotient under the original number, and the remainder at its side.  Continue dividing until the quotient is zero, noting the remainder at each stage.  Reverse the order of the remainders, that is the last remainder is the first digit, and the first remainder is the last digit.  This yields the binary, the octal, or the hexadecimal, or any other required radix  number.  An example is given below for converting 1492 to its binary equivalent.

        Quotients  Remainders

        1 4 9 2

            7 4 6                    0

            3 7 3                    0

            1 8 6                    1

                9 3                   0

                46                    1

                23                    0

                11                    1

                  5                    1

                   2                   1

                   1                   0

                   0                   1

Reversing the order of the remainders, the binary equivalent is 1 0 1 1 1 0 1 0 1 0 0
That is 1492₁₀ is equivalent to 10111010100₂

To convert from binary to decimal, sum up the powers of 2 corresponding to the 1 bit.  The rightmost bit is 2 ⁰ the one to its left is 2¹ and so on.  Ignore the 0 bit.  For example

        1    0    1    1    0

        2⁴  2³  2²   2¹  2⁰  The binary equivalent is 1 x 2⁴ + 0 x 2³ + 1 x 2² + 1 x 2¹ + 0 x 2⁰  This equals 16 + 4 + 2 = 22.
       16   8  4    2    1
To convert number from any other radix, use the powers of that radix.

To convert a binary  number to octal, divide it into groups of three bits starting from both sides of the decimal point. Each group is then converted into digits 0 to 7.  To convert from binary to hexadecimal, the process is the same except we use four rather than three bits in each group.  This is then converted to digits 0 to F.  In reverse conversion, each digit is replace by three (for octal) or four (for hexadecimal) binary digits.  In this case conversion of numbers after the decimal point is also easy.  See the two examples in the textbook on page 636.
 
 

Four different systems are used to represent negative numbers in binary.

1.    Signed magnitude.   The leftmost bit is the sign bit, o for positive and 1 for negative.  The remaining bits constitute the absolute magnitude of the number.

2.    One's complement.   To obtain the negative of a number replace every 0 bit by 1 and every 1 bit by 0.  This can also be achieved by subracting each number from one.  This system is now obsolete.

3.    Two's complement.    This is a two step.  The first step is the same as for one's complement, that is replacing every 0 by 1 and every 1 by 0 (or subtract each digit from 1).  The second step is to add 1 to the result.  Any carry occurring from the leftmost bit is discarded.

4.    Excess 2^{m - 1}.  A number is negated by adding 2^{m - 1}, where m is the number of bits.  If there are 8 bits in the number, m = 8 - 1 = 7.  2⁷ = 128.  So 128 is added to the negative of the number.  -3 is stored as -3 + 128 = 125.  In binary 125 is 1111101.  Using 8 bits, -3₁₀ is 01111101₂.

Both signed magnitude and one's complement have two representations for zero: a plus zero and a minus zero, which is undesirable.  In two's complement a negative of plus zero is also plus zero.  Also having 8 bits, there is no negative for 128 in binary, in signed magnitude and in one's complement.  The table below gives negative 8-bit numbers in all four systems.
 
 
 

N decimal	N binary	-N signed magnitude	-N one's complement	-N two's complement	-N excess 128
0	00000000	10000000	11111111	00000000	Non existent
1	00000001	10000001	11111110	11111111	01111111
2	00000010	10000010	11111101	11111110	01111110
3	00000011	10000011	11111100	11111101	01111101
4	00000100	10000100	11111011	11111100	01111100
5	00000101	10000101	11111010	11111011	01111011
6	00000110	10000110	11111001	11111010	01111010
7	00000111	10000111	11111000	11111001	01111001
8	00001000	10001000	11110111	11111000	01111000
9	00001001	10001001	11110110	11110111	01110111
10	00001010	10001010	11110101	11110110	01110110
20	00010100	10010100	11101011	11101100	01101100
30	00011110	10011110	11100001	11100010	01100010
40	00101000	10101000	11010111	11011000	01011000
50	00110010	10110010	11001101	11001110	01001110
60	00111100	10111100	11000011	11000100	01000100
70	01000110	11000110	10111001	10111010	00111010
80	01010000	11010000	10101111	10110000	00110000
90	01011010	11011010	10100101	10100110	00100110
100	01100100	11100100	10011011	10011100	00011100
127	01111111	11111111	10000000	10000001	00000001
128	Non existent	Non existent	Non existent	10000000	00000000

Note:  128 does not exist in signed 8-bit binary as 128₁₀ is equivalent to 10000000₂.  But since the first bit is a sign bit and since 1 is negative, this really is equivalent to -0.  However in unsigned 8-bit binary, 128₁₀ is equivalent to 10000000₂.  Similarly 0 does not exist in excess 128 as it equals 10000000.  But again the first bit is a sign bit.  In excess 128, all numbers are positive.  Also note that except for the sign bit, all magnitude bits are same in two's complement and excess 2^{m - 1}.

1.    Convert the following numbers to Binary:    1984, 4000, 8192

2.    What is 1001101001 (Binary) in Decimal?    In Octal?    In Hexadecimal?

3.    Which of the following are valid hexadecimal numbers?    BED,    CAB,    DEAD,    DECADE,    ACCEDED,    BAG,    DAD.

4.    Express the decimal number 100 in all radices from 2 to 9

5.    How many different positive integers can be expressed in k digits using radix r numbers?

6.    Most people can only count to 10 on their fingers; however, computer scientists can do better.  If you regard each finger as one binary bit, with fingers extended as 1 and finger touching palm as 0, how high can you count using both hands?  With both hands and both feet?    Now use both hands and both feet, with the big toe on your left foot as a sign bit for two's complement numbers.  What is the range of expressible numbers?

7.    Perform the following calculations on 8-bit two's complement numbers

       00101101           11111111           00000000          11110111
    + 01101111      + 11111111         - 11111111        - 11110111
 

8.    Repeat the calculations of the preceding problem but now in one's complement.

9.    Consider the following addition problems for 3-bit binary numbers .  For each sum, state
    a.    Whether the sign bit of the result is 0 or 1
    b.    Whether the low-order 3 bits are 0 or 1
    c.    Whether an overflow occurred.

        000            000            111            100            100
     +001          +111          +110          +111          +100

10.    Signed decimal numbers consisting of n digits can be represented in n + 1 digits without a sign.  positive numbers have 0 as the leftmost digit.  Negative numbers are formed by subtracting each digit from 9.  Thus the negative of 014725 is 985274.  Such numbers are called nine's complement numbers and are analogous to one's complement binary numbers.  Express the following as three-digit nine's complement numbers:    6, -2, 100, -14, -1, 0.

11.    Determine the rule for addition on nine's complement numbers and then perform the following additions.

           0001           0001            9997           9241
        + 9999     + 9998         + 9996      + 0802
.
14.    Multiply 0111 by 0011 in binary.

1.    The numbers are:    11111000000,    111110100000,    and 10000000000000

2.    In decimal it is 617, in octal it is 1151, and in hexadecimal it is 269.

3.    All strings containing the ten digits and the letters A, B, C, D, E, and F are valid hexadecimal numbers.  All the examples given are valid except BAG.

4.    The answers are:    1100100,  10201,  1210,  400, 244, 202, 144, and 121.

5.    The number of valid strings of length k with r possibilities in each position is r^k

6.    With just your hands you have 10 bits, so you can count to 2¹⁰ - 1 or 1023.  With hand and feet the limit is 2²⁰-1 or 1,048,575.  In two's complement the range is -2¹⁹ to 2¹⁹  - 1 or -524,288 to +524,287.

7.    The results are:    10011100,  11111110,  00000001,  00000000.  If the calculations results in an overflow, ignore the first digit.  Use only 8 bits not 9.

8.    The results are:    10011100,  11111110,   00000000,  11111111.  The overflow is added. Use only 8 bits.

9.    The five sums are (use only 3 bits):    001, 111, 101,  011, and 000.
    a.    The sign bits (first bits) are:    0,  1,  1,  0,  and 0.
    b.    The low order bits are:    1,  1,  1,  1,  and 0.
    c.    Overflow?    No,  No,  Yes,  Yes,  and Yes.

10.    The numbers are:    006,  997,  100,  985,  998, and 000.

11.    Add the numbers.  If there is an overflow, add 1.  The answers are:    0001, 9999, 9994, and 0044.

14.    The answer is 10101 (7 x 3 = 21).

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