## Number System

### Bit & Byte

Computer uses the binary system. Any physical system that can exist in two distinct states (e.g., 0-1, on-off, hi-lo, yes-no, up-down, north-south, etc.) has the potential of being used to represent numbers or characters.

A binary digit is called a bit. There are two possible states in a bit, usually expressed as 0 and 1.

A series of eight bits strung together makes a byte, much as 12 makes a dozen. With 8 bits, or 8 binary digits, there exist 2^8=256 possible combinations. The following table shows some of these combinations. (The number enclosed in parentheses represents the decimal equivalent.)

00000000 (  0)   00010000 ( 16)   00100000 ( 32)  ...  01110000 (112)
00000001 (  1)   00010001 ( 17)   00100001 ( 33)  ...  01110001 (113)
00000010 (  2)   00010010 ( 18)   00100010 ( 34)  ...  01110010 (114)
00000011 (  3)   00010011 ( 19)   00100011 ( 35)  ...  01110011 (115)
00000100 (  4)   00010100 ( 20)   00100100 ( 36)  ...  01110100 (116)
00000101 (  5)   00010101 ( 21)   00100101 ( 37)  ...  01110101 (117)
00000110 (  6)   00010110 ( 22)   00100110 ( 38)  ...  01110110 (118)
00000111 (  7)   00010111 ( 23)   00100111 ( 39)  ...  01110111 (119)
00001000 (  8)   00011000 ( 24)   00101000 ( 40)  ...  01111000 (120)
00001001 (  9)   00011001 ( 25)   00101001 ( 41)  ...  01111001 (121)
00001010 ( 10)   00011010 ( 26)   00101010 ( 42)  ...  01111010 (122)
00001011 ( 11)   00011011 ( 27)   00101011 ( 43)  ...  01111011 (123)
00001100 ( 12)   00011100 ( 28)   00101100 ( 44)  ...  01111100 (124)
00001101 ( 13)   00011101 ( 29)   00101101 ( 45)  ...  01111101 (125)
00001110 ( 14)   00011110 ( 30)   00101110 ( 46)  ...  01111110 (126)
00001111 ( 15)   00011111 ( 31)   00101111 ( 47)  ...  01111111 (127)

:
(continued)
:

10000000 (128)   10010000 (144)   10100000 (160)  ...  11110000 (240)
10000001 (129)   10010001 (145)   10100001 (161)  ...  11110001 (241)
10000010 (130)   10010010 (146)   10100010 (162)  ...  11110010 (242)
10000011 (131)   10010011 (147)   10100011 (163)  ...  11110011 (243)
10000100 (132)   10010100 (148)   10100100 (164)  ...  11110100 (244)
10000101 (133)   10010101 (149)   10100101 (165)  ...  11110101 (245)
10000110 (134)   10010110 (150)   10100110 (166)  ...  11110110 (246)
10000111 (135)   10010111 (151)   10100111 (167)  ...  11110111 (247)
10001000 (136)   10011000 (152)   10101000 (168)  ...  11111000 (248)
10001001 (137)   10011001 (153)   10101001 (169)  ...  11111001 (249)
10001010 (138)   10011010 (154)   10101010 (170)  ...  11111010 (250)
10001011 (139)   10011011 (155)   10101011 (171)  ...  11111011 (251)
10001100 (140)   10011100 (156)   10101100 (172)  ...  11111100 (252)
10001101 (141)   10011101 (157)   10101101 (173)  ...  11111101 (253)
10001110 (142)   10011110 (158)   10101110 (174)  ...  11111110 (254)
10001111 (143)   10011111 (159)   10101111 (175)  ...  11111111 (255)

### K & M

2^10=1024 is commonly referred to as a "K". It is approximately equal to one thousand. Thus, 1 Kbyte is 1024 bytes. Likewise, 1024K is referred to as a "Meg". It is approximately equal to a million. 1 Mega byte is 1024*1024=1,048,576 bytes. If you remember that 1 byte equals one alphabetical letter, you can develop a good feel for size.

### Number System

You may regard each digit as a box that can hold a number. In the binary system, there can be only two choices for this number -- either a "0" or a "1". In the octal system, there can be eight possibilities:
"0", "1", "2", "3", "4", "5", "6", "7".
In the decimal system, there are ten different numbers that can enter the digit box:
"0", "1", "2", "3", "4", "5", "6", "7", "8", "9".
In the hexadecimal system, we allow 16 numbers:
"0", "1", "2", "3", "4", "5", "6", "7", "8", "9", "A", "B", "C", "D", "E", and "F".
As demonstrated by the following table, there is a direct correspondence between the binary system and the octal system, with three binary digits corresponding to one octal digit. Likewise, four binary digits translate directly into one hexadecimal digit. In computer usage, hexadecimal notation is especially common because it easily replaces the binary notation, which is too long and human mistakes in transcribing the binary numbers are too easily made.

Base Conversion Table

BIN    OCT   HEX   DEC
----------------------
0000   00     0     0
0001   01     1     1
0010   02     2     2
0011   03     3     3
0100   04     4     4
0101   05     5     5
0110   06     6     6
0111   07     7     7
----------------------
1000   10     8     8
1001   11     9     9
1010   12     A    10
1011   13     B    11
1100   14     C    12
1101   15     D    13
1110   16     E    14
1111   17     F    15

### Convert From Any Base To Decimal

Let's think more carefully what a decimal number means. For example, 1234 means that there are four boxes (digits); and there are 4 one's in the right-most box (least significant digit), 3 ten's in the next box, 2 hundred's in the next box, and finally 1 thousand's in the left-most box (most significant digit). The total is 1234:
Original Number:      1     2     3    4
|     |     |    |
How Many Tokens:      1     2     3    4
Digit/Token Value: 1000   100    10    1
Value:             1000 + 200  + 30  + 4  = 1234

or simply,  1*1000 + 2*100 + 3*10 + 4*1 = 1234
Thus, each digit has a value: 10^0=1 for the least significant digit, increasing to 10^1=10, 10^2=100, 10^3=1000, and so forth.

Likewise, the least significant digit in a hexadecimal number has a value of 16^0=1 for the least significant digit, increasing to 16^1=16 for the next digit, 16^2=256 for the next, 16^3=4096 for the next, and so forth. Thus, 1234 means that there are four boxes (digits); and there are 4 one's in the right-most box (least significant digit), 3 sixteen's in the next box, 2 256's in the next, and 1 4096's in the left-most box (most significant digit). The total is:

1*4096 + 2*256 + 3*16 + 4*1 = 4660

Example. Convert the hexadecimal number 4B3 to decimal notation. What about the decimal equivalent of the hexadecimal number 4B3.3?

Solution:

Original Number:     4    B    3  .  3
|    |    |     |
How Many Tokens:     4   11    3     3
Digit/Token Value: 256   16    1     0.0625
Value:            1024 +176  + 3   + 0.1875  = 1203.1875

Example. Convert 234.14 expressed in an octal notation to decimal.

Solution:

Original Number:     2    3    4  .  1       4
|    |    |     |       |
How Many Tokens:     2    3    4     1       4
Digit/Token Value:  64    8    1     0.125   0.015625
Value:             128 + 24  + 4   + 0.125 + 0.0625    = 156.1875

Another way is to think of a cash register with different slots, each holding bills of a different denomination.

### Convert From Decimal to Any Base

Again, let's think about what you do to obtain each digit. As an example, let's start with a decimal number 1234 and convert it to decimal notation. To extract the last digit, you move the decimal point left by one digit, which means that you divide the given number by its base 10.
1234/10 = 123 + 4/10
The remainder of 4 is the last digit. To extract the next last digit, you again move the decimal point left by one digit and see what drops out.
123/10 = 12 + 3/10
The remainder of 3 is the next last digit. You repeat this process until there is nothing left. Then you stop. In summary, you do the following:
Quotient  Remainder
-----------------------------
1234/10 =   123        4 --------+
123/10 =    12        3 ------+ |
12/10 =     1        2 ----+ | |
1/10 =     0        1 --+ | | |   (Stop when the quotient is 0.)
| | | |
1 2 3 4   (Base 10)
Now, let's try a nontrivial example. Let's express a decimal number 1341 in binary notation. Note that the desired base is 2, so we repeatedly divide the given decimal number by 2.
Quotient  Remainder
-----------------------------
1341/2  =   670        1 ----------------------+
670/2  =   335        0 --------------------+ |
335/2  =   167        1 ------------------+ | |
167/2  =    83        1 ----------------+ | | |
83/2  =    41        1 --------------+ | | | |
41/2  =    20        1 ------------+ | | | | |
20/2  =    10        0 ----------+ | | | | | |
10/2  =     5        0 --------+ | | | | | | |
5/2  =     2        1 ------+ | | | | | | | |
2/2  =     1        0 ----+ | | | | | | | | |
1/2  =     0        1 --+ | | | | | | | | | |  (Stop when the quotient is 0)
| | | | | | | | | | |
1 0 1 0 0 1 1 1 1 0 1  (BIN; Base 2)
Let's express the same decimal number 1341 in octal notation.
Quotient  Remainder
-----------------------------
1341/8  =   167        5 --------+
167/8  =    20        7 ------+ |
20/8  =     2        4 ----+ | |
2/8  =     0        2 --+ | | |  (Stop when the quotient is 0)
| | | |
2 4 7 5  (OCT; Base 8)
Let's express the same decimal number 1341 in hexadecimal notation.
Quotient  Remainder
-----------------------------
1341/16 =    83       13 ------+
83/16 =     5        3 ----+ |
5/16 =     0        5 --+ | |  (Stop when the quotient is 0)
| | |
5 3 D  (HEX; Base 16)

Example. Convert the decimal number 3315 to hexadecimal notation. What about the hexadecimal equivalent of the decimal number 3315.3?

Solution:

Quotient  Remainder
-----------------------------
3315/16 =   207        3 ------+
207/16 =    12       15 ----+ |
12/16 =     0       12 --+ | |  (Stop when the quotient is 0)
| | |
C F 3  (HEX; Base 16)

(HEX; Base 16)
Product  Integer Part   0.4 C C C ...
--------------------------------     | | | |
0.3*16   =   4.8        4        ----+ | | | | |
0.8*16   =  12.8       12        ------+ | | | |
0.8*16   =  12.8       12        --------+ | | |
0.8*16   =  12.8       12        ----------+ | |
:                       ---------------------+
:
Thus, 3315.3 (DEC) --> CF3.4CCC... (HEX)

Note that from the Base Conversion Table, you can easily get the binary notation from the hexadecimal number by grouping four binary digits per hexadecimal digit, or from or the octal number by grouping three binary digits per octal digit, and vice versa.

HEX  5    3    D
BIN 0101 0011 1101

OCT  2   4   7   5
BIN 010 100 111 101
Finally, the fractional part in a decimal number can also be converted to any base by repeatedly multiplying the given number by the target base. Example: Convert a decimal number 0.1234 to binary notation
(BIN; Base 2)
Product  Integer Part   0.0 0 0 1 1 1 1 1 1 0 0 1 ...
--------------------------------     | | | | | | | | | | | | |
0.1234*2 =  0.2468      0        ----+ | | | | | | | | | | | |
0.2468*2 =  0.4936      0        ------+ | | | | | | | | | | |
0.4936*2 =  0.9872      0        --------+ | | | | | | | | | |
0.9872*2 =  1.9744      1        ----------+ | | | | | | | | |
0.9744*2 =  1.9488      1        ------------+ | | | | | | | |
0.9488*2 =  1.8976      1        --------------+ | | | | | | |
0.8976*2 =  1.7952      1        ----------------+ | | | | | |
0.7952*2 =  1.5904      1        ------------------+ | | | | |
0.5904*2 =  1.1808      1        --------------------+ | | | |
0.1808*2 =  0.3616      0        ----------------------+ | | |
0.3616*2 =  0.7232      0        ------------------------+ | |
0.7232*2 =  1.4464      1        --------------------------+ |
:                       ----------------------------+
:

You generate the addition tables in bases other then 10 by following the same rule you do in base 10. The resulting tables have the appearance of shifting the columns to the left by one in each subsequent rows. Note how simple the addition and multiplication tables are for the binary system; addition operation is simply the bit-wise XOR operation with carry, and multiplication is simply the logical AND operation.

| 0  1  2  3  4  5  6  7  8  9
---+-----------------------------
0 | 0  1  2  3  4  5  6  7  8  9
1 | 1  2  3  4  5  6  7  8  9 10
2 | 2  3  4  5  6  7  8  9 10 11
3 | 3  4  5  6  7  8  9 10 11 12
4 | 4  5  6  7  8  9 10 11 12 13
5 | 5  6  7  8  9 10 11 12 13 14
6 | 6  7  8  9 10 11 12 13 14 15
7 | 7  8  9 10 11 12 13 14 15 16
8 | 8  9 10 11 12 13 14 15 16 17
9 | 9 10 11 12 13 14 15 16 17 18
Binary Addition Table (equivalent to logical XOR operation with carry):

| 0  1
---+-----
0 | 0  1
1 | 1 10

| 0  1  2  3  4  5  6  7
---+-----------------------
0 | 0  1  2  3  4  5  6  7
1 | 1  2  3  4  5  6  7 10
2 | 2  3  4  5  6  7 10 11
3 | 3  4  5  6  7 10 11 12
4 | 4  5  6  7 10 11 12 13
5 | 5  6  7 10 11 12 13 14
6 | 6  7 10 11 12 13 14 15
7 | 7 10 11 12 13 14 15 16

| 0  1  2  3  4  5  6  7  8  9  A  B  C  D  E  F
---+-----------------------------------------------
0 | 0  1  2  3  4  5  6  7  8  9  A  B  C  D  E  F
1 | 1  2  3  4  5  6  7  8  9  A  B  C  D  E  F 10
2 | 2  3  4  5  6  7  8  9  A  B  C  D  E  F 10 11
3 | 3  4  5  6  7  8  9  A  B  C  D  E  F 10 11 12
4 | 4  5  6  7  8  9  A  B  C  D  E  F 10 11 12 13
5 | 5  6  7  8  9  A  B  C  D  E  F 10 11 12 13 14
6 | 6  7  8  9  A  B  C  D  E  F 10 11 12 13 14 15
7 | 7  8  9  A  B  C  D  E  F 10 11 12 13 14 15 16
8 | 8  9  A  B  C  D  E  F 10 11 12 13 14 15 16 17
9 | 9  A  B  C  D  E  F 10 11 12 13 14 15 16 17 18
A | A  B  C  D  E  F 10 11 12 13 14 15 16 17 18 19
B | B  C  D  E  F 10 11 12 13 14 15 16 17 18 19 1A
C | C  D  E  F 10 11 12 13 14 15 16 17 18 19 1A 1B
D | D  E  F 10 11 12 13 14 15 16 17 18 19 1A 1B 1C
E | E  F 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D
F | F 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E
Note (in any base, let it be 10, 2, 8, or 16), adding two single digit numbers can lead to the appearance of an additional digit, and that additional digit always has a value of 1.

You can also generate multiplication tables in bases other than 10 by following the same rule you do in base 10.

Decimal Multiplication Table:

| 0  1  2  3  4  5  6  7  8  9
---+-----------------------------
0 | 0  0  0  0  0  0  0  0  0  0
1 | 0  1  2  3  4  5  6  7  8  9
2 | 0  2  4  6  8 10 12 14 16 18
3 | 0  3  6  9 12 15 18 21 24 27
4 | 0  4  8 12 16 20 24 28 32 36
5 | 0  5 10 15 20 25 30 35 40 45
6 | 0  6 12 18 24 30 36 42 48 54
7 | 0  7 14 21 28 35 42 49 56 63
8 | 0  8 16 24 32 40 48 56 64 72
9 | 0  9 18 27 36 45 54 63 72 81
Binary Multiplication Table (equivalent to the logical AND operation):

| 0  1
---+-----
0 | 0  0
1 | 0  1
Octal Multiplication Table:

| 0  1  2  3  4  5  6  7
---+-----------------------
0 | 0  0  0  0  0  0  0  0
1 | 0  1  2  3  4  5  6  7
2 | 0  2  4  6 10 12 14 16
3 | 0  3  6 11 14 17 22 25
4 | 0  4 10 14 20 24 30 34
5 | 0  5 12 17 24 31 36 43
6 | 0  6 14 22 30 36 44 52
7 | 0  7 16 25 34 43 52 61

| 0  1  2  3  4  5  6  7  8  9  A  B  C  D  E  F
---+-----------------------------------------------
0 | 0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0
1 | 0  1  2  3  4  5  6  7  8  9  A  B  C  D  E  F
2 | 0  2  4  6  8  A  C  E 10 12 14 16 18 1A 1C 1E
3 | 0  3  6  9  C  F 12 15 18 1B 1E 21 24 27 2A 2D
4 | 0  4  8  C 10 14 18 1C 20 24 28 2C 30 34 38 3C
5 | 0  5  A  F 14 19 1E 23 28 2D 32 37 3C 41 46 4B
6 | 0  6  C 12 18 1E 24 2A 30 36 3C 42 48 4E 54 5A
7 | 0  7  E 15 1C 23 2A 31 38 3F 46 4D 54 5B 62 69
8 | 0  8 10 18 20 28 30 38 40 48 50 58 60 68 70 78
9 | 0  9 12 1B 24 2D 36 3F 48 51 5A 63 6C 75 7E 87
A | 0  A 14 1E 28 32 3C 46 50 5A 64 6E 78 82 8C 96
B | 0  B 16 21 2C 37 42 4D 58 63 6E 79 84 8F 9A A5
C | 0  C 18 24 30 3C 48 54 60 6C 78 84 90 9C A8 B4
D | 0  D 1A 27 34 41 4E 5B 68 75 82 8F 9C A9 B6 C3
E | 0  E 1C 2A 38 46 54 62 70 7E 8C 9A A8 B6 C4 D2
F | 0  F 1E 2D 3C 4B 5A 69 78 87 96 A5 B4 C3 D2 E1

### Arithmetic Operations

You do arithmetic with hexadecimal numbers or numbers in any base in exactly the same way you do with decimal numbers, except that the addition and multiplication tables you employ to base your calculations are a bit different. Subtraction is equivalent to adding a negative number, and division is equivalent to multiplying by the inverse.

Example. Find the sum of two decimal integers 123 and 789 (just to review arithmetic).

Solution:

From the above decimal addition table, we see that:
3+9=12, 2+8=10, and 1+7=8

123
+ 789
-----
12 <-- 3+9
10  <-- 2+8
+ 8   <-- 1+7
-----
sum     912

Or, we sometimes use the carry notation (depending where we learn our arithmetic)

123
+ 789
-----
carry   11
table + 802
-----
sum     912

Example. Find the sum of two hexadecimal integers 123 and DEF.

Solution:

3+F=12, 2+E=10, and 1+D=E

123
+ DEF
-----
12 <-- 3+F
10  <-- 2+E
+ E   <-- 1+D
-----
sum     F12

Or, we sometimes use the carry notation (depending where we learn our arithmetic)

123
+ DEF
-----
carry   11
table + E02
-----
sum     F12

Example. Find the sum of two binary integers 1 0010 0011 and 1101 1110 1111

Solution:

1 0010 0011
+ 1101 1110 1111
----------------
carry     1   1    11  ... AND operation followed by left-shift
table + 1100 1100 1100 ... XOR operation
----------------
carry        1    1
table + 1110 1000 1010
----------------
carry      1    1
table + 1110 0000 0010
----------------
carry                  <-- repeat until carry is all 0
sum     1111 0001 0010 (binary)
F    1    2    (hex)
15*256+1*16+2*1=3858 (dec)
A math processor repeats the addition process (i.e., perform AND & XOR operations) until the carry register is all 0. At that point, the 2nd register is the sum.

### Multiplication

Example. Find the product of two decimal integers 123 and 789 (just to review arithmetic).

Solution:

Step 1: We break down the second multiplier into single digits.
123*789 = 123*(700+80+9)
= (123*7)*100 + (123*8)*10 + (123*9)

Step 2: We find the product in parentheses, one parenthesis at a time.
From the above decimal multiplication table, we see that:
1*7=7, 2*7=14, 3*7=21; thus,
123*7 = (100+20+3)*7
= 1*7*100 + 2*7*10 + 3*7
= 7*100   + 14*10  + 21
= 700     + 140    + 21
= 861
Likewise,
123*8 = (100+20+3)*8
= 1*8*100 + 2*8*10 + 3*8
= 8*100   + 16*10  + 24
= 800     + 160    + 24
= 984
123*9 = (100+20+3)*9
= 1*9*100 + 2*9*10 + 3*9
= 9*100   + 18*10  + 27
= 900     + 180    + 27
= 1107

Or, in elementary school style:

123       123       123
x   7     x   8     x   9
-----     -----     -----
21        24        27
14        16        18
7         8         9
-----     -----     -----
861       984      1107

Step 3: We sum up the individual products.
123*789 = (123*7)*100 + (123*8)*10 + (123*9)
= 861*100     + 984*10     + 1107
= 86100       + 9840       + 1107
= 97047

Or, in elementary school style (in the order shown above):

123
x 789
-----
861   <-- 123*7
984  <-- 123*8
1107 <-- 123*9
-----
97047

Or, in elementary school style:

123
x 789
-----
1107 <-- 123*9
984  <-- 123*8
861   <-- 123*7
-----
97047

Example. Find the product of two hexadecimal integers 123 and DEF.

Solution:

Step 1: We break down the second multiplier into single digits.
123*DEF = 123*(D00+E0+F)
= (123*D)*100 + (123*E)*10 + (123*F)

Step 2: We find the product in parentheses.
From the above hexadecimal multiplication table, we see that:
1*D=D, 2*D=1A, 3*D=27; thus,
123*D = (100+20+3)*D
= 1*D*100 + 2*D*10 + 3*D
= D*100   + 1A*10  + 27
= D00     + 1A0    + 27
= EC7
Likewise,
123*E = (100+20+3)*E
= 1*E*100 + 2*E*10 + 3*E
= E*100   + 1C*10  + 2A
= E00     + 1C0    + 2A
= FEA
123*F = (100+20+3)*F
= 1*F*100 + 2*F*10 + 3*F
= F*100   + 1E*10  + 2D
= F00     + 1E0    + 2D
= 110D

Or, in elementary school style:

123       123       123
x   D     x   E     x   F
-----     -----     -----
27        2A        2D
1A        1C        1E
D         E         F
-----     -----     -----
EC7       FEA      110D

Step 3: We sum up the individual products.
123*DEF = (123*D)*100 + (123*E)*10 + (123*F)
= EC7*100     + FEA*10     + 110D
= EC700       + FEA0       + 110D

Or, in elementary school style (in the order shown above):

123
x DEF
-----
EC7   <-- 123*D
FEA  <-- 123*E
110D <-- 123*F
-----

Or, in elementary school style:

123
x DEF
-----
110D <-- 123*F
FEA  <-- 123*E
EC7   <-- 123*D
-----

Example. Find the product of two binary integers 1 0010 0011 and 1101 1110 1111

Solution:

Step 2:
1 0010 0011
x           1 (the 1st "1" in 1101 1110 1111)
-------------
1 0010 0011

1 0010 0011
x           1 (the 2nd "1" in 1101 1110 1111)
-------------
1 0010 0011

1 0010 0011
x           0 (the 1st "0" in 1101 1110 1111)
-------------
0 0000 0000

:
:

1 0010 0011
x           1 (the last "1" in 1101 1110 1111)
-------------
1 0010 0011

Step 3: We sum up the individual products.
1 0010 0011
x     1101 1110 1111
--------------------
100100011             <-- 100100011*1
100100011            <-- 100100011*1
000000000           <-- 100100011*0
100100011          <-- 100100011*1

100100011         <-- 100100011*1
100100011        <-- 100100011*1
100100011       <-- 100100011*1
000000000      <-- 100100011*0

100100011     <-- 100100011*1
100100011    <-- 100100011*1
100100011   <-- 100100011*1
100100011  <-- 100100011*1
--------------------
11111101011010101101 (binary, after summing up all the above 12 binary numbers)
F   D   6   A   D (hex)
Thus, in a computer, multiplication-division becomes a digit shift left-right operation.

### Subtraction

Subtraction is the same as addition of a negative counterpart. First, let us discuss how an integer can be expressed. An integer can be expressed as:
• Unsigned purely binary data. One byte can express an integer 0 through 255, similar to the base conversion table at the beginning of this web page; there are no negative numbers.
------------------------------
00000000      0           0
00000001      1           1
00000010      2           2
00000011      3           3
00000100      4           4
00000101      5           5
00000110      6           6
00000111      7           7
:          :           :
:          :           :
11111000     F8         248
11111001     F9         249
11111010     FA         250
11111011     FB         251
11111100     FC         252
11111101     FD         253
11111110     FE         254
11111111     FF         255

• Signed binary numbers. The most significant bit (sign bit) represents sign, where "0" denotes a positive number and "1" denotes a negative number. One byte can express an integer -127 through 0 through 127, similar to the base conversion table at the beginning of this web page, except for the sign bit. Note that there are 2 representations of the integer "0" because both +0 and -0 means zero. This coding scheme may be easier for a normal person to decipher, but not a computer.
------------------------------
00000000      0           0
00000001      1           1
00000010      2           2
00000011      3           3
00000100      4           4
00000101      5           5
00000110      6           6
00000111      7           7
:          :           :
:          :           :
01111000     78         120
01111001     79         121
01111010     7A         122
01111011     7B         123
01111100     7C         124
01111101     7D         125
01111110     7E         126
01111111     7F         127
10000000     80          -0
10000001     81          -1
10000010     82          -2
10000011     83          -3
10000100     84          -4
10000101     85          -5
10000110     86          -6
10000111     87          -7
:          :           :
:          :           :
11111000     F8        -120
11111001     F9        -121
11111010     FA        -122
11111011     FB        -123
11111100     FC        -124
11111101     FD        -125
11111110     FE        -126
11111111     FF        -127

• Two's complement. This is the scheme employed in programming and numerical packages.

To help understand this scheme, let us start with a decimal machine and consider a 8-digit number display in a tape deck or an odometer in a car. When we roll forward, numbers increment; when we roll backward, numbers decrement. If we were to roll backward past 00000000, all digits simultaneously turn into 9s (e.g., 99999999). If we wish to express both positive and negative numbers, we can agree that the lower half of the meter (00000000 through 49999999, i.e., the leading digit or the most significant digit being 0 through 4) denote positive numbers while the upper half of the meter (50000000 through 99999999, i.e., the leading digit or the most significant digits being 5 through 9) denote negative numbers. With this scheme, with 8 decimal digits, we can express a decimal integer from -50000000 through 49999999.

10's complement (for decimal system)
display      integer
-------------------------------
50000000    -5000000
50000001    -4999999
50000002    -4999998
50000003    -4999997
50000004    -4999996
:           :
:           :
99999995          -5
99999996          -4
99999997          -3
99999998          -2
99999999          -1
00000000           0
00000001           1
00000002           2
00000003           3
00000004           4
00000005           5
:           :
:           :
49999995    49999995
49999996    49999996
49999997    49999997
49999998    49999998
49999999    49999999

Note the symmetry of a positive number (say, 5) and its negative counter part (say, -5) around 0. Thus, subtraction of a number simply becomes addition of its counter part.
5-5 --> becomes --> 5+(-5) --> 0

stored as    integer
00000005           5
+ 99999995          -5
----------------------
100000000           0
(but "1" above is not stored, because the storage has only 8 decimal digits.

Another example: 125-125=?
125-125 --> becomes --> 125+(-125) --> 0

stored as    integer
00000125         125
+ 99999875        -125
----------------------
100000000           0

The above also suggests a quick way for us mortals to express a negative number in 10's complement notation.
Example: find 10's complement representation of -125 in decimal system.
Step 1. positive representation:      00000125 <-- +125
Step 2. find 10's complement          99999874 <-- complement that adds to "9"
Step 3. add 1                        +       1
---------
The resulting number is -125.         99999875 <-- -125

Now, having seen the decimal system, we apply the same idea/procedure to a binary system. With one byte, we first anchor 00000000 to mean zero (0). When we roll forward, numbers increment; when we roll backward, numbers decrement. If we were to roll backward past 00000000, all digits simultaneously turn into 1s (e.g., 11111111). If we wish to express both positive and negative numbers, we can agree that the lower half of the byte (00000000 through 01111111, i.e., the leading digit or the most significant digit being 0) denote positive numbers while the upper half of the byte (10000000 through 11111111, i.e., the leading digit being 1) denote negative numbers. With this scheme, with 8 binary digits (i.e., 1 byte), we can express a decimal integer from -128 through 127.
2's complement (for binary system)
display      integer
-------------------------------
10000000      -128
10000001      -127
10000010      -126
10000011      -125
10000100      -124
:            :
:            :
11111011        -5
11111100        -4
11111101        -3
11111110        -2
11111111        -1
00000000         0
00000001         1
00000010         2
00000011         3
00000100         4
00000101         5
:            :
:            :
01111011       123
01111100       124
01111101       125
01111110       126
01111111       127

Note that just like in signed integer notation, in two's complement notation, a negative number is indicated by "1" in the most significant bit. If we consider this to be the sign bit and also associated with it a value of -128, then, we can decompose the above table like the following.
The most significant bit has a value of -128.
display      integer
------------------------------------
10000000      -128 +   0 --> -128
10000001      -128 +   1 --> -127
10000010      -128 +   2 --> -126
10000011      -128 +   3 --> -125
10000100      -128 +   4 --> -124
:                      :
:                      :
11111011      -128 + 123 -->   -5
11111100      -128 + 124 -->   -4
11111101      -128 + 125 -->   -3
11111110      -128 + 126 -->   -2
11111111      -128 + 127 -->   -1
00000000         0 +   0 -->    0
00000001         0 +   1 -->    1
00000010         0 +   2 -->    2
00000011         0 +   3 -->    3
00000100         0 +   4 -->    4
00000101         0 +   5 -->    5
:                      :
:                      :
01111011         0 + 123 -->  123
01111100         0 + 124 -->  124
01111101         0 + 125 -->  125
01111110         0 + 126 -->  126
01111111         0 + 127 -->  127

Just as in the decimal system, subtraction of a binary number becomes addition of its counter part.
5-5 --> becomes > 5+(-5) > 0

stored as    integer
00000101           5
+ 11111011          -5
----------------------
100000000           0
(but "1" above is not stored, because the storage has only 8 binary digits.

Another example: 125-125=?
125-125 --> becomes > 125+(-125) > 0

stored as    integer
01111101         125
+ 10000011        -125
----------------------
100000000           0

The above also suggests a quick way for us mortals to express a negative number in 2's complement notation in binary system.
Example: find 2's complement representation of -125 in binary system.
Step 1. positive representation:      01111101 <-- +125
Step 2. find 2's complement           10000010 <-- complement that adds to "1", i.e., flip digits
Step 3. add 1                        +       1
---------
The resulting number is -125.         10000011 <-- -125

A even quicker way to express a negative number in 2's complement notation in binary system is consider the most significant bit to have a value of -2^7=-128 (for a 1-byte 8-bit number), or -2^15=-32768 (for a 2-byte 16-bit number), etc.
-125=-128+3
10000011

Example. Find the two's complement representation of the decimal number 1341. From the earlier part of this web page, we first find the hexadecimal notation of this number to be 54D (HEX), which in turn is easily converted to the binary notation. This demonstrates how to find a negative number in two's complement notation without resorting to a look-up table.
1341 (DEC) -->  53D (HEX)  -->  0101 0011 1101 (BIN)
One byte is not large enough to express 1341 --> need 2 bytes.
Step 1. positive representation: 0000 0101 0011 1101 <-- +1341
Step 2. find 2's complement      1111 1010 1100 0010 <-- complement that adds to "1", i.e., flip digits
Step 3. add 1                  +                   1
---------------------
The resulting number is -1341.   1111 1010 1100 0011 <-- -1341

An even quicker way is to recognize that -1341=-32768+31427
-1341 (DEC)
--> -32768 (DEC) + 31427 (DEC)
-->  -8000 (HEX) + 7AC3 (HEX)
-->  -8000 (HEX) + 0111 1010 1100 0011 (BIN)

Finally, we get back to subtraction, which is simply adding a negative counterpart.

Example. Find 1341-125. The computer simply adds -125.

0000 0101 0011 1101 <-- +1341 in 2-byte storage
+ 1111 1111 1000 0011 <--  -125 in 2-byte storage
---------------------
carry        1 1         1
+ 1111 1010 1011 1110
---------------------
carry      1  1         1
+ 1111 0000 1011 1100
---------------------
carry     1            1
+ 1110 0100 1011 1000
---------------------
carry    1           1
+ 1100 0100 1011 0000
---------------------
carry   1           1
+ 1000 0100 1010 0000
---------------------
carry 1            1        <-- the first "1" is pushed off, because the storage has only 2 bytes.
+ 0000 0100 1000 0000
---------------------
carry                       <-- repeat until carry is all 0;
+ 0000 0100 1100 0000
---------------------
sum     0000 0100 1100 0000 <-- answer in binary

### Real Numbers

A computer stores a real number (also known as floating point) in finite number of digits, typically 4 bytes = 32 bits for single precision, or 8 bytes = 64 bits for double precision. The following is an example of the IEEE format for a 4-byte single precision real number.
# bits   function
------------------------------------
1      sign
8      binary exponents (-126 through 127)
1 bit for exponent sign (1 for +, 0 for -; opposite of signed integer)
7 bits for exponent value (-126 through 127, expressed in two's complement, but shifted by 1)
23      1.0 ≤ mantissa < 2.0 (1.000...000
Because the higher order bit is always 1, there is no need to store this.
-----------------------------------------------
value     sign sign exp mantissa
bits        1 1    7    ----------23-----------
-----------------------------------------------
0           0 0 0000000 00000000000000000000000 all 0 gives 0 (the reason for inverted signed integer for exponent)
smallest    0 0 0000000 00000000000000000000001 smallest=~2^-126=1.175*10^-38
2^-125      0 0 0000001 00000000000000000000000
2^-124      0 0 0000010 00000000000000000000000
2^-123      0 0 0000011 00000000000000000000000
2^-122      0 0 0000100 00000000000000000000000
:
:
2^-3=0.125  0 0 1111100 00000000000000000000000
2^-2=0.25   0 0 1111101 00000000000000000000000
2^-1=0.5    0 0 1111110 00000000000000000000000
2^0 =1      0 0 1111111 00000000000000000000000
2^1 =2      0 1 0000000 00000000000000000000000
2^2 =4      0 1 0000001 00000000000000000000000
2^3 =8      0 1 0000010 00000000000000000000000
2^4 =16     0 1 0000011 00000000000000000000000
:
:
2^125       0 1 1111100 00000000000000000000000
2^126       0 1 1111101 00000000000000000000000
2^127       0 1 1111110 00000000000000000000000
largest     0 1 1111110 11111111111111111111111 largest=~2^128=3.40*10^38
inf         0 1 1111111 00000000000000000000000
Example. Find the real number expressed by a 4-byte sequence 4BD40000 HEX
HEX         4     B   D    4   0   0   0   0
BIN         0 1 0010111 10101000000000000000000
| | |       |
| | |       +-- binary mantissa: 1.10101000000000000000000
| | +---------- binary exponent value: 0010111+1
| +------------ binary exponent sign: +
+-------------- number sign: +
Binary value: 1.10101x2^10111+1 = 1.10101*2^11000
Decimal value of mantissa: 1.10101 (BIN) --> 1+0.5+0.125+0.03125=1.65625 (DEC)
Decimal value of exponent: 2^(16+8)=2^24 (DEC)
Decimal value: 1.65625*2^24=2.778726*10^7
Note that the number of bits available for the exponent determines the smallest and the largest number that can be expressed. And the number of bits available for the mantissa determines the number of significant digits.
exponent:  7 binary digits --> 2^(-126~+127) --> 10^(-38~+38)
mantissa: 23 binary digits --> ~7 decimal digits
Finally, a little math on digits for quick conversion/estimation:
2^BINdigit = 10^DECdigit                 2BINdigit = 10DECdigit
log (2^BINdigit) = log (10^DECdigit)     log (2BINdigit) = log (10DECdigit)
BINdigit*log(2) = DECdigit*log(10)
BINdigit*0.30 = DECdigit

Another approach:
2=10log(2) → 2BINdigit=10log(2)*BINdigit=10DECdigit
compare the exponent in 10 →  log(2)*BINdigit=DECdigit

Use the above factor of 0.3 for quick estimation
exponent of 127 in base 2 --> 127*0.3=~38 ... exponent of 38 in base 10
mantissa of 23 binary digits --> 23*0.3=~7 ... 7 significant digits in base 10

2^BINdigit=10^DECdigit --> BINdigit ~ 3.3*DECdigit ... 3.3 or ~3 binary digits = 1 decimal digit
2^BINdigit= 8^OCTdigit --> BINdigit = 3  *OCTdigit ... 3 binary digit ≡ 1 octal digit
2^BINdigit=16^HEXdigit --> BINdigit = 4  *OCTdigit ... 4 binary digit ≡ 1 hexadecimal digit