To get the total value of a number, say: 4563, you take each digit and multiply it by the appropriate power of 10. Then add all the products up to get the final value. The following table illustrates this process.

In programming in a high-level language, we don't usually worry about binary, because when the computer prints out a number for us, it has been translated into decimal. Similarly, when we input a decimal number into the computer, it is translated into binary. Nonetheless, an understanding of how a computer represents values as binary numbers is necessary, both to understand the limitations imposed by calculating using binary numbers and to understand character coding schemes, like ASCII and UNICODE.   (Incidentally, if you are interested in the subject of character coding schemes, you are encouraged to follow the following link:   "A tutorial on character code issues" at http://www.cs.tut.fi/~jkorpela/chars.html.)

To convert a binary number, say: 1101, to its decimal value, you take each digit and multiply it by the appropriate power of 2. Then add all the products up to get the final value. The following table illustrates this process.

The radix (or base) of a number system tells how many different numerals a digit can be in that number system (and consequently determines what powers the positions have.) The decimal system is radix 10; the binary system is radix 2. The radix of a number is sometimes written as a subscript to make clear what sort of a number is being represented. So, in our previous example, we might write "1011₂ is equal to 13₁₀".

• Count using only the numerals 0 and 1

• 0,1,10,11,100,101,110,111,1000,1001,1010   (That's zero through ten, in decimal.)
• Next 6 numbers are: 1011, 1100, 1101, 1110, 1111, and 10000   (11, ..., 16, in decimal)

• The decimal number 43,210
  equals   (4 × 10⁴) + (3 × 10³) + (2 × 10²) + (1 × 10¹) + (0 × 10⁰)
  equals   (4 × 10,000) + (3 × 1,000) + (2 × 100) + (1 × 10) + (0 × 1)

• The binary number 10110
  equals   (1 × 2⁴) + (0 × 2³) + (1 × 2²) + (1 × 2¹) + (0 × 2⁰)
  equals   (1 × 16) + (0 × 8) + (1 × 4) + (1 × 2) + (0 × 1) = 22

• Write the number down the page, starting with the smallest (rightmost) bit

• 11100100 gets written:
  0
  0
  1
  0
  0
  1
  1
  1

• Write consecutive powers of two beside each digit:
  0 · 1
  0 · 2
  1 · 4
  0 · 8
  0 · 16
  1 · 32
  1 · 64
  1 · 128

• Add up the decimal numbers with a binary 1 beside them
  0 · 1
  0 · 2
  1 · 4
  0 · 8
  0 · 16
  1 · 32
  1 · 64
  1 · 128
  

  or   4 + 32 + 64 + 128 = 228

• Repeatedly divide by 2, writing down the remainder (either 0 or 1)
• Stop when you do "1 divided by 2 is 0 with remainder 1"
• Reading the remainders from bottom to top gives you the binary number
• Example: 228₁₀ = 11100100₂

  228 div 2 = 114 with rem 0    top
  114 div 2 =  57 with rem 0     ^
   57 div 2 =  28 with rem 1     |
   28 div 2 =  14 with rem 0     |
   14 div 2 =   7 with rem 0     |
    7 div 2 =   3 with rem 1     |
    3 div 2 =   1 with rem 1     |
    1 div 2 =   0 with rem 1   bottom

• 0 + 0 = 0
• 0 + 1 = 1 + 0 = 1
• 1 + 1 = 10 (0 "carry the 1")
• 1 + 1 + 1 = 11 (1 "carry the 1")

  Example:

    111 
   00111010  |  58
  +00011100  | +28
  =01010110  | =86

• 0 - 0 = 0
• 1 - 0 = 1
• 1 - 1 = 0
• For 0 - 1, must borrow 1, do 10 - 1 = 1

  Example:

        1
   00111010  |  58
  -00010100  | -20
  =00100110  | =38

Just as 3215.12₁₀
equals   (3 × 10³) + (2 × 10²) + (1 × 10¹) + (5 × 10⁰) + (1 × 10^-1) + (2 × 10^-2)
equals   (3 × 1,000) + (2 × 100) + (1 × 10) + (5 × 1) + (1 × 0.1) + (2 × 0.01)

the binary number 1101.11₂
equals   (1 × 2³) + (1 × 2²) + (0 × 2¹) + (1 × 2⁰) + (1 × 2^-1) + (1 × 2^-2)
equals   (1 × 8) + (1 × 4) + (0 × 2) + (1 × 1) + (1 × 0.5) + (1 × 0.25)
equals   13.75₁₀

Each digit to the right of the binary point in a binary fraction is a multiple of an inverse power of 2. It is a standard convention to use negative exponents to represent inverse powers. Thus, for example,
2^-1 = 1/(2¹) = 1/2 = .5
2^-2 = 1/(2²) = 1/4 = .25
2^-3 = 1/(2³) = 1/8 = .125
2^-4 = 1/(2⁴) = 1/16 = .0625

Hence the binary fraction .1101₂ represents the decimal fraction .8125₁₀ because, multiplying by the appropriate column power, you get:
(1 × 2^-1) + (1 × 2^-2) + (0 × 2^-3) + (1 × 2^-4) = .8125₁₀

Putting whole numbers and fractions together, the decimal number 5.8125 is represented in binary as 101.1101₂

Using a fixed number of bits to store binary fractions may introduce rounding errors. For example, if we had four bits of fractional binary precision, the decimal fractions that can be represented include:

Binary .0000 is .0000 in decimal.
Binary .0001 is .0625 in decimal.
Binary .0010 is .1250 in decimal.
Binary .0011 is .1875 in decimal.
Binary .0100 is .2500 in decimal.
Binary .0101 is .3125 in decimal.

So if we had the decimal number .05 and we needed to represent it with four binary digits, we simply couldn't do it. Instead we would have to round it to the nearest decimal fraction that can be represented with four binary digits, which is .0625.

Another example: Assume we wanted to represent the number 5.6 in 8 bits, where the binary point was assumed to be after the third binary digit from the left. That is, 8 bits with the format:   ABC.DEFGH   i.e., eight bits with five places to the right of the binary point. The column weightings are:
A = 2² = 4
B = 2¹ = 2
C = 2⁰ = 1
D = 2^-1 = 1/2 = 0.5
E = 2^-2 = 1/4 = 0.25
F = 2^-3 = 1/8 = 0.125
G = 2^-4 = 1/16 = 0.0625
H = 2^-5 = 1/32 = 0.03125

So the answer would be 101.10011 = 5.59375   (Notice the rounding error!)