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The binary numeral system, or base-2 number system, is a numeral system that represents numeric values using two symbols, usually 0 and 1. More specifically, the usual base-2 system is a positional notation with a radix of 2. Owing to its straightforward implementation in digital electronic circuitry using logic gates, the binary system is used internally by all modern computers.

History

The ancient Indian writer Pingala developed advanced mathematical concepts for describing prosody, and in doing so presented the first known description of a binary numeral system, possibly as early as the 8th century BC.^[1] Others place him much later; R. Hall, Mathematics of Poetry, has "c. 200 BC". The numeration system was based on the Eye of Horus Old Kingdom numeration system.^{[citation needed]}

A full set of 8 trigrams and 64 hexagrams, analogous to the 3-bit and 6-bit binary numerals, were known to the ancient Chinese in the classic text I Ching. Similar sets of binary combinations have also been used in traditional African divination systems such as Ifá as well as in medieval Western geomancy.

An arrangement of the hexagrams of the I Ching, ordered according to the values of the corresponding binary numbers (from 0 to 63), and a method for generating the same, was developed by the Chinese scholar and philosopher Shao Yong in the 11th century. However, there is no evidence that Shao understood binary computation; the ordering is also the lexicographical order on sextuples of elements chosen from a two-element set.

In 1605 Francis Bacon discussed a system by which letters of the alphabet could be reduced to sequences of binary digits, which could then be encoded as scarcely visible variations in the font in any random text. Importantly for the general theory of binary encoding, he added that this method could be used with any objects at all: "provided those objects be capable of a twofold difference only; as by Bells, by Trumpets, by Lights and Torches, by the report of Muskets, and any instruments of like nature".^[2] (See Bacon's cipher.)

The modern binary number system was fully documented by Gottfried Leibniz in the 17th century in his article Explication de l'Arithmétique Binaire. Leibniz's system used 0 and 1, like the modern binary numeral system.

In 1854, British mathematician George Boole published a landmark paper detailing an algebraic system of logic that would become known as Boolean algebra. His logical calculus was to become instrumental in the design of digital electronic circuitry.

In 1937, Claude Shannon produced his master's thesis at MIT that implemented Boolean algebra and binary arithmetic using electronic relays and switches for the first time in history. Entitled A Symbolic Analysis of Relay and Switching Circuits, Shannon's thesis essentially founded practical digital circuit design.

In November of 1937, George Stibitz, then working at Bell Labs, completed a relay-based computer he dubbed the "Model K" (for "Kitchen", where he had assembled it), which calculated using binary addition. Bell Labs thus authorized a full research program in late 1938 with Stibitz at the helm. Their Complex Number Computer, completed January 8, 1940, was able to calculate complex numbers. In a demonstration to the American Mathematical Society conference at Dartmouth College on September 11, 1940, Stibitz was able to send the Complex Number Calculator remote commands over telephone lines by a teletype. It was the first computing machine ever used remotely over a phone line. Some participants of the conference who witnessed the demonstration were John Von Neumann, John Mauchly, and Norbert Wiener, who wrote about it in his memoirs.

Representation

A binary number can be represented by any sequence of bits (binary digits), which in turn may be represented by any mechanism capable of being in two mutually exclusive states. The following sequences of symbols could all be interpreted as the same binary numeric value of 667:

1 0 1 0 0 1 1 0 1 1
| - | - - | | - | |
x o x o o x x o x x
y n y n n y y n y y

The numeric value represented in each case is dependent upon the value assigned to each symbol. In a computer, the numeric values may be represented by two different voltages; on a magnetic disk, magnetic polarities may be used. A "positive", "yes", or "on" state is not necessarily equivalent to the numerical value of one; it depends on the architecture in use.

In keeping with customary representation of numerals using Arabic numerals, binary numbers are commonly written using the symbols 0 and 1. When written, binary numerals are often subscripted, prefixed or suffixed in order to indicate their base, or radix. The following notations are equivalent:

100101 binary (explicit statement of format)

100101b (a suffix indicating binary format)

100101B (a suffix indicating binary format)

bin 100101 (a prefix indicating binary format)

100101₂ (a subscript indicating base-2 (binary) notation)

%100101 (a prefix indicating binary format)

0b100101 (a prefix indicating binary format, common in programming languages)

When spoken, binary numerals are usually read digit-by-digit, in order to distinguish them from decimal numbers. For example, the binary numeral 100 is pronounced one zero zero, rather than one hundred, to make its binary nature explicit, and for purposes of correctness. Since the binary numeral 100 is equal to the decimal value four, it would be confusing, and numerically incorrect, to refer to the numeral as one hundred.

Counting in binary

Binary	Decimal
0	0
1	1
10	2
11	3
100	4
101	5
110	6
111	7
1000	8
1001	9
1010	10

Counting in binary is similar to counting in any other number system. Beginning with a single digit, counting proceeds through each symbol, in increasing order. Decimal counting uses the symbols 0 through 9, while binary only uses the symbols 0 and 1.

When the symbols for the first digit are exhausted, the next-higher digit (to the left) is incremented, and counting starts over at 0. In decimal, counting proceeds like so:

000, 001, 002, ... 007, 008, 009, (rightmost digit starts over, and next digit is incremented)

010, 011, 012, ...

...

090, 091, 092, ... 097, 098, 099, (rightmost two digits start over, and next digit is incremented)

100, 101, 102, ...

After a digit reaches 9, an increment resets it to 0 but also causes an increment of the next digit to the left. In binary, counting is the same except that only the two symbols 0 and 1 are used. Thus after a digit reaches 1 in binary, an increment resets it to 0 but also causes an increment of the next digit to the left:

000, 001, (rightmost digit starts over, and next digit is incremented)

010, 011, (rightmost two digits start over, and next digit is incremented)

100, 101, ...

Binary arithmetic

Arithmetic in binary is much like arithmetic in other numeral systems. Addition, subtraction, multiplication, and division can be performed on binary numerals.

Addition

The circuit diagram for a binary half adder, which adds two bits together, producing sum and carry bits.

The simplest arithmetic operation in binary is addition. Adding two single-digit binary numbers is relatively simple:

0 + 0 → 0

0 + 1 → 1

1 + 0 → 1

1 + 1 → 0, carry 1 (since 1 + 1 = 0 + 1 × 10 in binary)

Adding two "1" digits produces a digit "0", while 1 will have to be added to the next column. This is similar to what happens in decimal when certain single-digit numbers are added together; if the result equals or exceeds the value of the radix (10), the digit to the left is incremented:

5 + 5 → 0, carry 1 (since 5 + 5 = 0 + 1 × 10)

7 + 9 → 6, carry 1 (since 7 + 9 = 6 + 1 × 10)

This is known as carrying. When the result of an addition exceeds the value of a digit, the procedure is to "carry" the excess amount divided by the radix (that is, 10/10) to the left, adding it to the next positional value. This is correct since the next position has a weight that is higher by a factor equal to the radix. Carrying works the same way in binary:

  1 1 1 1 1  (carried digits)
    0 1 1 0 1
+   1 0 1 1 1
-------------
= 1 0 0 1 0 0

In this example, two numerals are being added together: 01101₂ (13 decimal) and 10111₂ (23 decimal). The top row shows the carry bits used. Starting in the rightmost column, 1 + 1 = 10₂. The 1 is carried to the left, and the 0 is written at the bottom of the rightmost column. The second column from the right is added: 1 + 0 + 1 = 10₂ again; the 1 is carried, and 0 is written at the bottom. The third column: 1 + 1 + 1 = 11₂. This time, a 1 is carried, and a 1 is written in the bottom row. Proceeding like this gives the final answer 100100₂ (36 decimal).

When computers must add two numbers, the rule that: x xor y = (x + y) mod 2 for any two bits x and y allows for very fast calculation, as well.

Subtraction

Subtraction works in much the same way:

0 − 0 → 0

0 − 1 → 1, borrow 1

1 − 0 → 1

1 − 1 → 0

Subtracting a "1" digit from a "0" digit produces the digit "1", while 1 will have to be subtracted from the next column. This is known as borrowing. The principle is the same as for carrying. When the result of a subtraction is less than 0, the least possible value of a digit, the procedure is to "borrow" the deficit divided by the radix (that is, 10/10) from the left, subtracting it from the next positional value.

    *   * * *   (starred columns are borrowed from)
  1 1 0 1 1 1 0
−     1 0 1 1 1
----------------
= 1 0 1 0 1 1 1

Subtracting a positive number is equivalent to adding a negative number of equal absolute value; computers typically use two's complement notation to represent negative values. This notation eliminates the need for a separate "subtract" operation. The subtraction can be summarized with this formula:

A - B = A + not B + 1

For further details, see two's complement.

Multiplication

Multiplication in binary is similar to its decimal counterpart. Two numbers A and B can be multiplied by partial products: for each digit in B, the product of that digit in A is calculated and written on a new line, shifted leftward so that its rightmost digit lines up with the digit in B that was used. The sum of all these partial products gives the final result.

Since there are only two digits in binary, there are only two possible outcomes of each partial multiplication:

If the digit in B is 0, the partial product is also 0
If the digit in B is 1, the partial product is equal to A

For example, the binary numbers 1011 and 1010 are multiplied as follows:

           1 0 1 1   (A)
         × 1 0 1 0   (B)
         ---------
           0 0 0 0   ← Corresponds to a zero in B
   +     1 0 1 1     ← Corresponds to a one in B
   +   0 0 0 0  
   + 1 0 1 1  
   ---------------
   = 1 1 0 1 1 1 0

Binary numbers can also be multiplied with bits after a binary point:

               1 0 1.1 0 1   (A) (5.625 in decimal)
             ×   1 1 0.0 1   (B) (6.25  in decimal)
             -------------
               1 0 1 1 0 1   ← Corresponds to a one in B
   +         0 0 0 0 0 0     ← Corresponds to a zero in B
   +       0 0 0 0 0 0
   +     1 0 1 1 0 1
   +   1 0 1 1 0 1
   -----------------------
   = 1 0 0 0 1 1.0 0 1 0 1   (35.15625 in decimal)

Division

Binary division is again similar to its decimal counterpart:

        __________
1 0 1  | 1 1 0 1 1

Here, the divisor is 101₂, or 5 decimal, while the dividend is 11011₂, or 27 decimal. The procedure is the same as that of decimal long division; here, the divisor 101₂ goes into the first three digits 110₂ of the dividend one time, so a "1" is written on the top line. This result is multiplied by the divisor, and subtracted from the first three digits of the dividend; the next digit (a "1") is included to obtain a new three-digit sequence:

             1
        __________
1 0 1  | 1 1 0 1 1
       − 1 0 1
         -----
           0 1 1

The procedure is then repeated with the new sequence, continuing until the digits in the dividend have been exhausted:

             1 0 1
        __________
1 0 1  | 1 1 0 1 1
       − 1 0 1
         -----
           0 1 1
         − 0 0 0
           -----
             1 1 1
           − 1 0 1
             -----
               1 0

Thus, the quotient of 11011₂ divided by 101₂ is 101₂, as shown on the top line, while the remainder, shown on the bottom line, is 10₂. In decimal, 27 divided by 5 is 5, with a remainder of 2.

Bitwise operations

Though not directly related to the numerical interpretation of binary symbols, sequences of bits may be manipulated using Boolean logical operators. When a string of binary symbols is manipulated in this way, it is called a bitwise operation; the logical operators AND, OR, and XOR may be performed on corresponding bits in two binary numerals provided as input. The logical NOT operation may be performed on individual bits in a single binary numeral provided as input. Sometimes, such operations may be used as arithmetic short-cuts, and may have other computational benefits as well. For example, an arithmetic shift left of a binary number is the equivalent of multiplication by a (positive, integral) power of 2.

Conversion to and from other numeral systems

Decimal

To convert from a base-10 integer numeral to its base-2 (binary) equivalent, the number is divided by two, and the remainder is the least-significant bit. The (integer) result is again divided by two, its remainder is the next most significant bit. This process repeats until the result of further division becomes zero.

For example, 118₁₀, in binary, is:

Operation	Remainder
118 ÷ 2 = 59	0
59 ÷ 2 = 29	1
29 ÷ 2 = 14	1
14 ÷ 2 = 7	0
7 ÷ 2 = 3	1
3 ÷ 2 = 1	1
1 ÷ 2 = 0	1

Reading the sequence of remainders from the bottom up gives the binary numeral $1110110_{2}$ .

This method works for conversion from any base, but there are better methods for bases which are powers of two, such as octal and hexadecimal given below.

To convert from base-2 to base-10 is the reverse algorithm. Write the binary number vertically from top to bottom, multiply the result from previous line by 2 and add the binary. For example to convert 110010101101₂ to decimal:

Prior result	× 2 +	Binary	Result
			= 0
0	× 2 +	1	= poohio
1	× 2 +	1	= 3
3	× 2 +	0	= 6
6	× 2 +	0	= 12
12	× 2 +	1	= 25
25	× 2 +	0	= 50
50	× 2 +	1	= 101
101	× 2 +	0	= 202
202	× 2 +	1	= 405
405	× 2 +	1	= 811
811	× 2 +	0	= 1622
1622	× 2 +	1	= 3245

The result is 3245₁₀.

   Binary:           1           1           0            0           1           0           1           0          1           1            0          1
  
  Decimal:      [(2^11)*1] + [(2^10)*1] + [(2^9)*0] + [(2^8)*0] + [(2^7)*1] + [(2^6)*0] + [(2^5)*1] + [(2^4)*0] + [(2^3)*1] + [(2^2)*1] + [(2^1)*0] + [(2^0)*1] = 3245

The fractional parts of a number are converted with similar methods. They are again based on the equivalence of shifting with doubling or halving.

In a fractional binary number such as .11010110101₂, the first digit is ${\begin{matrix}{\frac {1}{2}}\end{matrix}}$ , the second ${\begin{matrix}({\frac {1}{2}})^{2}={\frac {1}{4}}\end{matrix}}$ , etc. So if there is a 1 in the first place after the decimal, then the number is at least ${\begin{matrix}{\frac {1}{2}}\end{matrix}}$ , and vice versa. Double that number is at least 1. This suggests the algorithm: Repeatedly double the number to be converted, record if the result is at least 1, and then throw away the integer part.

For example, ${\begin{matrix}({\frac {1}{3}})\end{matrix}}$ ₁₀, in binary, is:

Converting	Result
${\begin{matrix}{\frac {1}{3}}\end{matrix}}$	0.
${\begin{matrix}{\frac {1}{3}}\times 2={\frac {2}{3}}<1\end{matrix}}$	0.0
${\begin{matrix}{\frac {2}{3}}\times 2=1{\frac {1}{3}}\geq 1\end{matrix}}$	0.01
${\begin{matrix}{\frac {1}{3}}\times 2={\frac {2}{3}}<1\end{matrix}}$	0.010
${\begin{matrix}{\frac {2}{3}}\times 2=1{\frac {1}{3}}\geq 1\end{matrix}}$	0.0101

Thus the repeating decimal fraction 0.3... is equivalent to the repeating binary fraction 0.01... .

Or for example, 0.1₁₀, in binary, is:

Converting	Result
0.1	0.
0.1 × 2 = 0.2 < 1	0.0
0.2 × 2 = 0.4 < 1	0.00
0.4 × 2 = 0.8 < 1	0.000
0.8 × 2 = 1.6 ≥ 1	0.0001
0.6 × 2 = 1.2 ≥ 1	0.00011
0.2 × 2 = 0.4 < 1	0.000110
0.4 × 2 = 0.8 < 1	0.0001100
0.8 × 2 = 1.6 ≥ 1	0.00011001
0.6 × 2 = 1.2 ≥ 1	0.000110011
0.2 × 2 = 0.4 < 1	0.0001100110

This is also a repeating binary fraction 0.000110011... . It may come as a surprise that terminating decimal fractions can have repeating expansions in binary. It is for this reason that many are surprised to discover that 0.1 + ... + 0.1, (10 additions) differs from 1 in floating point arithmetic. In fact, the only binary fractions with terminating expansions are of the form of an integer divided by a power of 2, which 1/10 is not.

The final conversion is from binary to decimal fractions. The only difficulty arises with repeating fractions, but otherwise the method is to shift the fraction to an integer, convert it as above, and then divide by the appropriate power of two in the decimal base. For example:

$x$	=	1100	.101110011100...
$x\times 2^{6}$	=	1100101110	.0111001110...
$x\times 2$	=	11001	.0111001110...
$x\times (2^{6}-2)$	=	1100010101
$x$	=	(789/62)₁₀

Another way of converting from binary to decimal, often quicker for a person familiar with hexadecimal, is to do so indirectly—first converting ( $x$ in binary) into ( $x$ in hexadecimal) and then converting ( $x$ in hexadecimal) into ( $x$ in decimal).

For very large numbers, these simple methods are inefficient because they perform a large number of multiplications or divisions where one operand is very large. A simple divide-and-conquer algorithm is more effective asymptotically: given a binary number, we divide by 10^k, where k is chosen so that the quotient roughly equals the remainder, then each of these pieces is converted to decimal and the two are concatenated. Given a decimal number, we split it into two pieces of about the same size, convert each to binary, then multiply the first piece by 10^k and add them, where k is the number of digits in the least-significant (rightmost) piece.

Hexadecimal

Binary may be converted to and from hexadecimal somewhat more easily. This is because the radix of the hexadecimal system (16) is a power of the radix of the binary system (2). More specifically, 16 = 2⁴, so it takes four digits of binary to represent one digit of hexadecimal.

The following table shows each hexadecimal digit along with the equivalent decimal value and four-digit binary sequence:

Hex	Dec	Binary
0	0	0000
1	1	0001
2	2	0010
3	3	0011

Hex	Dec	Binary
4	4	0100
5	5	0101
6	6	0110
7	7	0111

Hex	Dec	Binary
8	8	1000
9	9	1001
A	10	1010
B	11	1011

Hex	Dec	Binary
C	12	1100
D	13	1101
E	14	1110
F	15	1111

To convert a hexadecimal number into its binary equivalent, simply substitute the corresponding binary digits:

3A₁₆ = 0011 1010₂

E7₁₆ = 1110 0111₂

To convert a binary number into its hexadecimal equivalent, divide it into groups of four bits. If the number of bits isn't a multiple of four, simply insert extra 0 bits at the left (called padding). For example:

01010010₂ = 0101 0010 grouped with padding = 52₁₆

11011101₂ = 1101 1101 grouped = DD₁₆

To convert a hexadecimal number into its decimal equivalent, multiply the decimal equivalent of each hexadecimal digit by the corresponding power of 16 and add the resulting values:

C0E7₁₆ = (12 × 16³) + (0 × 16²) + (14 × 16¹) + (7 × 16⁰) = (12 × 4096) + (0 × 256) + (14 × 16) + (7 × 1) = 49,383₁₀

Octal

Binary is also easily converted to the octal numeral system, since octal uses a radix of 8, which is a power of two (namely, 2³, so it takes exactly three binary digits to represent an octal digit). The correspondence between octal and binary numerals is the same as for the first eight digits of hexadecimal in the table above. Binary 000 is equivalent to the octal digit 0, binary 111 is equivalent to octal 7, and so forth.

Octal	Binary
0	000
1	001
2	010
3	011
4	100
5	101
6	110
7	111

Converting from octal to binary proceeds in the same fashion as it does for hexadecimal:

65₈ = 110 101₂

17₈ = 001 111₂

And from binary to octal:

101100₂ = 101 100₂ grouped = 54₈

10011₂ = 010 011₂ grouped with padding = 23₈

And from octal to decimal:

65₈ = (6 × 8¹) + (5 × 8⁰) = (6 × 8) + (5 × 1) = 53₁₀

127₈ = (1 × 8²) + (2 × 8¹) + (7 × 8⁰) = (1 × 64) + (2 × 8) + (7 × 1) = 87₁₀

Representing real numbers

Non-integers can be represented by using negative powers, which are set off from the other digits by means of a radix point (called a decimal point in the decimal system). For example, the binary number 11.01₂ thus means:

1 × 2¹	(1 × 2 = 2)	plus
1 × 2⁰	(1 × 1 = 1)	plus
0 × 2^-1	(0 × ½ = 0)	plus
1 × 2^-2	(1 × ¼ = 0.25)

For a total of 3.25 decimal.

All dyadic rational numbers ${\frac {p}{2^{a}}}$ have a terminating binary numeral—the binary representation has a finite number of terms after the radix point. Other rational numbers have binary representation, but instead of terminating, they recur, with a finite sequence of digits repeating indefinitely. For instance

{\frac {1_{10}}{3_{10}}}

=

{\frac {1_{2}}{11_{2}}}

= 0.0101010101...₂

{\frac {12_{10}}{17_{10}}}

=

{\frac {1100_{2}}{10001_{2}}}

= 0.10110100 10110100 10110100...₂

The phenomenon that the binary representation of any rational is either terminating or recurring also occurs in other radix-based numeral systems. See, for instance, the explanation in decimal. Another similarity is the existence of alternative representations for any terminating representation, relying on the fact that 0.111111... is the sum of the geometric series 2^-1 + 2^-2 + 2^-3 + ... which is 1.

Binary numerals which neither terminate nor recur represent irrational numbers. For instance,

0.10100100010000100000100.... does have a pattern, but it is not a fixed-length recurring pattern, so the number is irrational
1.0110101000001001111001100110011111110... is the binary representation of ${\sqrt {2}}$ , the square root of 2, another irrational. It has no discernible pattern, although a proof that ${\sqrt {2}}$ is irrational requires more than this. See irrational number.

References

^ W. S. Anglin and J. Lambek, The Heritage of Thales, Springer, 1995, ISBN 038794544X
^ Bacon, Francis The Advancement of Learning, Book 6, Chapter 1, 1605. Online here.

External links

Template:Link FA Template:Link FA

[1] W. S. Anglin and J. Lambek, The Heritage of Thales, Springer, 1995, ISBN 038794544X

[2] Bacon, Francis The Advancement of Learning, Book 6, Chapter 1, 1605. Online here.

[1]

[2]

@@ Line 269: / Line 269: @@
 | || ||  || = 0
 |-
-|align="right"|0 ||× 2 +|| '''1''' || = 1
+|align="right"|0 ||× 2 +|| '''1''' || = poohio
 |-
 |align="right"|1 ||× 2 +|| '''1''' || = 3