Binary number

 ( Binary Arithmetic ) 

It is based on taoistic duality of yin and yang.

The Song dynasty scholar Shao Yong (1011–1077) rearranged the hexagrams in a format that resembles modern binary numbers, although he did not intend his arrangement to be used mathematically.

and reading along rows either from bottom right to top left with solid lines as 0 and broken lines as 1 or from top left to bottom right with solid lines as 1 and broken lines as 0 hexagrams can be interpreted as sequence from 0 to 63.

In 1605, Francis Bacon discussed a system whereby 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". (See Bacon's cipher.)

In 1617, John Napier described a system he called location arithmetic for doing binary calculations using a non-positional representation by letters. Thomas Harriot investigated several positional numbering systems, including binary, but did not publish his results; they were found later among his papers. Possibly the first publication of the system in Europe was by Juan Caramuel y Lobkowitz, in 1700.

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 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 authorized a full research program in late 1938 with Stibitz at the helm. Their Complex Number Computer, completed 8 January 1940, was able to calculate complex numbers. In a demonstration to the American Mathematical Society conference at Dartmouth College on 11 September 1940, Stibitz was able to send the Complex Number Calculator remote commands over telephone lines by a Teleprinter. 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.

The Z1 computer, which was designed and built by Konrad Zuse between 1935 and 1938, used Boolean logic and binary floating-point numbers. (12 pages)

The numeric value represented in each case depends on the value assigned to each symbol. In the earlier days of computing, switches, punched holes, and punched paper tapes were used to represent binary values. In a modern computer, the numeric values may be represented by two different ; on a Magnetic field Disk storage, 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 the 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 to indicate their base, or radix. The following notations are equivalent:

• 100101 binary (explicit statement of format)
• 100101b (a suffix indicating binary format; also known as Intel convention
  (2025). 9783528049522, Vieweg-Verlag, reprint: Springer-Verlag. . ISBN 9783528049522
  (2007). 9783834891914, Vieweg, reprint: Springer-Verlag. . ISBN 9783834891914
  )
• 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; also known as Motorola convention)
• 0b100101 (a prefix indicating binary format, common in programming languages)
• 6b100101 (a prefix indicating number of bits in binary format, common in programming languages)
• #b100101 (a prefix indicating binary format, common in Lisp programming languages)

When spoken, binary numerals are usually read digit-by-digit, to distinguish them from decimal numerals. 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 represents the value four, it would be confusing to refer to the numeral as one hundred (a word that represents a completely different value, or amount). Alternatively, the binary numeral 100 can be read out as "four" (the correct value), but this does not make its binary nature explicit.

000, 001, 002, ... 007, 008, 009, (rightmost digit is reset to zero, and the digit to its left is incremented)

0 10, 011, 012, ...

   ...

090, 091, 092, ... 097, 098, 099, (rightmost two digits are reset to zeroes, and next digit is incremented)

100, 101, 102, ...

0000,

000 1, (rightmost bit starts over, and the next bit is incremented)

00 10, 0011, (rightmost two bits start over, and the next bit is incremented)

0 100, 0101, 0110, 0111, (rightmost three bits start over, and the next bit is incremented)

1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111 ...

In the binary system, each bit represents an increasing power of 2, with the rightmost bit representing 2⁰, the next representing 2¹, then 2², and so on. The value of a binary number is the sum of the powers of 2 represented by each "1" bit. For example, the binary number 100101 is converted to decimal form as follows:

100101₂ = + + + + +

100101₂ = + + + + +

100101₂ = 37₁₀

0 + 0 → 0

0 + 1 → 1

1 + 0 → 1

1 + 1 → 0, carry 1 (since 1 + 1 = 2 = 0 + (1 × 2¹) )

5 + 5 → 0, carry 1 (since 5 + 5 = 10 = 0 + (1 × 10¹) )

7 + 9 → 6, carry 1 (since 7 + 9 = 16 = 6 + (1 × 10¹) )

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:

    0 1 1 0 1
+   1 0 1 1 1
-------------
= 1 0 0 1 0 0 = 36
     

In this example, two numerals are being added together: 01101₂ (13₁₀) and 10111₂ (23₁₀). 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 (36₁₀).

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

     Binary                        Decimal
    1 1 1 1 1     likewise        9 9 9 9 9
 +          1                  +          1
  ———————————                   ———————————
  1 0 0 0 0 0                   1 0 0 0 0 0
     

Such long strings are quite common in the binary system. From that one finds that large binary numbers can be added using two simple steps, without excessive carry operations. In the following example, two numerals are being added together: 1 1 1 0 1 1 1 1 1 0₂ (958₁₀) and 1 0 1 0 1 1 0 0 1 1₂ (691₁₀), using the traditional carry method on the left, and the long carry method on the right:

Traditional Carry Method                       Long Carry Method
                                vs.
              carry the 1 until it is one digit past the "string" below
    1 1 1 0 1 1 1 1 1 0                       1 1 1 0 1 1 1 1 1 0  cross out the "string",
+   1 0 1 0 1 1 0 0 1 1                   +   1 0 1 0 1 1 0 0 1 1  and cross out the digit that was added to it
———————————————————————                    ——————————————————————
= 1 1 0 0 1 1 1 0 0 0 1                     1 1 0 0 1 1 1 0 0 0 1
     

The top row shows the carry bits used. Instead of the standard carry from one column to the next, the lowest-ordered "1" with a "1" in the corresponding place value beneath it may be added and a "1" may be carried to one digit past the end of the series. The "used" numbers must be crossed off, since they are already added. Other long strings may likewise be cancelled using the same technique. Then, simply add together any remaining digits normally. Proceeding in this manner gives the final answer of 1 1 0 0 1 1 1 0 0 0 1₂ (1649₁₀). In our simple example using small numbers, the traditional carry method required eight carry operations, yet the long carry method required only two, representing a substantial reduction of effort.

The binary addition table is similar to, but not the same as, the truth table of the logical disjunction operation $\backslash lor$. The difference is that $1\; \backslash lor\; 1\; =\; 1$, while $1+1=10$.

0 − 0 → 0

0 − 1 → 1, borrow 1

1 − 0 → 1

1 − 1 → 0

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

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

Subtracting a positive number is equivalent to adding a negative number of equal absolute value. Computers use signed number representations to handle negative numbers—most commonly the two's complement notation. Such representations eliminate the need for a separate "subtract" operation. Using two's complement notation, subtraction can be summarized by the following formula:

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

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

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

           1 0 1 1   ()
         × 1 0 1 0   ()
         ---------
           0 0 0 0   ← to the rightmost 'zero' in
   +     1 0 1 1     ← to the next 'one' in
   +   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      (5.625 in decimal)
             × 1 1 0 . 0 1        (6.25 in decimal)
             -------------------
                   1 . 0 1 1 0 1   ← to a 'one' in
     +           0 0 . 0 0 0 0     ← to a 'zero' in
     +         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)
     

See also Booth's multiplication algorithm.

The binary multiplication table is the same as the truth table of the logical conjunction operation $\backslash land$.

In the example below, the divisor is 101₂, or 5 in decimal, while the dividend is 11011₂, or 27 in 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 0 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
         -----
             1 1 1
         −   1 0 1
             -----
             0 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, this corresponds to the fact that 27 divided by 5 is 5, with a remainder of 2.

Aside from long division, one can also devise the procedure so as to allow for over-subtracting from the partial remainder at each iteration, thereby leading to alternative methods which are less systematic, but more flexible as a result.

                             1                          1  1                       1  1  0                   1  1  0  1
 -------------             -------------              -------------              -------------             -------------
√ 10 10 10 01             √ 10 10 10 01              √ 10 10 10 01              √ 10 10 10 01             √ 10 10 10 01
                           - 1                        - 1                        - 1                       - 1
Answer so far is 0,        ----                       ----                       ----                      ----
extended by 01 is 001,       1 10                       1 10                       1 10                      1 10
this CAN be subtracted                                - 1 01                     - 1 01                    - 1 01
from first pair 10,       Answer so far is 1,         -------                    -------                   -------
so first digit of         extended by 01 is 101,           1 10                       1 10 01                   1 10 01
answer is 1.              this CAN be subtracted                                                              - 1 10 01
                          from remainder 110, so     Answer so far is 11,       Answer so far is 110,         ----------
                          next answer digit is 1.    extended by 01 is 1101,    extended by 01 is 11001,              0
                                                     this is TOO BIG to         this CAN be subtracted
                                                     subtract from remainder    from remainder 11001, so           Done!
                                                     110, so next digit of      next digit of answer is 1.
                                                     answer is 0.
     

The result is 1197₁₀. The first Prior Value of 0 is simply an initial decimal value. This method is an application of the Horner scheme.

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 0.11010110101₂, the first digit is $\backslash frac\{1\}\{2\}$, the second $(\backslash frac\{1\}\{2\})^2\; =\; \backslash frac\{1\}\{4\}$, etc. So if there is a 1 in the first place after the decimal, then the number is at least $\backslash frac\{1\}\{2\}$, 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, $(\backslash frac\{1\}\{3\})\_\{10\}$, in binary, is:

Thus the repeating decimal fraction 0.... is equivalent to the repeating binary fraction 0.... .

Or for example, 0.1₁₀, in binary, is:

This is also a repeating binary fraction 0.0... . 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 1/10 + ... + 1/10 (addition of 10 numbers) differs from 1 in binary 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:

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, it is divided 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 Concatenation. Given a decimal number, it can be split into two pieces of about the same size, each of which is converted to binary, whereupon the first converted piece is multiplied by 10 ^k and added to the second converted piece, where k is the number of decimal digits in the second, least-significant piece before conversion.

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:

1010010₂ = 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

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

For a total of 3.25 decimal.

All dyadic fraction $\backslash frac\{p\}\{2^a\}$ 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

$$\backslash frac\{1\_\{10\}\}\{3\_\{10\}\}\; =\; \backslash frac\{1\_2\}\{11\_2\}\; =\; 0.01010101\backslash overline\{01\}\backslash ldots\backslash ,\_2$$ $$\backslash frac\{12\_\{10\}\}\{17\_\{10\}\}\; =\; \backslash frac\{1100\_2\}\{10001\_2\}\; =\; 0.10110100\; 10110100\backslash overline\{10110100\}\backslash ldots\backslash ,\_2$$

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 that 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 $\backslash sqrt\{2\}$, the square root of 2, another irrational. It has no discernible pattern.

• ASCII
• Balanced ternary
• Bitwise operation
• Binary code
• Binary-coded decimal
• Finger binary
• Gray code
• IEEE 754
• Linear-feedback shift register
• Offset binary
• Quibinary
• Reduction of summands
• Redundant binary representation
• Repeating decimal
• Two's complement
• Unicode

• Binary System at cut-the-knot
• Conversion of Fractions at cut-the-knot
• Sir Francis Bacon's BiLiteral Cypher system , predates binary number system.