Numbering Systems Notes

Converting Fractions

• To convert a fractional decimal value to binary, repeatedly multiply the decimal fraction by 2.

  • Example: Convert 11.375 to binary.
    • First, convert 11 to binary: $11<em>{10} = 1011</em>2$
    • Then, convert 0.375 to binary:
      • 0.375 * 2 = 0.750 --> 0
      • 0.750 * 2 = 1.500 --> 1
      • 0.500 * 2 = 1.000 --> 1
      • Therefore, $.375<em>{10} = .011</em>2$
    • Finally, $11.375<em>{10} = 1011.011</em>2$

• Exercise:

  • Convert the following numbers to their binary equivalents. $<br /> (26.75<em>{10} ) = 11010.11</em>2$
    $<br /> (37.375_{10} ) = ?$

• Exercise:

  • Convert the following decimal number to binary? $(0.2)<em>{10} = (0.0011)</em>2$
  • $(0.3)<em>{10} = (0.01001)</em>2$

Adding Binary Fractions

• To add binary fractions, align the decimal points and add as usual.

  • Example:

  ```
  1011.0

• 0.011

  1011.011
  ```

• Example:

  ```
  1
  110.01

• 1.011
  --------
  111.101
  ```

Binary Subtraction

• Use 2's complement representation for subtraction.
  • Rewrite $A-B$ as $A + (-B)$
  • Example: Solve $01111111<em>2 – 76</em>{10} = ???$
    • Rewrite as $01111111<em>2 + (– 76)</em>{10}$
    • Convert 76 to binary: $76<em>{10} = 01001100</em>2$
    • Find the 1's complement: $10110011$
    • Find the 2's complement by adding 1: $10110011 + 1 = 10110100$
    • $(– 76)<em>{10} = 10110100</em>2$
    • Add the numbers:

  1 1
    01111111
+    10110100
    ---------
  1 00110011
    ```
        *   $01111111_2 + (– 76)_{10} = 00110011_2$
*   127 - 76 = 51

*   When computing, if the carry bit exists this represents overflow.

*   Example: $00110010_2 + (– 125)_{10}$
    *   $125_{10} = 01111101_2$
    *   1's complement: $10000010$
    *   2's complement: $10000011$

00110010

• 10000011 --------- 10110101 ```
  • $50-125=-75$
  • The 2’s comp for the result (10110101) is 01001011 equivalent to (75) 10

Data Representation

• Computers understand on and off states.
• Data is represented in binary form.
• Bit: The basic unit for storing data (0 = off, 1 = on).
• Byte: A group of 8 bits. Each byte has $2^8 = 256$ possible values.
• Word: Two bytes (16 bits).

Parity Bit

• Used for error detection.
  • Odd parity: The number of 1s is odd.
  • Even parity: The number of 1s is even.

Characters Representation

• ASCII (American Standard Code for Information Interchange) is used to represent characters.
  • A = 65, B = 66, …, a = 97, b = 98, …

Example using Even Parity

• Represent the character Q (81 in ASCII) in memory using even parity.
  • $(81)<em>{10} = (01010001)</em>2$
  • With even parity: D1 (hexadecimal)

Example using Odd Parity

• Represent letters using odd parity:
  • A 01000001
  • h 01101000
  • m 01101101

Integers Representation

• Representing integers in memory using 2 bytes.
  • Example: Represent 92
    • $92 = 1011100$
    • 0000 0000 01011100
    • 0 0 5 C
  • Example: Represent -94
    • $94 = 0000000001011110$
    • 1’s complement-> 1111111110100001
    • 2’s complement-> 1111111110100010
    • F F A 2

Floating Point Representation

• 32 bits are divided into three sections: sign, exponent, and mantissa.

  • 1 bit for sign (0 for positive, 1 for negative).
  • 8 bits for exponent.
  • 23 bits for mantissa.

• For the exponent:

  • Range: 0-255. To represent negative exponents, a bias is used.
  • Bias = 127 (integer part of 255/2 = 127.5).
  • So, the exponent range is -127 to 128.

• Example: Represent (26.75)10 using 32-bit floating-point representation.

  • Convert to binary: $(26.75)<em>{10} = (11010.11)</em>2$
  • Convert to scientific notation: $(11010.11)<em>2 = (1.101011 * 2^4)</em>2$
  • Exponent = 127 + 4 = 131. Convert to binary: $(131)<em>{10} = (10000011)</em>2$
  • Final representation:
    • 0 10000011 10101100000000000000000
    • 4 1 D 6 0 0 0 0