Computer Number Systems-BCA

A number system defines a way to represent numbers. Computers use different number systems for processing and representation. The most common systems are Binary, Decimal, Octal, and Hexadecimal.

1. Binary Number System

  • Base: 2
  • Digits Used: 0, 1
  • Example: 101121011_210112​
  • How it Works:
    Each binary digit (bit) represents a power of 2. For example:
    10112=(1×23)+(0×22)+(1×21)+(1×20)=11101011_2 = (1 \times 2^3) + (0 \times 2^2) + (1 \times 2^1) + (1 \times 2^0) = 11_{10}10112​=(1×23)+(0×22)+(1×21)+(1×20)=1110​

Use:
Widely used in computers for internal data representation and processing.

2. Decimal Number System

  • Base: 10
  • Digits Used: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
  • Example: 25610256_{10}25610​
  • How it Works:
    Each digit represents a power of 10. For example:
    25610=(2×102)+(5×101)+(6×100)256_{10} = (2 \times 10^2) + (5 \times 10^1) + (6 \times 10^0)25610​=(2×102)+(5×101)+(6×100)

Use:
The most commonly used number system in daily life.

3. Octal Number System

  • Base: 8
  • Digits Used: 0, 1, 2, 3, 4, 5, 6, 7
  • Example: 1578157_81578​
  • How it Works:
    Each digit represents a power of 8. For example:
    1578=(1×82)+(5×81)+(7×80)=11110157_8 = (1 \times 8^2) + (5 \times 8^1) + (7 \times 8^0) = 111_{10}1578​=(1×82)+(5×81)+(7×80)=11110​

Use:
Used in computer systems for compact binary representation.

4. Hexadecimal Number System

  • Base: 16
  • Digits Used: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F
    • A=10A = 10A=10, B=11B = 11B=11, C=12C = 12C=12, D=13D = 13D=13, E=14E = 14E=14, F=15F = 15F=15
  • Example: 2F3162F3_{16}2F316​
  • How it Works:
    Each digit represents a power of 16. For example:
    2F316=(2×162)+(15×161)+(3×160)=755102F3_{16} = (2 \times 16^2) + (15 \times 16^1) + (3 \times 16^0) = 755_{10}2F316​=(2×162)+(15×161)+(3×160)=75510​

Use:
Widely used in programming, memory addressing, and color codes.

Comparison of Number Systems

System

Base

Digits Used

Example

Binary

2

0, 1

101121011_210112​

Decimal

10

0–9

25610256_{10}25610​

Octal

8

0–7

1578157_81578​

Hexadecimal

16

0–9, A–F

2F3162F3_{16}2F316​

Conversions Between Systems

  1. Binary to Decimal
    Multiply each bit by 2n2^n2n and sum them.
    Example: 11012=(1×23)+(1×22)+(0×21)+(1×20)=13101101_2 = (1 \times 2^3) + (1 \times 2^2) + (0 \times 2^1) + (1 \times 2^0) = 13_{10}11012​=(1×23)+(1×22)+(0×21)+(1×20)=1310​
  2. Decimal to Binary
    Divide the decimal number by 2 and write the remainders in reverse.
    Example: 1310=1101213_{10} = 1101_21310​=11012​
  3. Binary to Hexadecimal
    Group bits in sets of 4 (starting from the right) and convert to hexadecimal digits.
    Example: 111011012=ED1611101101_2 = ED_{16}111011012​=ED16​
  4. Hexadecimal to Binary
    Convert each hexadecimal digit to its 4-bit binary equivalent.
    Example: 1A316=00011010001121A3_{16} = 0001 1010 0011_21A316​=0001101000112​

Practice Questions

  1. Convert 10110210110_2101102​ to decimal.
  2. Convert 34510345_{10}34510​ to binary.
  3. Convert 7F167F_{16}7F16​ to binary.
  4. Convert 12610126_{10}12610​ to hexadecimal.

1. Binary to Decimal

Rule: Multiply each bit by 2n2^n2n, where nnn is the position of the bit from the right (starting from 0).

Example: Convert 101121011_210112​ to Decimal

10112=(1×23)+(0×22)+(1×21)+(1×20)1011_2 = (1 \times 2^3) + (0 \times 2^2) + (1 \times 2^1) + (1 \times 2^0)10112​=(1×23)+(0×22)+(1×21)+(1×20) =8+0+2+1=1110= 8 + 0 + 2 + 1 = 11_{10}=8+0+2+1=1110​

2. Decimal to Binary

Rule: Repeatedly divide the decimal number by 2, recording the remainders. Write the remainders in reverse order.

Example: Convert 131013_{10}1310​ to Binary

  • 13÷2=613 \div 2 = 613÷2=6 remainder 1
  • 6÷2=36 \div 2 = 36÷2=3 remainder 0
  • 3÷2=13 \div 2 = 13÷2=1 remainder 1
  • 1÷2=01 \div 2 = 01÷2=0 remainder 1

Reading the remainders in reverse: 1310=1101213_{10} = 1101_21310​=11012​

3. Binary to Octal

Rule: Group bits in sets of 3 (from right to left) and convert each group to its octal equivalent.

Example: Convert 1101012110101_21101012​ to Octal

  • Group bits: 110110110 101101101
  • Convert: 1102=68110_2 = 6_81102​=68​, 1012=58101_2 = 5_81012​=58​

1101012=658110101_2 = 65_81101012​=658​

4. Octal to Binary

Rule: Convert each octal digit to its 3-bit binary equivalent.

Example: Convert 65865_8658​ to Binary

  • 68=11026_8 = 110_268​=1102​
  • 58=10125_8 = 101_258​=1012​

658=110101265_8 = 110101_2658​=1101012​

5. Binary to Hexadecimal

Rule: Group bits in sets of 4 (from right to left) and convert each group to its hexadecimal equivalent.

Example: Convert 11011011211011011_2110110112​ to Hexadecimal

  • Group bits: 110111011101 101110111011
  • Convert: 11012=D161101_2 = D_{16}11012​=D16​, 10112=B161011_2 = B_{16}10112​=B16​

110110112=DB1611011011_2 = DB_{16}110110112​=DB16​

6. Hexadecimal to Binary

Rule: Convert each hexadecimal digit to its 4-bit binary equivalent.

Example: Convert 3F163F_{16}3F16​ to Binary

  • 316=001123_{16} = 0011_2316​=00112​
  • F16=11112F_{16} = 1111_2F16​=11112​

3F16=0011111123F_{16} = 00111111_23F16​=001111112​

7. Decimal to Octal

Rule: Repeatedly divide the decimal number by 8, recording the remainders. Write the remainders in reverse order.

Example: Convert 11110111_{10}11110​ to Octal

  • 111÷8=13111 \div 8 = 13111÷8=13 remainder 7
  • 13÷8=113 \div 8 = 113÷8=1 remainder 5
  • 1÷8=01 \div 8 = 01÷8=0 remainder 1

Reading the remainders in reverse: 11110=1578111_{10} = 157_811110​=1578​

8. Octal to Decimal

Rule: Multiply each digit by 8n8^n8n, where nnn is the position from the right (starting from 0).

Example: Convert 1578157_81578​ to Decimal

1578=(1×82)+(5×81)+(7×80)157_8 = (1 \times 8^2) + (5 \times 8^1) + (7 \times 8^0)1578​=(1×82)+(5×81)+(7×80) =64+40+7=11110= 64 + 40 + 7 = 111_{10}=64+40+7=11110​

9. Decimal to Hexadecimal

Rule: Repeatedly divide the decimal number by 16, recording the remainders. Write the remainders in reverse order.

Example: Convert 75510755_{10}75510​ to Hexadecimal

  • 755÷16=47755 \div 16 = 47755÷16=47 remainder 3
  • 47÷16=247 \div 16 = 247÷16=2 remainder 15 (F)
  • 2÷16=02 \div 16 = 02÷16=0 remainder 2

Reading the remainders in reverse: 75510=2F316755_{10} = 2F3_{16}75510​=2F316​

10. Hexadecimal to Decimal

Rule: Multiply each digit by 16n16^n16n, where nnn is the position from the right (starting from 0).

Example: Convert 2F3162F3_{16}2F316​ to Decimal

2F316=(2×162)+(15×161)+(3×160)2F3_{16} = (2 \times 16^2) + (15 \times 16^1) + (3 \times 16^0)2F316​=(2×162)+(15×161)+(3×160) =512+240+3=75510= 512 + 240 + 3 = 755_{10}=512+240+3=75510​

Conversion Summary Table

From

To

How

Binary

Decimal

Multiply bits by 2n2^n2n and sum.

Decimal

Binary

Divide by 2, write remainders in reverse.

Binary

Octal

Group bits in 3's, convert to octal.

Octal

Binary

Convert each digit to 3-bit binary.

Binary

Hexadecimal

Group bits in 4's, convert to hexadecimal.

Hexadecimal

Binary

Convert each digit to 4-bit binary.

Decimal

Octal

Divide by 8, write remainders in reverse.

Octal

Decimal

Multiply digits by 8n8^n8n and sum.

Decimal

Hexadecimal

Divide by 16, write remainders in reverse.

Hexadecimal

Decimal

Multiply digits by 16n16^n16n and sum.

Steps to Convert Decimal to Another System

  1. Convert the Integer Part (e.g., 345434543454): Use the standard integer conversion method (e.g., divide by the base and record remainders for binary, octal, or hexadecimal).
  2. Convert the Fractional Part (e.g., 0.560.560.56):
    • Multiply the fractional part by the base of the target system.
    • Extract the integer part of the result as the next digit.
    • Repeat the process with the new fractional part until it becomes zero or you reach the desired precision.

Example: Convert 3454.56103454.56_{10}3454.5610​ to Binary

Step 1: Convert Integer Part (3454103454_{10}345410​)

  1. 3454÷2=17273454 \div 2 = 17273454÷2=1727 remainder 0
  2. 1727÷2=8631727 \div 2 = 8631727÷2=863 remainder 1
  3. 863÷2=431863 \div 2 = 431863÷2=431 remainder 1
  4. 431÷2=215431 \div 2 = 215431÷2=215 remainder 1
  5. 215÷2=107215 \div 2 = 107215÷2=107 remainder 1
  6. 107÷2=53107 \div 2 = 53107÷2=53 remainder 1
  7. 53÷2=2653 \div 2 = 2653÷2=26 remainder 1
  8. 26÷2=1326 \div 2 = 1326÷2=13 remainder 0
  9. 13÷2=613 \div 2 = 613÷2=6 remainder 1
  10. 6÷2=36 \div 2 = 36÷2=3 remainder 0
  11. 3÷2=13 \div 2 = 13÷2=1 remainder 1
  12. 1÷2=01 \div 2 = 01÷2=0 remainder 1

Reading remainders from bottom to top:

345410=11010111111023454_{10} = 110101111110_2345410​=1101011111102​

Step 2: Convert Fractional Part (0.56100.56_{10}0.5610​)

  1. 0.56×2=1.120.56 \times 2 = 1.120.56×2=1.12 → Integer part 1
    Fractional part = 0.120.120.12
  2. 0.12×2=0.240.12 \times 2 = 0.240.12×2=0.24 → Integer part 0
    Fractional part = 0.240.240.24
  3. 0.24×2=0.480.24 \times 2 = 0.480.24×2=0.48 → Integer part 0
    Fractional part = 0.480.480.48
  4. 0.48×2=0.960.48 \times 2 = 0.960.48×2=0.96 → Integer part 0
    Fractional part = 0.960.960.96
  5. 0.96×2=1.920.96 \times 2 = 1.920.96×2=1.92 → Integer part 1
    Fractional part = 0.920.920.92
    (Continue as needed for more precision.)

Combining the results:

3454.5610=110101111110.100012(approximation)3454.56_{10} = 110101111110.10001_2 \quad (\text{approximation})3454.5610​=110101111110.100012​(approximation)

Example: Convert 3454.56103454.56_{10}3454.5610​ to Hexadecimal

Step 1: Convert Integer Part (3454103454_{10}345410​)

  1. 3454÷16=2153454 \div 16 = 2153454÷16=215 remainder 14 (E)
  2. 215÷16=13215 \div 16 = 13215÷16=13 remainder 7
  3. 13÷16=013 \div 16 = 013÷16=0 remainder 13 (D)

Reading remainders from bottom to top:

345410=D7E163454_{10} = D7E_{16}345410​=D7E16​

Step 2: Convert Fractional Part (0.56100.56_{10}0.5610​)

  1. 0.56×16=8.960.56 \times 16 = 8.960.56×16=8.96 → Integer part 8
    Fractional part = 0.960.960.96
  2. 0.96×16=15.360.96 \times 16 = 15.360.96×16=15.36 → Integer part F
    Fractional part = 0.360.360.36
  3. 0.36×16=5.760.36 \times 16 = 5.760.36×16=5.76 → Integer part 5
    Fractional part = 0.760.760.76 (Continue as needed for more precision.)

Combining the results:

3454.5610=D7E.8F516(approximation)3454.56_{10} = D7E.8F5_{16} \quad (\text{approximation})3454.5610​=D7E.8F516​(approximation)

General Formula for Fractional Conversion

For a fractional decimal value XXX:

  1. Multiply XXX by the target base.
  2. Take the integer part as the next digit.
  3. Repeat with the fractional part.

This process works for any base (binary, octal, hexadecimal).

Step 1: Convert the Integer Part (1001102100110_21001102​)

Use the positional value of binary digits (2n2^n2n):

1001102=(1×25)+(0×24)+(0×23)+(1×22)+(1×21)+(0×20)100110_2 = (1 \times 2^5) + (0 \times 2^4) + (0 \times 2^3) + (1 \times 2^2) + (1 \times 2^1) + (0 \times 2^0)1001102​=(1×25)+(0×24)+(0×23)+(1×22)+(1×21)+(0×20) =32+0+0+4+2+0=38= 32 + 0 + 0 + 4 + 2 + 0 = 38=32+0+0+4+2+0=38

Step 2: Convert the Fractional Part (0.0110120.01101_20.011012​)

Each digit after the binary point represents a negative power of 2 (2−n2^{-n}2−n):

0.011012=(0×2−1)+(1×2−2)+(1×2−3)+(0×2−4)+(1×2−5)0.01101_2 = (0 \times 2^{-1}) + (1 \times 2^{-2}) + (1 \times 2^{-3}) + (0 \times 2^{-4}) + (1 \times 2^{-5})0.011012​=(0×2−1)+(1×2−2)+(1×2−3)+(0×2−4)+(1×2−5) =0+0.25+0.125+0+0.03125=0.40625= 0 + 0.25 + 0.125 + 0 + 0.03125 = 0.40625=0+0.25+0.125+0+0.03125=0.40625

Step 3: Add the Integer and Fractional Parts

100110.011012=38+0.40625=38.4062510100110.01101_2 = 38 + 0.40625 = 38.40625_{10}100110.011012​=38+0.40625=38.4062510​

Final Answer

100110.011012=38.4062510100110.01101_2 = 38.40625_{10}100110.011012​=38.4062510​

 

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