CBSE Class 11 Computer Science Chapter 3: Data Representation NCERT Solutions

The Octal number system, also known as base 8, uses eight unique digits: 0, 1, 2, 3, 4, 5, 6, and 7. It was historically popular for representing digital circuit numbers because it is more compact than binary. For example, counting in octal proceeds as: 0, 1, 2, 3, 4, 5, 6, 7, 10, 11, 12, and so on.

The Hexadecimal number system, or base 16, uses sixteen unique symbols: the digits 0 through 9 and the letters A through F, where A represents 10, B represents 11, C represents 12, D represents 13, E represents 14, and F represents 15. This system is widely used today for representing digital circuit numbers due to its compactness compared to binary. Hexadecimal numbers are often prefixed with '0x', such as 0x1B84. Counting in hexadecimal progresses as: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F, 10, 11, and so on.

The Decimal number system, commonly known as base 10, is the standard system used by humans for counting and calculations. It utilizes ten unique digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. Counting in the decimal system proceeds as: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, and so on.

The Binary number system, or base 2, is fundamental to digital computers. It uses only two digits: 0 and 1. This system is the natural way digital circuits represent and process information. Binary numbers are sometimes indicated with a '0b' prefix, like 0b1011. Counting in binary progresses as: 0, 1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111, 10000, 10001, and so on.

To convert the decimal number 259 to binary, we use the method of repeated division by 2 and record the remainders.

259 \div 2 = 129 \text{ remainder } 1

129 \div 2 = 64 \text{ remainder } 1

64 \div 2 = 32 \text{ remainder } 0

32 \div 2 = 16 \text{ remainder } 0

16 \div 2 = 8 \text{ remainder } 0

8 \div 2 = 4 \text{ remainder } 0

4 \div 2 = 2 \text{ remainder } 0

2 \div 2 = 1 \text{ remainder } 0

1 \div 2 = 0 \text{ remainder } 1

The remainders are read from bottom to top to form the binary number. The last remainder obtained is the Most Significant Bit (MSB), and the first remainder is the Least Significant Bit (LSB).

Therefore, the binary representation of (259)10 is (100000011)2.

To convert the decimal number (17.35)10 to binary, we convert the integer part (17) and the fractional part (0.35) separately.

Integer Part Conversion (17 to binary):

17 \div 2 = 8 \text{ remainder } 1

8 \div 2 = 4 \text{ remainder } 0

4 \div 2 = 2 \text{ remainder } 0

2 \div 2 = 1 \text{ remainder } 0

1 \div 2 = 0 \text{ remainder } 1

Reading the remainders from bottom to top, we get (17)10 = (10001)2.

Fractional Part Conversion (0.35 to binary):

To convert the fractional part, we repeatedly multiply by 2 and take the integer part of the result.

0.35 \times 2 = 0.70 \quad \rightarrow \text{ integer part is } 0

0.70 \times 2 = 1.40 \quad \rightarrow \text{ integer part is } 1

0.40 \times 2 = 0.80 \quad \rightarrow \text{ integer part is } 0

0.80 \times 2 = 1.60 \quad \rightarrow \text{ integer part is } 1

0.60 \times 2 = 1.20 \quad \rightarrow \text{ integer part is } 1

0.20 \times 2 = 0.40 \quad \rightarrow \text{ integer part is } 0

The process can continue, but we observe that the fractional part 0.40 repeats, indicating a recurring binary fraction. Reading the integer parts from top to bottom, we get (0.35)10 = (0.010110...)2. The repeating part is 0110.

Combining the integer and fractional parts, we get:

(17.35)_{10} = (10001.010110...)_2 = (10001.01\overline{0110})_2 (recurring fraction)

To convert the decimal number 39286 to octal (base 8), we perform repeated division by 8 and record the remainders.

39286 \div 8 = 4910 \text{ remainder } 6

4910 \div 8 = 613 \text{ remainder } 6

613 \div 8 = 76 \text{ remainder } 5

76 \div 8 = 9 \text{ remainder } 4

9 \div 8 = 1 \text{ remainder } 1

1 \div 8 = 0 \text{ remainder } 1

Reading the remainders from bottom to top (from MSB to LSB), we get the octal representation.

Therefore, the octal representation of (39286)10 is (114566)8.