Answer :
Certainly! Let's perform each of the given conversions step-by-step:
### (i) Convert \(198.625_{10}\) to binary
1. Convert the integer part (198) to binary:
[tex]\[ 198_{10} = 198 \div 2 = 99 \text{ remainder } 0 \quad (\text{Least Significant Bit}) \][/tex]
[tex]\[ 99 \div 2 = 49 \text{ remainder } 1 \][/tex]
[tex]\[ 49 \div 2 = 24 \text{ remainder } 1 \][/tex]
[tex]\[ 24 \div 2 = 12 \text{ remainder } 0 \][/tex]
[tex]\[ 12 \div 2 = 6 \text{ remainder } 0 \][/tex]
[tex]\[ 6 \div 2 = 3 \text{ remainder } 0 \][/tex]
[tex]\[ 3 \div 2 = 1 \text{ remainder } 1 \][/tex]
[tex]\[ 1 \div 2 = 0 \text{ remainder } 1 \quad (\text{Most Significant Bit}) \][/tex]
Reading the remainders from bottom to top, we get:
[tex]\[ 198_{10} = 11000110_2 \][/tex]
2. Convert the fractional part (0.625) to binary:
[tex]\[ 0.625 \times 2 = 1.25 \quad (\text{integer part is } 1) \][/tex]
[tex]\[ 0.25 \times 2 = 0.5 \quad (\text{integer part is } 0) \][/tex]
[tex]\[ 0.5 \times 2 = 1.0 \quad (\text{integer part is } 1) \][/tex]
Combining these, the binary fractional part is \(0.101\).
Thus, the binary representation of \(198.625_{10}\) is:
[tex]\[ 198.625_{10} = 11000110.101_2 \][/tex]
### (ii) Convert \(64.125_{10}\) to octal
1. Convert the integer part (64) to octal:
[tex]\[ 64_{10} = 64 \div 8 = 8 \text{ remainder } 0 \][/tex]
[tex]\[ 8 \div 8 = 1 \text{ remainder } 0 \][/tex]
[tex]\[ 1 \div 8 = 0 \text{ remainder } 1 \][/tex]
Reading the remainders from bottom to top, we get:
[tex]\[ 64_{10} = 100_8 \][/tex]
2. Convert the fractional part (0.125) to octal:
[tex]\[ 0.125 \times 8 = 1.0 \quad (\text{integer part is } 1) \][/tex]
Thus, the octal fractional part is \(0.1\).
Therefore, the octal representation of \(64.125_{10}\) is:
[tex]\[ 64.125_{10} = 100.1_8 \][/tex]
### (iii) Convert \(65.375_{10}\) to hexadecimal
1. Convert the integer part (65) to hexadecimal:
[tex]\[ 65_{10} = 65 \div 16 = 4 \text{ remainder } 1 \][/tex]
[tex]\[ 4 \div 16 = 0 \text{ remainder } 4 \][/tex]
Reading the remainders from bottom to top, we get:
[tex]\[ 65_{10} = 41_{16} \][/tex]
2. Convert the fractional part (0.375) to hexadecimal:
[tex]\[ 0.375 \times 16 = 6.0 \quad (\text{integer part is } 6) \][/tex]
Thus, the hexadecimal fractional part is \(0.6\).
Therefore, the hexadecimal representation of \(65.375_{10}\) is:
[tex]\[ 65.375_{10} = 41.6_{16} \][/tex]
### (iv) Convert \(1000001.001_2\) to decimal
1. Convert the integer part (\(1000001_2\)) to decimal:
[tex]\[ 1000001_2 = 1 \times 2^6 + 0 \times 2^5 + 0 \times 2^4 + 0 \times 2^3 + 0 \times 2^2 + 0 \times 2^1 + 1 \times 2^0 \][/tex]
[tex]\[ = 1 \times 64 + 0 \times 32 + 0 \times 16 + 0 \times 8 + 0 \times 4 + 0 \times 2 + 1 \times 1 \][/tex]
[tex]\[ = 64 + 1 = 65 \][/tex]
2. Convert the fractional part (\(0.001_2\)) to decimal:
[tex]\[ 0.001_2 = 0 \times 2^{-1} + 0 \times 2^{-2} + 1 \times 2^{-3} \][/tex]
[tex]\[ = 0 + 0 + \frac{1}{8} \][/tex]
[tex]\[ = 0.125 \][/tex]
Thus, the decimal representation of \(1000001.001_2\) is:
[tex]\[ 1000001.001_2 = 65 + 0.125 = 65.125_{10} \][/tex]
### (v) Convert \(110001110011001100101011110011_2\) to hexadecimal
1. Group the binary digits into sets of 4 (starting from the right):
[tex]\[ 1100 0111 0011 0011 0010 1011 1100 11 \][/tex]
2. Convert each group of 4 binary digits to hexadecimal:
[tex]\[ 1100 (12) = C \][/tex]
[tex]\[ 0111 (7) = 7 \][/tex]
[tex]\[ 0011 (3) = 3 \][/tex]
[tex]\[ 0011 (3) = 3 \][/tex]
[tex]\[ 0010 (2) = 2 \][/tex]
[tex]\[ 1011 (11) = B \][/tex]
[tex]\[ 1100 (12) = C \][/tex]
[tex]\[ 0011 (3) = 3 \][/tex]
Thus, the hexadecimal representation of \(110001110011001100101011110011_2\) is:
[tex]\[ 110001110011001100101011110011_2 = 31cccaf3_{16} \][/tex]
### (i) Convert \(198.625_{10}\) to binary
1. Convert the integer part (198) to binary:
[tex]\[ 198_{10} = 198 \div 2 = 99 \text{ remainder } 0 \quad (\text{Least Significant Bit}) \][/tex]
[tex]\[ 99 \div 2 = 49 \text{ remainder } 1 \][/tex]
[tex]\[ 49 \div 2 = 24 \text{ remainder } 1 \][/tex]
[tex]\[ 24 \div 2 = 12 \text{ remainder } 0 \][/tex]
[tex]\[ 12 \div 2 = 6 \text{ remainder } 0 \][/tex]
[tex]\[ 6 \div 2 = 3 \text{ remainder } 0 \][/tex]
[tex]\[ 3 \div 2 = 1 \text{ remainder } 1 \][/tex]
[tex]\[ 1 \div 2 = 0 \text{ remainder } 1 \quad (\text{Most Significant Bit}) \][/tex]
Reading the remainders from bottom to top, we get:
[tex]\[ 198_{10} = 11000110_2 \][/tex]
2. Convert the fractional part (0.625) to binary:
[tex]\[ 0.625 \times 2 = 1.25 \quad (\text{integer part is } 1) \][/tex]
[tex]\[ 0.25 \times 2 = 0.5 \quad (\text{integer part is } 0) \][/tex]
[tex]\[ 0.5 \times 2 = 1.0 \quad (\text{integer part is } 1) \][/tex]
Combining these, the binary fractional part is \(0.101\).
Thus, the binary representation of \(198.625_{10}\) is:
[tex]\[ 198.625_{10} = 11000110.101_2 \][/tex]
### (ii) Convert \(64.125_{10}\) to octal
1. Convert the integer part (64) to octal:
[tex]\[ 64_{10} = 64 \div 8 = 8 \text{ remainder } 0 \][/tex]
[tex]\[ 8 \div 8 = 1 \text{ remainder } 0 \][/tex]
[tex]\[ 1 \div 8 = 0 \text{ remainder } 1 \][/tex]
Reading the remainders from bottom to top, we get:
[tex]\[ 64_{10} = 100_8 \][/tex]
2. Convert the fractional part (0.125) to octal:
[tex]\[ 0.125 \times 8 = 1.0 \quad (\text{integer part is } 1) \][/tex]
Thus, the octal fractional part is \(0.1\).
Therefore, the octal representation of \(64.125_{10}\) is:
[tex]\[ 64.125_{10} = 100.1_8 \][/tex]
### (iii) Convert \(65.375_{10}\) to hexadecimal
1. Convert the integer part (65) to hexadecimal:
[tex]\[ 65_{10} = 65 \div 16 = 4 \text{ remainder } 1 \][/tex]
[tex]\[ 4 \div 16 = 0 \text{ remainder } 4 \][/tex]
Reading the remainders from bottom to top, we get:
[tex]\[ 65_{10} = 41_{16} \][/tex]
2. Convert the fractional part (0.375) to hexadecimal:
[tex]\[ 0.375 \times 16 = 6.0 \quad (\text{integer part is } 6) \][/tex]
Thus, the hexadecimal fractional part is \(0.6\).
Therefore, the hexadecimal representation of \(65.375_{10}\) is:
[tex]\[ 65.375_{10} = 41.6_{16} \][/tex]
### (iv) Convert \(1000001.001_2\) to decimal
1. Convert the integer part (\(1000001_2\)) to decimal:
[tex]\[ 1000001_2 = 1 \times 2^6 + 0 \times 2^5 + 0 \times 2^4 + 0 \times 2^3 + 0 \times 2^2 + 0 \times 2^1 + 1 \times 2^0 \][/tex]
[tex]\[ = 1 \times 64 + 0 \times 32 + 0 \times 16 + 0 \times 8 + 0 \times 4 + 0 \times 2 + 1 \times 1 \][/tex]
[tex]\[ = 64 + 1 = 65 \][/tex]
2. Convert the fractional part (\(0.001_2\)) to decimal:
[tex]\[ 0.001_2 = 0 \times 2^{-1} + 0 \times 2^{-2} + 1 \times 2^{-3} \][/tex]
[tex]\[ = 0 + 0 + \frac{1}{8} \][/tex]
[tex]\[ = 0.125 \][/tex]
Thus, the decimal representation of \(1000001.001_2\) is:
[tex]\[ 1000001.001_2 = 65 + 0.125 = 65.125_{10} \][/tex]
### (v) Convert \(110001110011001100101011110011_2\) to hexadecimal
1. Group the binary digits into sets of 4 (starting from the right):
[tex]\[ 1100 0111 0011 0011 0010 1011 1100 11 \][/tex]
2. Convert each group of 4 binary digits to hexadecimal:
[tex]\[ 1100 (12) = C \][/tex]
[tex]\[ 0111 (7) = 7 \][/tex]
[tex]\[ 0011 (3) = 3 \][/tex]
[tex]\[ 0011 (3) = 3 \][/tex]
[tex]\[ 0010 (2) = 2 \][/tex]
[tex]\[ 1011 (11) = B \][/tex]
[tex]\[ 1100 (12) = C \][/tex]
[tex]\[ 0011 (3) = 3 \][/tex]
Thus, the hexadecimal representation of \(110001110011001100101011110011_2\) is:
[tex]\[ 110001110011001100101011110011_2 = 31cccaf3_{16} \][/tex]
