What are the mantissa and exponent values if 6.75 is represented in &-bit binary floating-point representation

What are the mantissa and exponent values if 6.75 is represented in &-bit binary floating-point representation? Mantissa is 101 and exponent is 101 Mantissa is 001 and exponent is 100 Mantissa is 101 and exponent is 100 Mantissa is 001 and exponent is 101′

The correct answer and explanation is:

To find the mantissa and exponent for 6.75 in an 8-bit binary floating-point representation, we must follow a process similar to IEEE 754 floating-point format, though simplified for 8 bits.

Step 1: Convert 6.75 to binary

Break it into integer and decimal parts:

• 6 → binary: 110
• 0.75 → binary:
  • 0.75 × 2 = 1.5 → 1
  • 0.5 × 2 = 1.0 → 1
    So, 6.75 = 110.11 in binary.

Step 2: Normalize the binary number

Normalized form is of the type 1.xxxxx × 2^n.

• 110.11 = 1.1011 × 2^2

Step 3: Identify mantissa and exponent

Assuming an 8-bit layout:

• 1 sign bit
• 3 exponent bits (with bias, commonly 3)
• 4 mantissa bits (excluding the leading 1 which is implied)

Exponent:

• True exponent is 2
• With bias 3 → exponent field = 2 + 3 = 5 → binary: 101

Mantissa:

• From 1.1011, we take digits after the decimal: 1011
• Only 3 bits are used (in most simplified 8-bit versions)
• So mantissa = 101

Final Answer:

Mantissa is 101 and exponent is 101

Explanation (300 words):

Floating-point representation is used to encode real numbers in binary using three parts: sign, exponent, and mantissa (or significand). In an 8-bit floating-point system, due to limited space, the format is usually divided into 1 sign bit, 3 exponent bits, and 4 bits for the mantissa, with one of the bits being implied (hidden bit).

To represent the number 6.75:

1. Convert it to binary:
   • The integer part 6 is 110
   • The fractional part 0.75 becomes .11
     So, 6.75 in binary is 110.11.
2. Normalize this binary number to the form 1.xxxx × 2^n. For 110.11, this is 1.1011 × 2^2.
3. The exponent is 2, but to store this in 3 bits using a bias (commonly 3 in small formats), we store 2 + 3 = 5. Binary 5 is 101.
4. The mantissa is the fractional part after the leading 1: from 1.1011, we take 1011. Truncating to 3 bits (for a total of 4 mantissa bits with one implied), we get 101.

Therefore, the correct floating-point representation uses an exponent of 101 and a mantissa of 101.

✅ Correct answer: Mantissa is 101 and exponent is 101.