To convert the value $CCCC_{16}$ in the textbook floating-point format to its corresponding value in decimal, we first need to understand the floating-point format used in the textbook. Typically, floating-point formats include a sign bit, an exponent, and a mantissa (or significand).

For this problem, I will assume the standard IEEE 754 single-precision floating-point format (32 bits), which is commonly used in textbooks. This format is structured as follows:

• 1 bit for the sign (S)
• 8 bits for the exponent (E)
• 23 bits for the mantissa (M)

Let's break down $CCCC_{16}$:

• $C$ in hexadecimal is $1100$ in binary.

Thus, $CCCC_{16}$ in binary is: $CCCC_{16} = 1100 1100 1100 1100_{2}$

Next, we will pad this to fit into a 32-bit floating-point format: $1100 1100 1100 1100 0000 0000 0000 0000$

Now, we divide it into the three parts of the IEEE 754 format:

• Sign bit (S): The first bit is 1.
• Exponent (E): The next 8 bits are 10011001.
• Mantissa (M): The last 23 bits are 10011001100000000000000.

Step-by-step Conversion

1. Sign bit (S):

   • $S = 1$ (which means the number is negative)

2. Exponent (E):

   • $E = 10011001_{2}$
   • Convert this to decimal: $10011001_{2} = 153_{10}$

3. Bias adjustment:

   • IEEE 754 uses a bias of 127 for single-precision numbers.
   • Actual exponent $e$ is calculated as $E - 127$: $e = 153 - 127 = 26$

4. Mantissa (M):

   • Mantissa is $1.M$ in binary where $M$ is the fraction part.
   • $M = 10011001100000000000000_{2}$
   • Converting $M$ to decimal: $1 + 2^{-1} + 0 \cdot 2^{-2} + 0 \cdot 2^{-3} + 1 \cdot 2^{-4} + 1 \cdot 2^{-5} + 0 \cdot 2^{-6} + 0 \cdot 2^{-7} + 1 \cdot 2^{-8} + 1 \cdot 2^{-9} + 0 \cdot 2^{-10} + \cdots$ $1 + 0.5 + 0.0625 + 0.03125 + 0.00390625 = 1.59765625$

5. Combine the parts:

   • The value of the floating-point number is: $(-1)^S \times 1.M \times 2^e = (-1)^1 \times 1.59765625 \times 2^{26}$

6. Calculate the final value: $-1.59765625 \times 2^{26}$

Perform the calculation: $1.59765625 \times 2^{26} = 1.59765625 \times 67108864 = 107374182.4$ $-107374182.4$

So, the value $CCCC_{16}$ in the textbook floating-point format corresponds to $-107374182.4$ in decimal.

Would you like further details or have any questions?

Here are 8 related questions that expand on this topic:

1. How does the IEEE 754 standard define the double-precision floating-point format?
2. How is the bias value determined in the IEEE 754 standard?
3. What are the differences between single-precision and double-precision formats?
4. How does the IEEE 754 standard handle special values like infinity and NaN?
5. How do you convert a decimal number to its IEEE 754 floating-point representation?
6. What are the limitations of floating-point arithmetic in computers?
7. How do rounding errors occur in floating-point calculations?
8. What are the benefits of using floating-point representation over fixed-point representation?

Tip: When converting between different numerical bases, always ensure to keep track of place values and biases correctly to avoid calculation errors.