Determine the real number represented by the 16 hexadecimal digit number 6074548800000000 considering it as a 64-bit IEEE 754 format double precision number.

To determine the real number represented by the hexadecimal value \(6074548800000000\) in the IEEE 754 double-precision floating-point format, we need to analyze its structure as per the IEEE 754 standard.

### Step-by-Step Solution

1. **Convert the Hexadecimal Value to Binary**

   The hexadecimal number \(6074548800000000\) can be broken down into binary:

   \[
   6074548800000000_{16} = 0110\ 0000\ 0111\ 0100\ 0101\ 0100\ 1000\ 1000\ 0000\ 0000\ 0000\ 0000\ 0000\ 0000\ 0000\ 0000_2
   \]

   So, in binary, we have:

   \[
   0110000001110100010101001000100000000000000000000000000000000000
   \]

2. **Break Down the IEEE 754 Double-Precision Format**

   IEEE 754 double-precision format has:
   
   - **1 sign bit** (the first bit),
   - **11 exponent bits** (the next 11 bits),
   - **52 fraction (mantissa) bits** (the remaining 52 bits).

   For this binary number:
   
   - **Sign bit**: `0` (indicating a positive number),
   - **Exponent bits**: `11000000111`,
   - **Mantissa bits**: `0100010101001000100000000000000000000000000000000000`.

3. **Interpret the Exponent**

   The exponent in binary, `11000000111`, converts to decimal:

   \[
   11000000111_2 = 775_{10}
   \]

   In IEEE 754 format, the exponent is stored with a bias of 1023. Thus, the actual exponent is:

   \[
   775 - 1023 = -248
   \]

4. **Interpret the Mantissa (Fraction)**

   The mantissa in IEEE 754 format represents the fractional part. Since the exponent is not all 0s or 1s, we have a **normalized** number, and the mantissa has an implicit leading 1. So the mantissa is:

   \[
   1.0100010101001000100000000000000000000000000000000000_2
   \]

   Converting this binary fraction to decimal:

   \[
   1.0100010101001000100000000000000000000000000000000000_2 = 1.2646484375
   \]

5. **Calculate the Final Value**

   Putting it all together, the value of the number is:

   \[
   (-1)^{\text{sign}} \times \text{mantissa} \times 2^{\text{exponent}}
   \]

   Substituting in our values:

   \[
   = (+1) \times 1.2646484375 \times 2^{-248}
   \]

6. **Expressing in Decimal Notation**

   Calculating \(1.2646484375 \times 2^{-248}\) gives a very small positive number. This is approximately:

   \[
   \approx 2.2284624111932685 \times 10^{-75}
   \]

### Final Answer

The real number represented by the hexadecimal value \(6074548800000000\) in IEEE 754 double precision is approximately:

\[
2.2284624111932685 \times 10^{-75}
\]

Would you like a more detailed breakdown of any step, or do you have further questions?

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### Follow-up Questions

1. How is the bias determined in IEEE 754 for different formats (single, double)?
2. Why does IEEE 754 include an implicit leading 1 in normalized numbers?
3. How does the IEEE 754 format handle extremely large or small numbers, including underflow?
4. What is the difference between normalized and denormalized numbers in IEEE 754?
5. How does floating-point rounding work, and what impact can it have on calculations?

### Tip
For quick calculations involving IEEE 754 formats, you can use online IEEE 754 converters to save time on manual conversion.

Learn how to determine the real number represented by the hexadecimal value 6074548800000000 in IEEE 754 double precision format. This problem involves binary conversion, floating-point arithmetic, and understanding of IEEE 754 format structure. Suitable for college-level students.

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