Math Problem Statement

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

Solution

To determine the real number represented by the hexadecimal value 60745488000000006074548800000000 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 60745488000000006074548800000000 can be broken down into binary:

    607454880000000016=0110 0000 0111 0100 0101 0100 1000 1000 0000 0000 0000 0000 0000 0000 0000 000026074548800000000_{16} = 0110\ 0000\ 0111\ 0100\ 0101\ 0100\ 1000\ 1000\ 0000\ 0000\ 0000\ 0000\ 0000\ 0000\ 0000\ 0000_2

    So, in binary, we have:

    01100000011101000101010010001000000000000000000000000000000000000110000001110100010101001000100000000000000000000000000000000000

  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:

    110000001112=7751011000000111_2 = 775_{10}

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

    7751023=248775 - 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:

    1.010001010100100010000000000000000000000000000000000021.0100010101001000100000000000000000000000000000000000_2

    Converting this binary fraction to decimal:

    1.01000101010010001000000000000000000000000000000000002=1.26464843751.0100010101001000100000000000000000000000000000000000_2 = 1.2646484375

  5. Calculate the Final Value

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

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

    Substituting in our values:

    =(+1)×1.2646484375×2248= (+1) \times 1.2646484375 \times 2^{-248}

  6. Expressing in Decimal Notation

    Calculating 1.2646484375×22481.2646484375 \times 2^{-248} gives a very small positive number. This is approximately:

    2.2284624111932685×1075\approx 2.2284624111932685 \times 10^{-75}

Final Answer

The real number represented by the hexadecimal value 60745488000000006074548800000000 in IEEE 754 double precision is approximately:

2.2284624111932685×10752.2284624111932685 \times 10^{-75}

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


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.

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Math Problem Analysis

Mathematical Concepts

IEEE 754 Floating-Point Representation
Binary to Decimal Conversion
Floating-Point Arithmetic

Formulas

(-1)^sign * mantissa * 2^(exponent - bias)
IEEE 754 bias for double-precision: 1023

Theorems

IEEE 754 Standard for Floating-Point Arithmetic

Suitable Grade Level

College Level