0 - 110 0100 0110 - 0001 0110 1101 0110 1000 1000 0001 0000 0000 0000 0000 0001 1101 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard Converted to Decimal

0 - 110 0100 0110 - 0001 0110 1101 0110 1000 1000 0001 0000 0000 0000 0000 0001 1101: 64 bit double precision IEEE 754 binary floating point representation standard converted to decimal

What are the steps to convert
0 - 110 0100 0110 - 0001 0110 1101 0110 1000 1000 0001 0000 0000 0000 0000 0001 1101, a 64 bit double precision IEEE 754 binary floating point representation standard to decimal?

1. Identify the elements that make up the binary representation of the number:

The first bit (the leftmost) indicates the sign,
1 = negative, 0 = positive.
0

The next 11 bits contain the exponent:
110 0100 0110

The last 52 bits contain the mantissa:
0001 0110 1101 0110 1000 1000 0001 0000 0000 0000 0000 0001 1101


2. Convert the exponent from binary (from base 2) to decimal (in base 10).

The exponent is allways a positive integer.

110 0100 0110(2) =

1 × 210 + 1 × 29 + 0 × 28 + 0 × 27 + 1 × 26 + 0 × 25 + 0 × 24 + 0 × 23 + 1 × 22 + 1 × 21 + 0 × 20 =

1,024 + 512 + 0 + 0 + 64 + 0 + 0 + 0 + 4 + 2 + 0 =

1,024 + 512 + 64 + 4 + 2 =

1,606(10)

3. Adjust the exponent.

Subtract the excess bits: 2(11 - 1) - 1 = 1023,

that is due to the 11 bit excess/bias notation.

The exponent, adjusted = 1,606 - 1023 = 583


4. Convert the mantissa from binary (from base 2) to decimal (in base 10).

The mantissa represents the fractional part of the number (what comes after the whole part of the number, separated from it by a comma).


0001 0110 1101 0110 1000 1000 0001 0000 0000 0000 0000 0001 1101(2) =

0 × 2-1 + 0 × 2-2 + 0 × 2-3 + 1 × 2-4 + 0 × 2-5 + 1 × 2-6 + 1 × 2-7 + 0 × 2-8 + 1 × 2-9 + 1 × 2-10 + 0 × 2-11 + 1 × 2-12 + 0 × 2-13 + 1 × 2-14 + 1 × 2-15 + 0 × 2-16 + 1 × 2-17 + 0 × 2-18 + 0 × 2-19 + 0 × 2-20 + 1 × 2-21 + 0 × 2-22 + 0 × 2-23 + 0 × 2-24 + 0 × 2-25 + 0 × 2-26 + 0 × 2-27 + 1 × 2-28 + 0 × 2-29 + 0 × 2-30 + 0 × 2-31 + 0 × 2-32 + 0 × 2-33 + 0 × 2-34 + 0 × 2-35 + 0 × 2-36 + 0 × 2-37 + 0 × 2-38 + 0 × 2-39 + 0 × 2-40 + 0 × 2-41 + 0 × 2-42 + 0 × 2-43 + 0 × 2-44 + 0 × 2-45 + 0 × 2-46 + 0 × 2-47 + 1 × 2-48 + 1 × 2-49 + 1 × 2-50 + 0 × 2-51 + 1 × 2-52 =

0 + 0 + 0 + 0.062 5 + 0 + 0.015 625 + 0.007 812 5 + 0 + 0.001 953 125 + 0.000 976 562 5 + 0 + 0.000 244 140 625 + 0 + 0.000 061 035 156 25 + 0.000 030 517 578 125 + 0 + 0.000 007 629 394 531 25 + 0 + 0 + 0 + 0.000 000 476 837 158 203 125 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 000 003 725 290 298 461 914 062 5 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 000 000 000 003 552 713 678 800 500 929 355 621 337 890 625 + 0.000 000 000 000 001 776 356 839 400 250 464 677 810 668 945 312 5 + 0.000 000 000 000 000 888 178 419 700 125 232 338 905 334 472 656 25 + 0 + 0.000 000 000 000 000 222 044 604 925 031 308 084 726 333 618 164 062 5 =

0.062 5 + 0.015 625 + 0.007 812 5 + 0.001 953 125 + 0.000 976 562 5 + 0.000 244 140 625 + 0.000 061 035 156 25 + 0.000 030 517 578 125 + 0.000 007 629 394 531 25 + 0.000 000 476 837 158 203 125 + 0.000 000 003 725 290 298 461 914 062 5 + 0.000 000 000 000 003 552 713 678 800 500 929 355 621 337 890 625 + 0.000 000 000 000 001 776 356 839 400 250 464 677 810 668 945 312 5 + 0.000 000 000 000 000 888 178 419 700 125 232 338 905 334 472 656 25 + 0.000 000 000 000 000 222 044 604 925 031 308 084 726 333 618 164 062 5 =

0.089 210 990 816 361 190 880 456 888 407 934 457 063 674 926 757 812 5(10)

5. Put all the numbers into expression to calculate the double precision floating point decimal value:

(-1)Sign × (1 + Mantissa) × 2(Adjusted exponent) =

(-1)0 × (1 + 0.089 210 990 816 361 190 880 456 888 407 934 457 063 674 926 757 812 5) × 2583 =

1.089 210 990 816 361 190 880 456 888 407 934 457 063 674 926 757 812 5 × 2583 = ...

= 34 482 558 930 883 659 392 646 886 006 659 574 037 037 368 543 987 849 373 382 411 864 271 343 389 969 856 992 797 290 591 871 745 756 643 690 618 019 939 468 853 420 674 555 967 193 050 884 461 896 706 970 729 382 838 041 128 730 624

0 - 110 0100 0110 - 0001 0110 1101 0110 1000 1000 0001 0000 0000 0000 0000 0001 1101, a 64 bit double precision IEEE 754 binary floating point representation standard to a decimal number, written in base ten (double) = 34 482 558 930 883 659 392 646 886 006 659 574 037 037 368 543 987 849 373 382 411 864 271 343 389 969 856 992 797 290 591 871 745 756 643 690 618 019 939 468 853 420 674 555 967 193 050 884 461 896 706 970 729 382 838 041 128 730 624(10)

Spaces were used to group digits: for binary, by 4, for decimal, by 3.

Share

» More ops: Convert 0 - 110 0100 0110 - 0001 0110 1101 0110 1000 1000 0001 0000 0000 0000 0000 0001 1100, 64 bit double precision IEEE 754 binary floating point representation standard to decimal

» Calculations Performed by Our Visitors: 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard Numbers Converted to Decimal. Data organized on a Monthly Basis

» Month 07, 2026 [July]: 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard Numbers Converted to Decimal. Calculations performed during the month of: July - by our visitors

» Improve your social skills: Your Greatest Competitor, Your Most Powerful Ally. Gemini Notebook YouTube Video of 7.54 minutes

How to convert numbers from 64 bit double precision IEEE 754 binary floating point standard to decimal system in base 10

Follow the steps below to convert a number from 64 bit double precision IEEE 754 binary floating point representation to base 10 decimal system:

  • 1. Identify the elements that make up the binary representation of the number:
    First bit (leftmost) indicates the sign, 1 = negative, 0 = pozitive.
    The next 11 bits contain the exponent.
    The last 52 bits contain the mantissa.
  • 2. Convert the exponent, that is allways a positive integer, from binary (base 2) to decimal (base 10).
  • 3. Adjust the exponent, subtract the excess bits, 2(11 - 1) - 1 = 1,023, that is due to the 11 bit excess/bias notation.
  • 4. Convert the mantissa, that represents the number's fractional part (the excess beyond the number's integer part, comma delimited), from binary (base 2) to decimal (base 10).
  • 5. Put all the numbers into expression to calculate the double precision floating point decimal value:
    (-1)Sign × (1 + Mantissa) × 2(Exponent adjusted)

Example: convert the number 1 - 100 0011 1101 - 1000 0000 0010 0001 0100 0000 0100 1110 0000 0100 0000 1010 1000 from 64 bit double precision IEEE 754 binary floating point system to base ten decimal (double):

  • 1. Identify the elements that make up the binary representation of the number:
    First bit (leftmost) indicates the sign, 1 = negative, 0 = pozitive.
    The next 11 bits contain the exponent: 100 0011 1101
    The last 52 bits contain the mantissa:
    1000 0000 0010 0001 0100 0000 0100 1110 0000 0100 0000 1010 1000
  • 2. Convert the exponent, that is allways a positive integer, from binary (base 2) to decimal (base 10):
    100 0011 1101(2) =
    1 × 210 + 0 × 29 + 0 × 28 + 0 × 27 + 0 × 26 + 1 × 25 + 1 × 24 + 1 × 23 + 1 × 22 + 0 × 21 + 1 × 20 =
    1,024 + 0 + 0 + 0 + 0 + 32 + 16 + 8 + 4 + 0 + 1 =
    1,024 + 32 + 16 + 8 + 4 + 1 =
    1,085(10)
  • 3. Adjust the exponent, subtract the excess bits, 2(11 - 1) - 1 = 1,023, that is due to the 11 bit excess/bias notation:
    Exponent adjusted = 1,085 - 1,023 = 62
  • 4. Convert the mantissa, that represents the number's fractional part (the excess beyond the number's integer part, comma delimited), from binary (base 2) to decimal (base 10):
    1000 0000 0010 0001 0100 0000 0100 1110 0000 0100 0000 1010 1000(2) =
    1 × 2-1 + 0 × 2-2 + 0 × 2-3 + 0 × 2-4 + 0 × 2-5 + 0 × 2-6 + 0 × 2-7 + 0 × 2-8 + 0 × 2-9 + 0 × 2-10 + 1 × 2-11 + 0 × 2-12 + 0 × 2-13 + 0 × 2-14 + 0 × 2-15 + 1 × 2-16 + 0 × 2-17 + 1 × 2-18 + 0 × 2-19 + 0 × 2-20 + 0 × 2-21 + 0 × 2-22 + 0 × 2-23 + 0 × 2-24 + 0 × 2-25 + 1 × 2-26 + 0 × 2-27 + 0 × 2-28 + 1 × 2-29 + 1 × 2-30 + 1 × 2-31 + 0 × 2-32 + 0 × 2-33 + 0 × 2-34 + 0 × 2-35 + 0 × 2-36 + 0 × 2-37 + 1 × 2-38 + 0 × 2-39 + 0 × 2-40 + 0 × 2-41 + 0 × 2-42 + 0 × 2-43 + 0 × 2-44 + 1 × 2-45 + 0 × 2-46 + 1 × 2-47 + 0 × 2-48 + 1 × 2-49 + 0 × 2-50 + 0 × 2-51 + 0 × 2-52 =
    0.5 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 488 281 25 + 0 + 0 + 0 + 0 + 0.000 015 258 789 062 5 + 0 + 0.000 003 814 697 265 625 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 000 014 901 161 193 847 656 25 + 0 + 0 + 0.000 000 001 862 645 149 230 957 031 25 + 0.000 000 000 931 322 574 615 478 515 625 + 0.000 000 000 465 661 287 307 739 257 812 5 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 000 000 003 637 978 807 091 712 951 660 156 25 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 000 000 000 028 421 709 430 404 007 434 844 970 703 125 + 0 + 0.000 000 000 000 007 105 427 357 601 001 858 711 242 675 781 25 + 0 + 0.000 000 000 000 001 776 356 839 400 250 464 677 810 668 945 312 5 + 0 + 0 + 0 =
    0.5 + 0.000 488 281 25 + 0.000 015 258 789 062 5 + 0.000 003 814 697 265 625 + 0.000 000 014 901 161 193 847 656 25 + 0.000 000 001 862 645 149 230 957 031 25 + 0.000 000 000 931 322 574 615 478 515 625 + 0.000 000 000 465 661 287 307 739 257 812 5 + 0.000 000 000 003 637 978 807 091 712 951 660 156 25 + 0.000 000 000 000 028 421 709 430 404 007 434 844 970 703 125 + 0.000 000 000 000 007 105 427 357 601 001 858 711 242 675 781 25 + 0.000 000 000 000 001 776 356 839 400 250 464 677 810 668 945 312 5 =
    0.500 507 372 900 793 612 302 550 172 898 918 390 274 047 851 562 5(10)
  • 5. Put all the numbers into expression to calculate the double precision floating point decimal value:
    (-1)Sign × (1 + Mantissa) × 2(Exponent adjusted) =
    (-1)1 × (1 + 0.500 507 372 900 793 612 302 550 172 898 918 390 274 047 851 562 5) × 262 =
    -1.500 507 372 900 793 612 302 550 172 898 918 390 274 047 851 562 5 × 262 =
    -6 919 868 872 153 800 704(10)

  • 1 - 100 0011 1101 - 1000 0000 0010 0001 0100 0000 0100 1110 0000 0100 0000 1010 1000 converted from 64 bit double precision IEEE 754 binary floating point representation to a decimal number (float) in decimal system (in base 10) = -6 919 868 872 153 800 704(10)