Binary ↘ Double: The 64 Bit Double Precision IEEE 754 Binary Floating Point Standard Representation Number 1 - 011 1111 1101 - 1000 0111 0010 0010 0001 1001 0001 0000 0010 1101 0110 0000 0001 Converted and Written as a Base Ten Decimal System Number (as a Double)

1 - 011 1111 1101 - 1000 0111 0010 0010 0001 1001 0001 0000 0010 1101 0110 0000 0001: 64 bit double precision IEEE 754 binary floating point standard representation number converted to decimal system (base ten)

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.
1


The next 11 bits contain the exponent:
011 1111 1101


The last 52 bits contain the mantissa:
1000 0111 0010 0010 0001 1001 0001 0000 0010 1101 0110 0000 0001


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

The exponent is allways a positive integer.

011 1111 1101(2) =


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


0 + 512 + 256 + 128 + 64 + 32 + 16 + 8 + 4 + 0 + 1 =


512 + 256 + 128 + 64 + 32 + 16 + 8 + 4 + 1 =


1,021(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 = 1,021 - 1023 = -2


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).


1000 0111 0010 0010 0001 1001 0001 0000 0010 1101 0110 0000 0001(2) =

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


0.5 + 0 + 0 + 0 + 0 + 0.015 625 + 0.007 812 5 + 0.003 906 25 + 0 + 0 + 0.000 488 281 25 + 0 + 0 + 0 + 0.000 030 517 578 125 + 0 + 0 + 0 + 0 + 0.000 000 953 674 316 406 25 + 0.000 000 476 837 158 203 125 + 0 + 0 + 0.000 000 059 604 644 775 390 625 + 0 + 0 + 0 + 0.000 000 003 725 290 298 461 914 062 5 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 000 000 029 103 830 456 733 703 613 281 25 + 0 + 0.000 000 000 007 275 957 614 183 425 903 320 312 5 + 0.000 000 000 003 637 978 807 091 712 951 660 156 25 + 0 + 0.000 000 000 000 909 494 701 772 928 237 915 039 062 5 + 0 + 0.000 000 000 000 227 373 675 443 232 059 478 759 765 625 + 0.000 000 000 000 113 686 837 721 616 029 739 379 882 812 5 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 000 000 000 000 222 044 604 925 031 308 084 726 333 618 164 062 5 =


0.5 + 0.015 625 + 0.007 812 5 + 0.003 906 25 + 0.000 488 281 25 + 0.000 030 517 578 125 + 0.000 000 953 674 316 406 25 + 0.000 000 476 837 158 203 125 + 0.000 000 059 604 644 775 390 625 + 0.000 000 003 725 290 298 461 914 062 5 + 0.000 000 000 029 103 830 456 733 703 613 281 25 + 0.000 000 000 007 275 957 614 183 425 903 320 312 5 + 0.000 000 000 003 637 978 807 091 712 951 660 156 25 + 0.000 000 000 000 909 494 701 772 928 237 915 039 062 5 + 0.000 000 000 000 227 373 675 443 232 059 478 759 765 625 + 0.000 000 000 000 113 686 837 721 616 029 739 379 882 812 5 + 0.000 000 000 000 000 222 044 604 925 031 308 084 726 333 618 164 062 5 =


0.527 864 042 710 803 227 365 090 606 326 702 982 187 271 118 164 062 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)1 × (1 + 0.527 864 042 710 803 227 365 090 606 326 702 982 187 271 118 164 062 5) × 2-2 =


-1.527 864 042 710 803 227 365 090 606 326 702 982 187 271 118 164 062 5 × 2-2 =


-0.381 966 010 677 700 806 841 272 651 581 675 745 546 817 779 541 015 625

1 - 011 1111 1101 - 1000 0111 0010 0010 0001 1001 0001 0000 0010 1101 0110 0000 0001 converted from a 64 bit double precision IEEE 754 binary floating point standard representation number to a decimal system number, written in base ten (double) = -0.381 966 010 677 700 806 841 272 651 581 675 745 546 817 779 541 015 625(10)

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

The latest 64 bit double precision IEEE 754 floating point binary standard numbers converted and written as decimal system numbers (in base ten, double)

The number 0 - 000 0000 0111 - 0111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110 0101 converted from 64 bit double precision IEEE 754 binary floating point system and written as a decimal number (double) written in base ten = ? Jul 13 14:01 UTC (GMT)
The number 0 - 100 0001 1101 - 0110 0011 1000 1000 1001 0110 1100 0000 0010 0001 0101 0011 1101 converted from 64 bit double precision IEEE 754 binary floating point system and written as a decimal number (double) written in base ten = ? Jul 13 14:00 UTC (GMT)
The number 0 - 100 0001 0011 - 1010 0001 0001 0001 0000 0100 0000 0111 1111 1111 1111 1100 1001 converted from 64 bit double precision IEEE 754 binary floating point system and written as a decimal number (double) written in base ten = ? Jul 13 14:00 UTC (GMT)
The number 1 - 001 1011 1000 - 0111 0011 1011 1011 1001 1101 1101 1100 1110 1110 1101 1100 0100 converted from 64 bit double precision IEEE 754 binary floating point system and written as a decimal number (double) written in base ten = ? Jul 13 14:00 UTC (GMT)
The number 1 - 100 0011 0011 - 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1010 0111 converted from 64 bit double precision IEEE 754 binary floating point system and written as a decimal number (double) written in base ten = ? Jul 13 13:59 UTC (GMT)
The number 0 - 100 0000 0100 - 0000 0110 1110 0100 0101 1011 0110 1111 1110 1011 0010 1010 0000 converted from 64 bit double precision IEEE 754 binary floating point system and written as a decimal number (double) written in base ten = ? Jul 13 13:59 UTC (GMT)
The number 1 - 101 1110 1010 - 1101 1011 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 0000 converted from 64 bit double precision IEEE 754 binary floating point system and written as a decimal number (double) written in base ten = ? Jul 13 13:59 UTC (GMT)
The number 0 - 100 0011 0000 - 1111 0100 0011 1000 1101 1010 1010 0000 0110 0000 0000 0100 1001 converted from 64 bit double precision IEEE 754 binary floating point system and written as a decimal number (double) written in base ten = ? Jul 13 13:59 UTC (GMT)
The number 1 - 100 0010 0011 - 0110 0110 0110 0011 0000 0000 0111 0011 0101 1010 1010 1001 1010 converted from 64 bit double precision IEEE 754 binary floating point system and written as a decimal number (double) written in base ten = ? Jul 13 13:59 UTC (GMT)
The number 0 - 011 1111 1010 - 0101 0011 1111 1110 0110 0111 1110 1111 1001 1101 1011 0011 1111 converted from 64 bit double precision IEEE 754 binary floating point system and written as a decimal number (double) written in base ten = ? Jul 13 13:59 UTC (GMT)
All 64 bit double precision IEEE 754 binary floating point representation numbers converted to base ten decimal numbers (double)

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)