64 bit double precision IEEE 754 binary floating point number 0 - 100 0000 0000 - 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 converted to decimal base ten (double)

How to convert 64 bit double precision IEEE 754 binary floating point:
0 - 100 0000 0000 - 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000.

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

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


The next 11 bits contain the exponent:
100 0000 0000


The last 52 bits contain the mantissa:
0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000

2. Convert the exponent, that is allways a positive integer, from binary (base 2) to decimal (base 10):

100 0000 0000(2) =


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


1,024 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 =


1,024 =


1,024(10)

3. Adjust the exponent, subtract the excess bits, 2(11 - 1) - 1 = 1023, that is due to the 11 bit excess/bias notation:

Exponent adjusted = 1,024 - 1023 = 1

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):

0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000(2) =

0 × 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 + 0 × 2-11 + 0 × 2-12 + 0 × 2-13 + 0 × 2-14 + 0 × 2-15 + 0 × 2-16 + 0 × 2-17 + 0 × 2-18 + 0 × 2-19 + 0 × 2-20 + 0 × 2-21 + 0 × 2-22 + 0 × 2-23 + 0 × 2-24 + 0 × 2-25 + 0 × 2-26 + 0 × 2-27 + 0 × 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 + 0 × 2-48 + 0 × 2-49 + 0 × 2-50 + 0 × 2-51 + 0 × 2-52 =


0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 =


0(10)

Conclusion:

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

(-1)Sign × (1 + Mantissa) × 2(Exponent adjusted) =


(-1)0 × (1 + 0) × 21 =


1 × 21 =


2

0 - 100 0000 0000 - 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000
converted from
64 bit double precision IEEE 754 binary floating point
to
base ten decimal system (double) =


2(10)

Convert 64 bit double precision IEEE 754 floating point standard binary numbers to base ten decimal system (double)

64 bit double precision IEEE 754 binary floating point standard representation of numbers requires three building blocks: sign (it takes 1 bit and it's either 0 for positive or 1 for negative numbers), exponent (11 bits), mantissa (52 bits)

Latest 64 bit double precision IEEE 754 floating point binary standard numbers converted to decimal base ten (double)

0 - 100 0000 0000 - 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 = 2 Dec 12 00:24 UTC (GMT)
0 - 000 0000 0000 - 0111 1000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 = 0 Dec 12 00:23 UTC (GMT)
0 - 100 0001 1011 - 1000 0010 1101 1010 1101 1010 1101 1000 1111 0000 0011 0111 1011 = 405 646 765.558 646 857 738 494 873 046 875 Dec 12 00:21 UTC (GMT)
0 - 100 0000 0011 - 1011 1001 0001 1000 0000 0000 0000 0000 0000 0000 0000 0000 0000 = 27.568 359 375 Dec 12 00:20 UTC (GMT)
0 - 100 0001 1001 - 1111 1000 1001 0000 1101 0011 0111 1001 0110 1010 1100 1101 1001 = 132 268 877.897 143 736 481 666 564 941 406 25 Dec 12 00:20 UTC (GMT)
0 - 100 0000 0011 - 1011 1001 0001 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 = 27.566 406 250 000 014 210 854 715 202 003 717 422 485 351 562 5 Dec 12 00:19 UTC (GMT)
1 - 100 0000 0100 - 1111 1010 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 = -63.25 Dec 12 00:18 UTC (GMT)
1 - 100 0100 0010 - 1000 0010 0101 0011 1010 1010 0111 1101 0101 0111 1101 1010 1000 = -222 702 249 416 229 912 576 Dec 12 00:16 UTC (GMT)
0 - 100 0001 1001 - 0000 0101 1111 0000 0110 1111 0011 0001 0011 0100 1011 0110 0110 = 68 665 788.768 842 309 713 363 647 460 937 5 Dec 12 00:14 UTC (GMT)
0 - 100 1010 1000 - 0110 1100 0011 0111 1010 1000 1010 0000 0000 0000 0000 0000 0000 = 1 064 608 701 888 006 151 252 602 686 964 999 266 829 445 191 696 384 Dec 12 00:13 UTC (GMT)
1 - 100 0000 0110 - 0100 0000 0110 1010 0000 0101 1001 1010 0111 0011 1011 0100 0011 = -160.207 074 000 000 005 753 463 483 415 544 033 050 537 109 375 Dec 12 00:09 UTC (GMT)
0 - 100 0001 0001 - 1000 1011 0110 1010 1111 0100 0001 1000 0100 0010 1001 1101 0000 = 404 907.813 980 725 593 864 917 755 126 953 125 Dec 12 00:08 UTC (GMT)
0 - 000 0000 0000 - 1101 1100 1000 0100 0110 1011 0101 1011 0000 0000 0000 0000 0000 = 0 Dec 12 00:06 UTC (GMT)
All base ten decimal numbers converted to 64 bit double precision IEEE 754 binary floating point

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)