64 bit double precision IEEE 754 binary floating point number 1 - 100 0000 1001 - 1000 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:
1 - 100 0000 1001 - 1000 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 1001


The last 52 bits contain the mantissa:
1000 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 1001(2) =


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


1,024 + 0 + 0 + 0 + 0 + 0 + 0 + 8 + 0 + 0 + 1 =


1,024 + 8 + 1 =


1,033(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 = 1,033 - 1023 = 10

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 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000(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 + 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 =


0.5 + 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 =


0.5 =


0.5(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)1 × (1 + 0.5) × 210 =


-1.5 × 210 =


-1 536

1 - 100 0000 1001 - 1000 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) =


-1 536(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)

1 - 100 0000 1001 - 1000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 = -1 536 Oct 20 19:10 UTC (GMT)
1 - 101 0101 0101 - 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 = -14 931 631 614 518 693 410 854 318 111 025 249 439 791 641 590 042 089 964 783 525 213 498 032 633 297 134 587 447 191 255 055 046 017 024 Oct 20 19:07 UTC (GMT)
0 - 100 0000 1110 - 0010 1100 0001 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 = 38 414.399 999 999 994 179 233 908 653 259 277 343 75 Oct 20 19:06 UTC (GMT)
1 - 111 1111 1000 - 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 = -1 404 447 761 611 184 302 913 519 680 303 925 573 139 044 514 798 677 009 948 672 509 044 786 529 730 476 274 474 284 979 081 308 875 165 000 889 686 495 260 606 709 295 068 862 629 863 225 370 551 870 891 596 701 311 667 381 761 603 721 111 090 634 735 110 308 227 210 563 164 107 569 048 052 205 800 491 261 514 946 176 100 212 790 338 675 864 723 330 454 999 587 858 894 372 783 631 526 221 325 189 251 072 Oct 20 19:06 UTC (GMT)
0 - 100 1000 0000 - 0000 0001 1100 1001 0011 1000 0000 0000 0000 0000 0000 0000 0000 = 685 312 764 800 453 368 366 313 570 600 587 624 448 Oct 20 19:03 UTC (GMT)
0 - 100 0001 1001 - 0000 1001 0111 1100 0110 1000 1111 0010 1010 0101 0101 1000 1110 = 69 595 555.791 341 990 232 467 651 367 187 5 Oct 20 19:02 UTC (GMT)
1 - 001 1000 0000 - 0000 0001 1100 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 = -0.000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 441 358 484 146 028 033 981 450 506 535 643 667 664 524 178 224 586 280 975 401 435 791 156 993 868 421 155 304 813 603 810 027 612 857 098 684 506 064 Oct 20 19:01 UTC (GMT)
0 - 100 0000 0011 - 1011 1001 0001 1000 0000 0000 0000 0000 0000 0000 0000 0000 0000 = 27.568 359 375 Oct 20 18:58 UTC (GMT)
0 - 100 0000 0000 - 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 1010 = 3.333 333 333 333 333 037 273 860 099 958 255 887 031 555 175 781 25 Oct 20 18:58 UTC (GMT)
0 - 100 0000 0111 - 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 = 256 Oct 20 18:54 UTC (GMT)
0 - 111 1111 1111 - 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 = QNaN, Quiet Not a Number Oct 20 18:54 UTC (GMT)
0 - 000 0000 0000 - 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 = 0 Oct 20 18:53 UTC (GMT)
0 - 101 0111 1000 - 1110 0111 1110 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 = 586 647 148 836 241 349 660 969 137 233 202 074 742 226 495 256 907 037 143 715 626 360 973 556 801 428 272 309 684 300 204 528 143 867 753 101 524 992 Oct 20 18:50 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)