64 bit double precision IEEE 754 binary floating point number 0 - 000 0000 0101 - 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 converted to decimal base ten (double)

How to convert 64 bit double precision IEEE 754 binary floating point:
0 - 000 0000 0101 - 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001.

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:
000 0000 0101


The last 52 bits contain the mantissa:
0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001

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

000 0000 0101(2) =


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


0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 4 + 0 + 1 =


4 + 1 =


5(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 = 5 - 1023 = -1018

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 0001(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 + 1 × 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 + 0.000 000 000 000 000 222 044 604 925 031 308 084 726 333 618 164 062 5 =


0.000 000 000 000 000 222 044 604 925 031 308 084 726 333 618 164 062 5 =


0.000 000 000 000 000 222 044 604 925 031 308 084 726 333 618 164 062 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)0 × (1 + 0.000 000 000 000 000 222 044 604 925 031 308 084 726 333 618 164 062 5) × 2-1018 =


1.000 000 000 000 000 222 044 604 925 031 308 084 726 333 618 164 062 5 × 2-1018 =


0.000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 3

0 - 000 0000 0101 - 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001
converted from
64 bit double precision IEEE 754 binary floating point
to
base ten decimal system (double) =


0.000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 3(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 - 000 0000 0101 - 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 = 0.000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 3 May 20 04:23 UTC (GMT)
0 - 100 0001 1000 - 1001 1101 0100 0111 1011 0001 1000 0100 0000 1011 0011 0011 1010 = 54 169 443.031 591 847 538 948 059 082 031 25 May 20 04:22 UTC (GMT)
0 - 011 1101 1000 - 0101 1100 0010 1000 1111 0110 0000 0000 0000 0000 0000 0000 0000 = 0.000 000 000 002 473 825 614 843 216 946 781 012 666 178 867 220 878 601 074 218 75 May 20 04:17 UTC (GMT)
0 - 011 1111 1011 - 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 = 0.099 999 999 999 999 991 673 327 315 311 325 946 822 762 489 318 847 656 25 May 20 04:13 UTC (GMT)
1 - 100 0001 1111 - 1000 0000 0101 0011 1010 1010 0111 1101 0101 0111 1101 1010 1011 = -6 447 934 077.343 180 656 433 105 468 75 May 20 04:11 UTC (GMT)
1 - 100 1001 1000 - 1101 0010 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 = -20 784 294 525 030 540 767 911 289 430 108 272 923 573 747 712 May 20 04:11 UTC (GMT)
0 - 101 0001 1111 - 0100 1000 0110 0111 1110 0100 1101 0011 1110 1010 0000 0010 0010 = 637 983 800 427 785 974 806 369 964 000 476 768 484 450 855 262 159 104 551 416 935 437 007 798 656 940 940 197 888 May 20 04:08 UTC (GMT)
0 - 101 0110 0101 - 0101 0101 0100 0101 0011 0101 0010 0101 0001 0101 0000 0100 1000 = 782 703 057 459 310 695 076 594 449 605 207 922 339 929 801 346 732 656 759 580 479 187 340 370 042 035 954 241 518 837 135 406 804 146 061 312 May 20 04:04 UTC (GMT)
0 - 100 0000 0100 - 0000 0110 1110 0100 0101 1011 0110 1111 1110 1011 0001 0101 1100 = 32.861 502 527 581 905 042 097 787 372 767 925 262 451 171 875 May 20 04:04 UTC (GMT)
0 - 011 1111 0010 - 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 = 0.000 122 070 312 5 May 20 04:03 UTC (GMT)
0 - 010 0000 1000 - 0100 0001 0101 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 = 0.000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 047 929 162 123 875 328 584 073 565 858 493 678 572 440 550 199 225 864 804 312 344 688 766 846 472 056 847 757 096 093 889 435 749 171 351 913 940 422 816 746 964 993 543 921 941 601 221 949 216 736 166 134 2 May 20 04:01 UTC (GMT)
1 - 100 0011 0000 - 0010 0101 1000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 = -645 413 325 504 512 May 20 04:00 UTC (GMT)
0 - 100 0001 1010 - 0011 1111 1101 0011 0101 1001 1101 0011 1100 1100 1010 1000 0010 = 167 680 718.618 732 511 997 222 900 390 625 May 20 03:58 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)