64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 0.234 234 234 234 234 234 234 234 234 234 238 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 0.234 234 234 234 234 234 234 234 234 234 238(10) converted and written in 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

2. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.


0(10) =

0(2)


3. Convert to binary (base 2) the fractional part: 0.234 234 234 234 234 234 234 234 234 234 238.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.234 234 234 234 234 234 234 234 234 234 238 × 2 = 0 + 0.468 468 468 468 468 468 468 468 468 468 476;
  • 2) 0.468 468 468 468 468 468 468 468 468 468 476 × 2 = 0 + 0.936 936 936 936 936 936 936 936 936 936 952;
  • 3) 0.936 936 936 936 936 936 936 936 936 936 952 × 2 = 1 + 0.873 873 873 873 873 873 873 873 873 873 904;
  • 4) 0.873 873 873 873 873 873 873 873 873 873 904 × 2 = 1 + 0.747 747 747 747 747 747 747 747 747 747 808;
  • 5) 0.747 747 747 747 747 747 747 747 747 747 808 × 2 = 1 + 0.495 495 495 495 495 495 495 495 495 495 616;
  • 6) 0.495 495 495 495 495 495 495 495 495 495 616 × 2 = 0 + 0.990 990 990 990 990 990 990 990 990 991 232;
  • 7) 0.990 990 990 990 990 990 990 990 990 991 232 × 2 = 1 + 0.981 981 981 981 981 981 981 981 981 982 464;
  • 8) 0.981 981 981 981 981 981 981 981 981 982 464 × 2 = 1 + 0.963 963 963 963 963 963 963 963 963 964 928;
  • 9) 0.963 963 963 963 963 963 963 963 963 964 928 × 2 = 1 + 0.927 927 927 927 927 927 927 927 927 929 856;
  • 10) 0.927 927 927 927 927 927 927 927 927 929 856 × 2 = 1 + 0.855 855 855 855 855 855 855 855 855 859 712;
  • 11) 0.855 855 855 855 855 855 855 855 855 859 712 × 2 = 1 + 0.711 711 711 711 711 711 711 711 711 719 424;
  • 12) 0.711 711 711 711 711 711 711 711 711 719 424 × 2 = 1 + 0.423 423 423 423 423 423 423 423 423 438 848;
  • 13) 0.423 423 423 423 423 423 423 423 423 438 848 × 2 = 0 + 0.846 846 846 846 846 846 846 846 846 877 696;
  • 14) 0.846 846 846 846 846 846 846 846 846 877 696 × 2 = 1 + 0.693 693 693 693 693 693 693 693 693 755 392;
  • 15) 0.693 693 693 693 693 693 693 693 693 755 392 × 2 = 1 + 0.387 387 387 387 387 387 387 387 387 510 784;
  • 16) 0.387 387 387 387 387 387 387 387 387 510 784 × 2 = 0 + 0.774 774 774 774 774 774 774 774 775 021 568;
  • 17) 0.774 774 774 774 774 774 774 774 775 021 568 × 2 = 1 + 0.549 549 549 549 549 549 549 549 550 043 136;
  • 18) 0.549 549 549 549 549 549 549 549 550 043 136 × 2 = 1 + 0.099 099 099 099 099 099 099 099 100 086 272;
  • 19) 0.099 099 099 099 099 099 099 099 100 086 272 × 2 = 0 + 0.198 198 198 198 198 198 198 198 200 172 544;
  • 20) 0.198 198 198 198 198 198 198 198 200 172 544 × 2 = 0 + 0.396 396 396 396 396 396 396 396 400 345 088;
  • 21) 0.396 396 396 396 396 396 396 396 400 345 088 × 2 = 0 + 0.792 792 792 792 792 792 792 792 800 690 176;
  • 22) 0.792 792 792 792 792 792 792 792 800 690 176 × 2 = 1 + 0.585 585 585 585 585 585 585 585 601 380 352;
  • 23) 0.585 585 585 585 585 585 585 585 601 380 352 × 2 = 1 + 0.171 171 171 171 171 171 171 171 202 760 704;
  • 24) 0.171 171 171 171 171 171 171 171 202 760 704 × 2 = 0 + 0.342 342 342 342 342 342 342 342 405 521 408;
  • 25) 0.342 342 342 342 342 342 342 342 405 521 408 × 2 = 0 + 0.684 684 684 684 684 684 684 684 811 042 816;
  • 26) 0.684 684 684 684 684 684 684 684 811 042 816 × 2 = 1 + 0.369 369 369 369 369 369 369 369 622 085 632;
  • 27) 0.369 369 369 369 369 369 369 369 622 085 632 × 2 = 0 + 0.738 738 738 738 738 738 738 739 244 171 264;
  • 28) 0.738 738 738 738 738 738 738 739 244 171 264 × 2 = 1 + 0.477 477 477 477 477 477 477 478 488 342 528;
  • 29) 0.477 477 477 477 477 477 477 478 488 342 528 × 2 = 0 + 0.954 954 954 954 954 954 954 956 976 685 056;
  • 30) 0.954 954 954 954 954 954 954 956 976 685 056 × 2 = 1 + 0.909 909 909 909 909 909 909 913 953 370 112;
  • 31) 0.909 909 909 909 909 909 909 913 953 370 112 × 2 = 1 + 0.819 819 819 819 819 819 819 827 906 740 224;
  • 32) 0.819 819 819 819 819 819 819 827 906 740 224 × 2 = 1 + 0.639 639 639 639 639 639 639 655 813 480 448;
  • 33) 0.639 639 639 639 639 639 639 655 813 480 448 × 2 = 1 + 0.279 279 279 279 279 279 279 311 626 960 896;
  • 34) 0.279 279 279 279 279 279 279 311 626 960 896 × 2 = 0 + 0.558 558 558 558 558 558 558 623 253 921 792;
  • 35) 0.558 558 558 558 558 558 558 623 253 921 792 × 2 = 1 + 0.117 117 117 117 117 117 117 246 507 843 584;
  • 36) 0.117 117 117 117 117 117 117 246 507 843 584 × 2 = 0 + 0.234 234 234 234 234 234 234 493 015 687 168;
  • 37) 0.234 234 234 234 234 234 234 493 015 687 168 × 2 = 0 + 0.468 468 468 468 468 468 468 986 031 374 336;
  • 38) 0.468 468 468 468 468 468 468 986 031 374 336 × 2 = 0 + 0.936 936 936 936 936 936 937 972 062 748 672;
  • 39) 0.936 936 936 936 936 936 937 972 062 748 672 × 2 = 1 + 0.873 873 873 873 873 873 875 944 125 497 344;
  • 40) 0.873 873 873 873 873 873 875 944 125 497 344 × 2 = 1 + 0.747 747 747 747 747 747 751 888 250 994 688;
  • 41) 0.747 747 747 747 747 747 751 888 250 994 688 × 2 = 1 + 0.495 495 495 495 495 495 503 776 501 989 376;
  • 42) 0.495 495 495 495 495 495 503 776 501 989 376 × 2 = 0 + 0.990 990 990 990 990 991 007 553 003 978 752;
  • 43) 0.990 990 990 990 990 991 007 553 003 978 752 × 2 = 1 + 0.981 981 981 981 981 982 015 106 007 957 504;
  • 44) 0.981 981 981 981 981 982 015 106 007 957 504 × 2 = 1 + 0.963 963 963 963 963 964 030 212 015 915 008;
  • 45) 0.963 963 963 963 963 964 030 212 015 915 008 × 2 = 1 + 0.927 927 927 927 927 928 060 424 031 830 016;
  • 46) 0.927 927 927 927 927 928 060 424 031 830 016 × 2 = 1 + 0.855 855 855 855 855 856 120 848 063 660 032;
  • 47) 0.855 855 855 855 855 856 120 848 063 660 032 × 2 = 1 + 0.711 711 711 711 711 712 241 696 127 320 064;
  • 48) 0.711 711 711 711 711 712 241 696 127 320 064 × 2 = 1 + 0.423 423 423 423 423 424 483 392 254 640 128;
  • 49) 0.423 423 423 423 423 424 483 392 254 640 128 × 2 = 0 + 0.846 846 846 846 846 848 966 784 509 280 256;
  • 50) 0.846 846 846 846 846 848 966 784 509 280 256 × 2 = 1 + 0.693 693 693 693 693 697 933 569 018 560 512;
  • 51) 0.693 693 693 693 693 697 933 569 018 560 512 × 2 = 1 + 0.387 387 387 387 387 395 867 138 037 121 024;
  • 52) 0.387 387 387 387 387 395 867 138 037 121 024 × 2 = 0 + 0.774 774 774 774 774 791 734 276 074 242 048;
  • 53) 0.774 774 774 774 774 791 734 276 074 242 048 × 2 = 1 + 0.549 549 549 549 549 583 468 552 148 484 096;
  • 54) 0.549 549 549 549 549 583 468 552 148 484 096 × 2 = 1 + 0.099 099 099 099 099 166 937 104 296 968 192;
  • 55) 0.099 099 099 099 099 166 937 104 296 968 192 × 2 = 0 + 0.198 198 198 198 198 333 874 208 593 936 384;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (losing precision...)


4. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.234 234 234 234 234 234 234 234 234 234 238(10) =

0.0011 1011 1111 0110 1100 0110 0101 0111 1010 0011 1011 1111 0110 110(2)


5. Positive number before normalization:

0.234 234 234 234 234 234 234 234 234 234 238(10) =

0.0011 1011 1111 0110 1100 0110 0101 0111 1010 0011 1011 1111 0110 110(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 3 positions to the right, so that only one non zero digit remains to the left of it:


0.234 234 234 234 234 234 234 234 234 234 238(10) =

0.0011 1011 1111 0110 1100 0110 0101 0111 1010 0011 1011 1111 0110 110(2) =

0.0011 1011 1111 0110 1100 0110 0101 0111 1010 0011 1011 1111 0110 110(2) × 20 =

1.1101 1111 1011 0110 0011 0010 1011 1101 0001 1101 1111 1011 0110(2) × 2-3


7. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign 0 (a positive number)

Exponent (unadjusted): -3

Mantissa (not normalized):
1.1101 1111 1011 0110 0011 0010 1011 1101 0001 1101 1111 1011 0110


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-3 + 2(11-1) - 1 =

(-3 + 1 023)(10) =

1 020(10)


9. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 1 020 ÷ 2 = 510 + 0;
  • 510 ÷ 2 = 255 + 0;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1020(10) =

011 1111 1100(2)


11. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 52 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 1101 1111 1011 0110 0011 0010 1011 1101 0001 1101 1111 1011 0110 =

1101 1111 1011 0110 0011 0010 1011 1101 0001 1101 1111 1011 0110


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
011 1111 1100

Mantissa (52 bits) =
1101 1111 1011 0110 0011 0010 1011 1101 0001 1101 1111 1011 0110


The base ten decimal number 0.234 234 234 234 234 234 234 234 234 234 238 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 011 1111 1100 - 1101 1111 1011 0110 0011 0010 1011 1101 0001 1101 1111 1011 0110

More operations with decimal numbers converted to 64 bit double precision IEEE 754 binary floating point representation:

» Number 0.234 234 234 234 234 234 234 234 234 234 237 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point representation = ?

» Number 0.234 234 234 234 234 234 234 234 234 234 239 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point representation = ?

The latest decimal numbers converted from base ten to 64 bit double precision IEEE 754 floating point binary standard representation

Number 3 599 973 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard May 03 09:10 UTC (GMT)
Number 144 728 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard May 03 09:10 UTC (GMT)
Number 1 234 567 890 123 456 788 999 999 960 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard May 03 09:10 UTC (GMT)
Number -22.313 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard May 03 09:10 UTC (GMT)
Number 0.000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 002 321 432 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard May 03 09:10 UTC (GMT)
Number 120 359 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard May 03 09:09 UTC (GMT)
Number 101 205 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard May 03 09:09 UTC (GMT)
Number 170 420 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard May 03 09:09 UTC (GMT)
Number -8 360 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard May 03 09:09 UTC (GMT)
Number 101 966 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard May 03 09:09 UTC (GMT)
All base ten decimal numbers converted to 64 bit double precision IEEE 754 binary floating point

How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

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

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders from the previous operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the list constructed above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100