-0.000 000 000 000 176 557 375 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 375(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
-0.000 000 000 000 176 557 375(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 000 000 000 176 557 375| = 0.000 000 000 000 176 557 375


2. 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;

3. 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)


4. Convert to binary (base 2) the fractional part: 0.000 000 000 000 176 557 375.

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) 0.000 000 000 000 176 557 375 × 2 = 0 + 0.000 000 000 000 353 114 75;
  • 2) 0.000 000 000 000 353 114 75 × 2 = 0 + 0.000 000 000 000 706 229 5;
  • 3) 0.000 000 000 000 706 229 5 × 2 = 0 + 0.000 000 000 001 412 459;
  • 4) 0.000 000 000 001 412 459 × 2 = 0 + 0.000 000 000 002 824 918;
  • 5) 0.000 000 000 002 824 918 × 2 = 0 + 0.000 000 000 005 649 836;
  • 6) 0.000 000 000 005 649 836 × 2 = 0 + 0.000 000 000 011 299 672;
  • 7) 0.000 000 000 011 299 672 × 2 = 0 + 0.000 000 000 022 599 344;
  • 8) 0.000 000 000 022 599 344 × 2 = 0 + 0.000 000 000 045 198 688;
  • 9) 0.000 000 000 045 198 688 × 2 = 0 + 0.000 000 000 090 397 376;
  • 10) 0.000 000 000 090 397 376 × 2 = 0 + 0.000 000 000 180 794 752;
  • 11) 0.000 000 000 180 794 752 × 2 = 0 + 0.000 000 000 361 589 504;
  • 12) 0.000 000 000 361 589 504 × 2 = 0 + 0.000 000 000 723 179 008;
  • 13) 0.000 000 000 723 179 008 × 2 = 0 + 0.000 000 001 446 358 016;
  • 14) 0.000 000 001 446 358 016 × 2 = 0 + 0.000 000 002 892 716 032;
  • 15) 0.000 000 002 892 716 032 × 2 = 0 + 0.000 000 005 785 432 064;
  • 16) 0.000 000 005 785 432 064 × 2 = 0 + 0.000 000 011 570 864 128;
  • 17) 0.000 000 011 570 864 128 × 2 = 0 + 0.000 000 023 141 728 256;
  • 18) 0.000 000 023 141 728 256 × 2 = 0 + 0.000 000 046 283 456 512;
  • 19) 0.000 000 046 283 456 512 × 2 = 0 + 0.000 000 092 566 913 024;
  • 20) 0.000 000 092 566 913 024 × 2 = 0 + 0.000 000 185 133 826 048;
  • 21) 0.000 000 185 133 826 048 × 2 = 0 + 0.000 000 370 267 652 096;
  • 22) 0.000 000 370 267 652 096 × 2 = 0 + 0.000 000 740 535 304 192;
  • 23) 0.000 000 740 535 304 192 × 2 = 0 + 0.000 001 481 070 608 384;
  • 24) 0.000 001 481 070 608 384 × 2 = 0 + 0.000 002 962 141 216 768;
  • 25) 0.000 002 962 141 216 768 × 2 = 0 + 0.000 005 924 282 433 536;
  • 26) 0.000 005 924 282 433 536 × 2 = 0 + 0.000 011 848 564 867 072;
  • 27) 0.000 011 848 564 867 072 × 2 = 0 + 0.000 023 697 129 734 144;
  • 28) 0.000 023 697 129 734 144 × 2 = 0 + 0.000 047 394 259 468 288;
  • 29) 0.000 047 394 259 468 288 × 2 = 0 + 0.000 094 788 518 936 576;
  • 30) 0.000 094 788 518 936 576 × 2 = 0 + 0.000 189 577 037 873 152;
  • 31) 0.000 189 577 037 873 152 × 2 = 0 + 0.000 379 154 075 746 304;
  • 32) 0.000 379 154 075 746 304 × 2 = 0 + 0.000 758 308 151 492 608;
  • 33) 0.000 758 308 151 492 608 × 2 = 0 + 0.001 516 616 302 985 216;
  • 34) 0.001 516 616 302 985 216 × 2 = 0 + 0.003 033 232 605 970 432;
  • 35) 0.003 033 232 605 970 432 × 2 = 0 + 0.006 066 465 211 940 864;
  • 36) 0.006 066 465 211 940 864 × 2 = 0 + 0.012 132 930 423 881 728;
  • 37) 0.012 132 930 423 881 728 × 2 = 0 + 0.024 265 860 847 763 456;
  • 38) 0.024 265 860 847 763 456 × 2 = 0 + 0.048 531 721 695 526 912;
  • 39) 0.048 531 721 695 526 912 × 2 = 0 + 0.097 063 443 391 053 824;
  • 40) 0.097 063 443 391 053 824 × 2 = 0 + 0.194 126 886 782 107 648;
  • 41) 0.194 126 886 782 107 648 × 2 = 0 + 0.388 253 773 564 215 296;
  • 42) 0.388 253 773 564 215 296 × 2 = 0 + 0.776 507 547 128 430 592;
  • 43) 0.776 507 547 128 430 592 × 2 = 1 + 0.553 015 094 256 861 184;
  • 44) 0.553 015 094 256 861 184 × 2 = 1 + 0.106 030 188 513 722 368;
  • 45) 0.106 030 188 513 722 368 × 2 = 0 + 0.212 060 377 027 444 736;
  • 46) 0.212 060 377 027 444 736 × 2 = 0 + 0.424 120 754 054 889 472;
  • 47) 0.424 120 754 054 889 472 × 2 = 0 + 0.848 241 508 109 778 944;
  • 48) 0.848 241 508 109 778 944 × 2 = 1 + 0.696 483 016 219 557 888;
  • 49) 0.696 483 016 219 557 888 × 2 = 1 + 0.392 966 032 439 115 776;
  • 50) 0.392 966 032 439 115 776 × 2 = 0 + 0.785 932 064 878 231 552;
  • 51) 0.785 932 064 878 231 552 × 2 = 1 + 0.571 864 129 756 463 104;
  • 52) 0.571 864 129 756 463 104 × 2 = 1 + 0.143 728 259 512 926 208;
  • 53) 0.143 728 259 512 926 208 × 2 = 0 + 0.287 456 519 025 852 416;
  • 54) 0.287 456 519 025 852 416 × 2 = 0 + 0.574 913 038 051 704 832;
  • 55) 0.574 913 038 051 704 832 × 2 = 1 + 0.149 826 076 103 409 664;
  • 56) 0.149 826 076 103 409 664 × 2 = 0 + 0.299 652 152 206 819 328;
  • 57) 0.299 652 152 206 819 328 × 2 = 0 + 0.599 304 304 413 638 656;
  • 58) 0.599 304 304 413 638 656 × 2 = 1 + 0.198 608 608 827 277 312;
  • 59) 0.198 608 608 827 277 312 × 2 = 0 + 0.397 217 217 654 554 624;
  • 60) 0.397 217 217 654 554 624 × 2 = 0 + 0.794 434 435 309 109 248;
  • 61) 0.794 434 435 309 109 248 × 2 = 1 + 0.588 868 870 618 218 496;
  • 62) 0.588 868 870 618 218 496 × 2 = 1 + 0.177 737 741 236 436 992;
  • 63) 0.177 737 741 236 436 992 × 2 = 0 + 0.355 475 482 472 873 984;
  • 64) 0.355 475 482 472 873 984 × 2 = 0 + 0.710 950 964 945 747 968;
  • 65) 0.710 950 964 945 747 968 × 2 = 1 + 0.421 901 929 891 495 936;
  • 66) 0.421 901 929 891 495 936 × 2 = 0 + 0.843 803 859 782 991 872;
  • 67) 0.843 803 859 782 991 872 × 2 = 1 + 0.687 607 719 565 983 744;
  • 68) 0.687 607 719 565 983 744 × 2 = 1 + 0.375 215 439 131 967 488;
  • 69) 0.375 215 439 131 967 488 × 2 = 0 + 0.750 430 878 263 934 976;
  • 70) 0.750 430 878 263 934 976 × 2 = 1 + 0.500 861 756 527 869 952;
  • 71) 0.500 861 756 527 869 952 × 2 = 1 + 0.001 723 513 055 739 904;
  • 72) 0.001 723 513 055 739 904 × 2 = 0 + 0.003 447 026 111 479 808;
  • 73) 0.003 447 026 111 479 808 × 2 = 0 + 0.006 894 052 222 959 616;
  • 74) 0.006 894 052 222 959 616 × 2 = 0 + 0.013 788 104 445 919 232;
  • 75) 0.013 788 104 445 919 232 × 2 = 0 + 0.027 576 208 891 838 464;
  • 76) 0.027 576 208 891 838 464 × 2 = 0 + 0.055 152 417 783 676 928;
  • 77) 0.055 152 417 783 676 928 × 2 = 0 + 0.110 304 835 567 353 856;
  • 78) 0.110 304 835 567 353 856 × 2 = 0 + 0.220 609 671 134 707 712;
  • 79) 0.220 609 671 134 707 712 × 2 = 0 + 0.441 219 342 269 415 424;
  • 80) 0.441 219 342 269 415 424 × 2 = 0 + 0.882 438 684 538 830 848;
  • 81) 0.882 438 684 538 830 848 × 2 = 1 + 0.764 877 369 077 661 696;
  • 82) 0.764 877 369 077 661 696 × 2 = 1 + 0.529 754 738 155 323 392;
  • 83) 0.529 754 738 155 323 392 × 2 = 1 + 0.059 509 476 310 646 784;
  • 84) 0.059 509 476 310 646 784 × 2 = 0 + 0.119 018 952 621 293 568;
  • 85) 0.119 018 952 621 293 568 × 2 = 0 + 0.238 037 905 242 587 136;
  • 86) 0.238 037 905 242 587 136 × 2 = 0 + 0.476 075 810 485 174 272;
  • 87) 0.476 075 810 485 174 272 × 2 = 0 + 0.952 151 620 970 348 544;
  • 88) 0.952 151 620 970 348 544 × 2 = 1 + 0.904 303 241 940 697 088;
  • 89) 0.904 303 241 940 697 088 × 2 = 1 + 0.808 606 483 881 394 176;
  • 90) 0.808 606 483 881 394 176 × 2 = 1 + 0.617 212 967 762 788 352;
  • 91) 0.617 212 967 762 788 352 × 2 = 1 + 0.234 425 935 525 576 704;
  • 92) 0.234 425 935 525 576 704 × 2 = 0 + 0.468 851 871 051 153 408;
  • 93) 0.468 851 871 051 153 408 × 2 = 0 + 0.937 703 742 102 306 816;
  • 94) 0.937 703 742 102 306 816 × 2 = 1 + 0.875 407 484 204 613 632;
  • 95) 0.875 407 484 204 613 632 × 2 = 1 + 0.750 814 968 409 227 264;

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 - the converted number we get in the end will be just a very good approximation of the initial one).


5. 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:


0.000 000 000 000 176 557 375(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1100 1011 0110 0000 0000 1110 0001 1110 011(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 375(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1100 1011 0110 0000 0000 1110 0001 1110 011(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 43 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 176 557 375(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1100 1011 0110 0000 0000 1110 0001 1110 011(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1100 1011 0110 0000 0000 1110 0001 1110 011(2) × 20 =

1.1000 1101 1001 0010 0110 0101 1011 0000 0000 0111 0000 1111 0011(2) × 2-43


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): -43

Mantissa (not normalized):
1.1000 1101 1001 0010 0110 0101 1011 0000 0000 0111 0000 1111 0011


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-43 + 2(11-1) - 1 =

(-43 + 1 023)(10) =

980(10)


10. 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;
  • 980 ÷ 2 = 490 + 0;
  • 490 ÷ 2 = 245 + 0;
  • 245 ÷ 2 = 122 + 1;
  • 122 ÷ 2 = 61 + 0;
  • 61 ÷ 2 = 30 + 1;
  • 30 ÷ 2 = 15 + 0;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

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

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


Exponent (adjusted) =

980(10) =

011 1101 0100(2)


12. 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. 1000 1101 1001 0010 0110 0101 1011 0000 0000 0111 0000 1111 0011 =

1000 1101 1001 0010 0110 0101 1011 0000 0000 0111 0000 1111 0011


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

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

Exponent (11 bits) =
011 1101 0100

Mantissa (52 bits) =
1000 1101 1001 0010 0110 0101 1011 0000 0000 0111 0000 1111 0011


Decimal number -0.000 000 000 000 176 557 375 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 0110 0101 1011 0000 0000 0111 0000 1111 0011

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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