64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 0.000 000 000 003 637 977 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 0.000 000 000 003 637 977(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: 0.000 000 000 003 637 977.

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) 0.000 000 000 003 637 977 × 2 = 0 + 0.000 000 000 007 275 954;
  • 2) 0.000 000 000 007 275 954 × 2 = 0 + 0.000 000 000 014 551 908;
  • 3) 0.000 000 000 014 551 908 × 2 = 0 + 0.000 000 000 029 103 816;
  • 4) 0.000 000 000 029 103 816 × 2 = 0 + 0.000 000 000 058 207 632;
  • 5) 0.000 000 000 058 207 632 × 2 = 0 + 0.000 000 000 116 415 264;
  • 6) 0.000 000 000 116 415 264 × 2 = 0 + 0.000 000 000 232 830 528;
  • 7) 0.000 000 000 232 830 528 × 2 = 0 + 0.000 000 000 465 661 056;
  • 8) 0.000 000 000 465 661 056 × 2 = 0 + 0.000 000 000 931 322 112;
  • 9) 0.000 000 000 931 322 112 × 2 = 0 + 0.000 000 001 862 644 224;
  • 10) 0.000 000 001 862 644 224 × 2 = 0 + 0.000 000 003 725 288 448;
  • 11) 0.000 000 003 725 288 448 × 2 = 0 + 0.000 000 007 450 576 896;
  • 12) 0.000 000 007 450 576 896 × 2 = 0 + 0.000 000 014 901 153 792;
  • 13) 0.000 000 014 901 153 792 × 2 = 0 + 0.000 000 029 802 307 584;
  • 14) 0.000 000 029 802 307 584 × 2 = 0 + 0.000 000 059 604 615 168;
  • 15) 0.000 000 059 604 615 168 × 2 = 0 + 0.000 000 119 209 230 336;
  • 16) 0.000 000 119 209 230 336 × 2 = 0 + 0.000 000 238 418 460 672;
  • 17) 0.000 000 238 418 460 672 × 2 = 0 + 0.000 000 476 836 921 344;
  • 18) 0.000 000 476 836 921 344 × 2 = 0 + 0.000 000 953 673 842 688;
  • 19) 0.000 000 953 673 842 688 × 2 = 0 + 0.000 001 907 347 685 376;
  • 20) 0.000 001 907 347 685 376 × 2 = 0 + 0.000 003 814 695 370 752;
  • 21) 0.000 003 814 695 370 752 × 2 = 0 + 0.000 007 629 390 741 504;
  • 22) 0.000 007 629 390 741 504 × 2 = 0 + 0.000 015 258 781 483 008;
  • 23) 0.000 015 258 781 483 008 × 2 = 0 + 0.000 030 517 562 966 016;
  • 24) 0.000 030 517 562 966 016 × 2 = 0 + 0.000 061 035 125 932 032;
  • 25) 0.000 061 035 125 932 032 × 2 = 0 + 0.000 122 070 251 864 064;
  • 26) 0.000 122 070 251 864 064 × 2 = 0 + 0.000 244 140 503 728 128;
  • 27) 0.000 244 140 503 728 128 × 2 = 0 + 0.000 488 281 007 456 256;
  • 28) 0.000 488 281 007 456 256 × 2 = 0 + 0.000 976 562 014 912 512;
  • 29) 0.000 976 562 014 912 512 × 2 = 0 + 0.001 953 124 029 825 024;
  • 30) 0.001 953 124 029 825 024 × 2 = 0 + 0.003 906 248 059 650 048;
  • 31) 0.003 906 248 059 650 048 × 2 = 0 + 0.007 812 496 119 300 096;
  • 32) 0.007 812 496 119 300 096 × 2 = 0 + 0.015 624 992 238 600 192;
  • 33) 0.015 624 992 238 600 192 × 2 = 0 + 0.031 249 984 477 200 384;
  • 34) 0.031 249 984 477 200 384 × 2 = 0 + 0.062 499 968 954 400 768;
  • 35) 0.062 499 968 954 400 768 × 2 = 0 + 0.124 999 937 908 801 536;
  • 36) 0.124 999 937 908 801 536 × 2 = 0 + 0.249 999 875 817 603 072;
  • 37) 0.249 999 875 817 603 072 × 2 = 0 + 0.499 999 751 635 206 144;
  • 38) 0.499 999 751 635 206 144 × 2 = 0 + 0.999 999 503 270 412 288;
  • 39) 0.999 999 503 270 412 288 × 2 = 1 + 0.999 999 006 540 824 576;
  • 40) 0.999 999 006 540 824 576 × 2 = 1 + 0.999 998 013 081 649 152;
  • 41) 0.999 998 013 081 649 152 × 2 = 1 + 0.999 996 026 163 298 304;
  • 42) 0.999 996 026 163 298 304 × 2 = 1 + 0.999 992 052 326 596 608;
  • 43) 0.999 992 052 326 596 608 × 2 = 1 + 0.999 984 104 653 193 216;
  • 44) 0.999 984 104 653 193 216 × 2 = 1 + 0.999 968 209 306 386 432;
  • 45) 0.999 968 209 306 386 432 × 2 = 1 + 0.999 936 418 612 772 864;
  • 46) 0.999 936 418 612 772 864 × 2 = 1 + 0.999 872 837 225 545 728;
  • 47) 0.999 872 837 225 545 728 × 2 = 1 + 0.999 745 674 451 091 456;
  • 48) 0.999 745 674 451 091 456 × 2 = 1 + 0.999 491 348 902 182 912;
  • 49) 0.999 491 348 902 182 912 × 2 = 1 + 0.998 982 697 804 365 824;
  • 50) 0.998 982 697 804 365 824 × 2 = 1 + 0.997 965 395 608 731 648;
  • 51) 0.997 965 395 608 731 648 × 2 = 1 + 0.995 930 791 217 463 296;
  • 52) 0.995 930 791 217 463 296 × 2 = 1 + 0.991 861 582 434 926 592;
  • 53) 0.991 861 582 434 926 592 × 2 = 1 + 0.983 723 164 869 853 184;
  • 54) 0.983 723 164 869 853 184 × 2 = 1 + 0.967 446 329 739 706 368;
  • 55) 0.967 446 329 739 706 368 × 2 = 1 + 0.934 892 659 479 412 736;
  • 56) 0.934 892 659 479 412 736 × 2 = 1 + 0.869 785 318 958 825 472;
  • 57) 0.869 785 318 958 825 472 × 2 = 1 + 0.739 570 637 917 650 944;
  • 58) 0.739 570 637 917 650 944 × 2 = 1 + 0.479 141 275 835 301 888;
  • 59) 0.479 141 275 835 301 888 × 2 = 0 + 0.958 282 551 670 603 776;
  • 60) 0.958 282 551 670 603 776 × 2 = 1 + 0.916 565 103 341 207 552;
  • 61) 0.916 565 103 341 207 552 × 2 = 1 + 0.833 130 206 682 415 104;
  • 62) 0.833 130 206 682 415 104 × 2 = 1 + 0.666 260 413 364 830 208;
  • 63) 0.666 260 413 364 830 208 × 2 = 1 + 0.332 520 826 729 660 416;
  • 64) 0.332 520 826 729 660 416 × 2 = 0 + 0.665 041 653 459 320 832;
  • 65) 0.665 041 653 459 320 832 × 2 = 1 + 0.330 083 306 918 641 664;
  • 66) 0.330 083 306 918 641 664 × 2 = 0 + 0.660 166 613 837 283 328;
  • 67) 0.660 166 613 837 283 328 × 2 = 1 + 0.320 333 227 674 566 656;
  • 68) 0.320 333 227 674 566 656 × 2 = 0 + 0.640 666 455 349 133 312;
  • 69) 0.640 666 455 349 133 312 × 2 = 1 + 0.281 332 910 698 266 624;
  • 70) 0.281 332 910 698 266 624 × 2 = 0 + 0.562 665 821 396 533 248;
  • 71) 0.562 665 821 396 533 248 × 2 = 1 + 0.125 331 642 793 066 496;
  • 72) 0.125 331 642 793 066 496 × 2 = 0 + 0.250 663 285 586 132 992;
  • 73) 0.250 663 285 586 132 992 × 2 = 0 + 0.501 326 571 172 265 984;
  • 74) 0.501 326 571 172 265 984 × 2 = 1 + 0.002 653 142 344 531 968;
  • 75) 0.002 653 142 344 531 968 × 2 = 0 + 0.005 306 284 689 063 936;
  • 76) 0.005 306 284 689 063 936 × 2 = 0 + 0.010 612 569 378 127 872;
  • 77) 0.010 612 569 378 127 872 × 2 = 0 + 0.021 225 138 756 255 744;
  • 78) 0.021 225 138 756 255 744 × 2 = 0 + 0.042 450 277 512 511 488;
  • 79) 0.042 450 277 512 511 488 × 2 = 0 + 0.084 900 555 025 022 976;
  • 80) 0.084 900 555 025 022 976 × 2 = 0 + 0.169 801 110 050 045 952;
  • 81) 0.169 801 110 050 045 952 × 2 = 0 + 0.339 602 220 100 091 904;
  • 82) 0.339 602 220 100 091 904 × 2 = 0 + 0.679 204 440 200 183 808;
  • 83) 0.679 204 440 200 183 808 × 2 = 1 + 0.358 408 880 400 367 616;
  • 84) 0.358 408 880 400 367 616 × 2 = 0 + 0.716 817 760 800 735 232;
  • 85) 0.716 817 760 800 735 232 × 2 = 1 + 0.433 635 521 601 470 464;
  • 86) 0.433 635 521 601 470 464 × 2 = 0 + 0.867 271 043 202 940 928;
  • 87) 0.867 271 043 202 940 928 × 2 = 1 + 0.734 542 086 405 881 856;
  • 88) 0.734 542 086 405 881 856 × 2 = 1 + 0.469 084 172 811 763 712;
  • 89) 0.469 084 172 811 763 712 × 2 = 0 + 0.938 168 345 623 527 424;
  • 90) 0.938 168 345 623 527 424 × 2 = 1 + 0.876 336 691 247 054 848;
  • 91) 0.876 336 691 247 054 848 × 2 = 1 + 0.752 673 382 494 109 696;

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:


0.000 000 000 003 637 977(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 1111 1111 1111 1111 1101 1110 1010 1010 0100 0000 0010 1011 011(2)


5. Positive number before normalization:

0.000 000 000 003 637 977(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 1111 1111 1111 1111 1101 1110 1010 1010 0100 0000 0010 1011 011(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 39 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 003 637 977(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 1111 1111 1111 1111 1101 1110 1010 1010 0100 0000 0010 1011 011(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 1111 1111 1111 1111 1101 1110 1010 1010 0100 0000 0010 1011 011(2) × 20 =

1.1111 1111 1111 1111 1110 1111 0101 0101 0010 0000 0001 0101 1011(2) × 2-39


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): -39

Mantissa (not normalized):
1.1111 1111 1111 1111 1110 1111 0101 0101 0010 0000 0001 0101 1011


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-39 + 2(11-1) - 1 =

(-39 + 1 023)(10) =

984(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;
  • 984 ÷ 2 = 492 + 0;
  • 492 ÷ 2 = 246 + 0;
  • 246 ÷ 2 = 123 + 0;
  • 123 ÷ 2 = 61 + 1;
  • 61 ÷ 2 = 30 + 1;
  • 30 ÷ 2 = 15 + 0;
  • 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) =

984(10) =

011 1101 1000(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. 1111 1111 1111 1111 1110 1111 0101 0101 0010 0000 0001 0101 1011 =

1111 1111 1111 1111 1110 1111 0101 0101 0010 0000 0001 0101 1011


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 1101 1000

Mantissa (52 bits) =
1111 1111 1111 1111 1110 1111 0101 0101 0010 0000 0001 0101 1011


The base ten decimal number 0.000 000 000 003 637 977 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 011 1101 1000 - 1111 1111 1111 1111 1110 1111 0101 0101 0010 0000 0001 0101 1011

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

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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