0.009 234 567 810 987 234 567 098 341 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.009 234 567 810 987 234 567 098 341 5(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
0.009 234 567 810 987 234 567 098 341 5(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. 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: 0.009 234 567 810 987 234 567 098 341 5.

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) 0.009 234 567 810 987 234 567 098 341 5 × 2 = 0 + 0.018 469 135 621 974 469 134 196 683;
  • 2) 0.018 469 135 621 974 469 134 196 683 × 2 = 0 + 0.036 938 271 243 948 938 268 393 366;
  • 3) 0.036 938 271 243 948 938 268 393 366 × 2 = 0 + 0.073 876 542 487 897 876 536 786 732;
  • 4) 0.073 876 542 487 897 876 536 786 732 × 2 = 0 + 0.147 753 084 975 795 753 073 573 464;
  • 5) 0.147 753 084 975 795 753 073 573 464 × 2 = 0 + 0.295 506 169 951 591 506 147 146 928;
  • 6) 0.295 506 169 951 591 506 147 146 928 × 2 = 0 + 0.591 012 339 903 183 012 294 293 856;
  • 7) 0.591 012 339 903 183 012 294 293 856 × 2 = 1 + 0.182 024 679 806 366 024 588 587 712;
  • 8) 0.182 024 679 806 366 024 588 587 712 × 2 = 0 + 0.364 049 359 612 732 049 177 175 424;
  • 9) 0.364 049 359 612 732 049 177 175 424 × 2 = 0 + 0.728 098 719 225 464 098 354 350 848;
  • 10) 0.728 098 719 225 464 098 354 350 848 × 2 = 1 + 0.456 197 438 450 928 196 708 701 696;
  • 11) 0.456 197 438 450 928 196 708 701 696 × 2 = 0 + 0.912 394 876 901 856 393 417 403 392;
  • 12) 0.912 394 876 901 856 393 417 403 392 × 2 = 1 + 0.824 789 753 803 712 786 834 806 784;
  • 13) 0.824 789 753 803 712 786 834 806 784 × 2 = 1 + 0.649 579 507 607 425 573 669 613 568;
  • 14) 0.649 579 507 607 425 573 669 613 568 × 2 = 1 + 0.299 159 015 214 851 147 339 227 136;
  • 15) 0.299 159 015 214 851 147 339 227 136 × 2 = 0 + 0.598 318 030 429 702 294 678 454 272;
  • 16) 0.598 318 030 429 702 294 678 454 272 × 2 = 1 + 0.196 636 060 859 404 589 356 908 544;
  • 17) 0.196 636 060 859 404 589 356 908 544 × 2 = 0 + 0.393 272 121 718 809 178 713 817 088;
  • 18) 0.393 272 121 718 809 178 713 817 088 × 2 = 0 + 0.786 544 243 437 618 357 427 634 176;
  • 19) 0.786 544 243 437 618 357 427 634 176 × 2 = 1 + 0.573 088 486 875 236 714 855 268 352;
  • 20) 0.573 088 486 875 236 714 855 268 352 × 2 = 1 + 0.146 176 973 750 473 429 710 536 704;
  • 21) 0.146 176 973 750 473 429 710 536 704 × 2 = 0 + 0.292 353 947 500 946 859 421 073 408;
  • 22) 0.292 353 947 500 946 859 421 073 408 × 2 = 0 + 0.584 707 895 001 893 718 842 146 816;
  • 23) 0.584 707 895 001 893 718 842 146 816 × 2 = 1 + 0.169 415 790 003 787 437 684 293 632;
  • 24) 0.169 415 790 003 787 437 684 293 632 × 2 = 0 + 0.338 831 580 007 574 875 368 587 264;
  • 25) 0.338 831 580 007 574 875 368 587 264 × 2 = 0 + 0.677 663 160 015 149 750 737 174 528;
  • 26) 0.677 663 160 015 149 750 737 174 528 × 2 = 1 + 0.355 326 320 030 299 501 474 349 056;
  • 27) 0.355 326 320 030 299 501 474 349 056 × 2 = 0 + 0.710 652 640 060 599 002 948 698 112;
  • 28) 0.710 652 640 060 599 002 948 698 112 × 2 = 1 + 0.421 305 280 121 198 005 897 396 224;
  • 29) 0.421 305 280 121 198 005 897 396 224 × 2 = 0 + 0.842 610 560 242 396 011 794 792 448;
  • 30) 0.842 610 560 242 396 011 794 792 448 × 2 = 1 + 0.685 221 120 484 792 023 589 584 896;
  • 31) 0.685 221 120 484 792 023 589 584 896 × 2 = 1 + 0.370 442 240 969 584 047 179 169 792;
  • 32) 0.370 442 240 969 584 047 179 169 792 × 2 = 0 + 0.740 884 481 939 168 094 358 339 584;
  • 33) 0.740 884 481 939 168 094 358 339 584 × 2 = 1 + 0.481 768 963 878 336 188 716 679 168;
  • 34) 0.481 768 963 878 336 188 716 679 168 × 2 = 0 + 0.963 537 927 756 672 377 433 358 336;
  • 35) 0.963 537 927 756 672 377 433 358 336 × 2 = 1 + 0.927 075 855 513 344 754 866 716 672;
  • 36) 0.927 075 855 513 344 754 866 716 672 × 2 = 1 + 0.854 151 711 026 689 509 733 433 344;
  • 37) 0.854 151 711 026 689 509 733 433 344 × 2 = 1 + 0.708 303 422 053 379 019 466 866 688;
  • 38) 0.708 303 422 053 379 019 466 866 688 × 2 = 1 + 0.416 606 844 106 758 038 933 733 376;
  • 39) 0.416 606 844 106 758 038 933 733 376 × 2 = 0 + 0.833 213 688 213 516 077 867 466 752;
  • 40) 0.833 213 688 213 516 077 867 466 752 × 2 = 1 + 0.666 427 376 427 032 155 734 933 504;
  • 41) 0.666 427 376 427 032 155 734 933 504 × 2 = 1 + 0.332 854 752 854 064 311 469 867 008;
  • 42) 0.332 854 752 854 064 311 469 867 008 × 2 = 0 + 0.665 709 505 708 128 622 939 734 016;
  • 43) 0.665 709 505 708 128 622 939 734 016 × 2 = 1 + 0.331 419 011 416 257 245 879 468 032;
  • 44) 0.331 419 011 416 257 245 879 468 032 × 2 = 0 + 0.662 838 022 832 514 491 758 936 064;
  • 45) 0.662 838 022 832 514 491 758 936 064 × 2 = 1 + 0.325 676 045 665 028 983 517 872 128;
  • 46) 0.325 676 045 665 028 983 517 872 128 × 2 = 0 + 0.651 352 091 330 057 967 035 744 256;
  • 47) 0.651 352 091 330 057 967 035 744 256 × 2 = 1 + 0.302 704 182 660 115 934 071 488 512;
  • 48) 0.302 704 182 660 115 934 071 488 512 × 2 = 0 + 0.605 408 365 320 231 868 142 977 024;
  • 49) 0.605 408 365 320 231 868 142 977 024 × 2 = 1 + 0.210 816 730 640 463 736 285 954 048;
  • 50) 0.210 816 730 640 463 736 285 954 048 × 2 = 0 + 0.421 633 461 280 927 472 571 908 096;
  • 51) 0.421 633 461 280 927 472 571 908 096 × 2 = 0 + 0.843 266 922 561 854 945 143 816 192;
  • 52) 0.843 266 922 561 854 945 143 816 192 × 2 = 1 + 0.686 533 845 123 709 890 287 632 384;
  • 53) 0.686 533 845 123 709 890 287 632 384 × 2 = 1 + 0.373 067 690 247 419 780 575 264 768;
  • 54) 0.373 067 690 247 419 780 575 264 768 × 2 = 0 + 0.746 135 380 494 839 561 150 529 536;
  • 55) 0.746 135 380 494 839 561 150 529 536 × 2 = 1 + 0.492 270 760 989 679 122 301 059 072;
  • 56) 0.492 270 760 989 679 122 301 059 072 × 2 = 0 + 0.984 541 521 979 358 244 602 118 144;
  • 57) 0.984 541 521 979 358 244 602 118 144 × 2 = 1 + 0.969 083 043 958 716 489 204 236 288;
  • 58) 0.969 083 043 958 716 489 204 236 288 × 2 = 1 + 0.938 166 087 917 432 978 408 472 576;
  • 59) 0.938 166 087 917 432 978 408 472 576 × 2 = 1 + 0.876 332 175 834 865 956 816 945 152;

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).


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:


0.009 234 567 810 987 234 567 098 341 5(10) =

0.0000 0010 0101 1101 0011 0010 0101 0110 1011 1101 1010 1010 1001 1010 111(2)

5. Positive number before normalization:

0.009 234 567 810 987 234 567 098 341 5(10) =

0.0000 0010 0101 1101 0011 0010 0101 0110 1011 1101 1010 1010 1001 1010 111(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 7 positions to the right, so that only one non zero digit remains to the left of it:


0.009 234 567 810 987 234 567 098 341 5(10) =

0.0000 0010 0101 1101 0011 0010 0101 0110 1011 1101 1010 1010 1001 1010 111(2) =

0.0000 0010 0101 1101 0011 0010 0101 0110 1011 1101 1010 1010 1001 1010 111(2) × 20 =

1.0010 1110 1001 1001 0010 1011 0101 1110 1101 0101 0100 1101 0111(2) × 2-7


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): -7

Mantissa (not normalized):
1.0010 1110 1001 1001 0010 1011 0101 1110 1101 0101 0100 1101 0111


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-7 + 2(11-1) - 1 =

(-7 + 1 023)(10) =

1 016(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;
  • 1 016 ÷ 2 = 508 + 0;
  • 508 ÷ 2 = 254 + 0;
  • 254 ÷ 2 = 127 + 0;
  • 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) =

1016(10) =

011 1111 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. 0010 1110 1001 1001 0010 1011 0101 1110 1101 0101 0100 1101 0111 =

0010 1110 1001 1001 0010 1011 0101 1110 1101 0101 0100 1101 0111


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 1000

Mantissa (52 bits) =
0010 1110 1001 1001 0010 1011 0101 1110 1101 0101 0100 1101 0111


Decimal number 0.009 234 567 810 987 234 567 098 341 5 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1000 - 0010 1110 1001 1001 0010 1011 0101 1110 1101 0101 0100 1101 0111

» Convert decimal 0.009 234 567 810 987 234 567 098 344 2 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 10, 2026 [October]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: October - by our visitors

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