0.019 932 208 134 231 138 919 244 739 99 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.019 932 208 134 231 138 919 244 739 99(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
0.019 932 208 134 231 138 919 244 739 99(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: 0.019 932 208 134 231 138 919 244 739 99.

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) 0.019 932 208 134 231 138 919 244 739 99 × 2 = 0 + 0.039 864 416 268 462 277 838 489 479 98;
  • 2) 0.039 864 416 268 462 277 838 489 479 98 × 2 = 0 + 0.079 728 832 536 924 555 676 978 959 96;
  • 3) 0.079 728 832 536 924 555 676 978 959 96 × 2 = 0 + 0.159 457 665 073 849 111 353 957 919 92;
  • 4) 0.159 457 665 073 849 111 353 957 919 92 × 2 = 0 + 0.318 915 330 147 698 222 707 915 839 84;
  • 5) 0.318 915 330 147 698 222 707 915 839 84 × 2 = 0 + 0.637 830 660 295 396 445 415 831 679 68;
  • 6) 0.637 830 660 295 396 445 415 831 679 68 × 2 = 1 + 0.275 661 320 590 792 890 831 663 359 36;
  • 7) 0.275 661 320 590 792 890 831 663 359 36 × 2 = 0 + 0.551 322 641 181 585 781 663 326 718 72;
  • 8) 0.551 322 641 181 585 781 663 326 718 72 × 2 = 1 + 0.102 645 282 363 171 563 326 653 437 44;
  • 9) 0.102 645 282 363 171 563 326 653 437 44 × 2 = 0 + 0.205 290 564 726 343 126 653 306 874 88;
  • 10) 0.205 290 564 726 343 126 653 306 874 88 × 2 = 0 + 0.410 581 129 452 686 253 306 613 749 76;
  • 11) 0.410 581 129 452 686 253 306 613 749 76 × 2 = 0 + 0.821 162 258 905 372 506 613 227 499 52;
  • 12) 0.821 162 258 905 372 506 613 227 499 52 × 2 = 1 + 0.642 324 517 810 745 013 226 454 999 04;
  • 13) 0.642 324 517 810 745 013 226 454 999 04 × 2 = 1 + 0.284 649 035 621 490 026 452 909 998 08;
  • 14) 0.284 649 035 621 490 026 452 909 998 08 × 2 = 0 + 0.569 298 071 242 980 052 905 819 996 16;
  • 15) 0.569 298 071 242 980 052 905 819 996 16 × 2 = 1 + 0.138 596 142 485 960 105 811 639 992 32;
  • 16) 0.138 596 142 485 960 105 811 639 992 32 × 2 = 0 + 0.277 192 284 971 920 211 623 279 984 64;
  • 17) 0.277 192 284 971 920 211 623 279 984 64 × 2 = 0 + 0.554 384 569 943 840 423 246 559 969 28;
  • 18) 0.554 384 569 943 840 423 246 559 969 28 × 2 = 1 + 0.108 769 139 887 680 846 493 119 938 56;
  • 19) 0.108 769 139 887 680 846 493 119 938 56 × 2 = 0 + 0.217 538 279 775 361 692 986 239 877 12;
  • 20) 0.217 538 279 775 361 692 986 239 877 12 × 2 = 0 + 0.435 076 559 550 723 385 972 479 754 24;
  • 21) 0.435 076 559 550 723 385 972 479 754 24 × 2 = 0 + 0.870 153 119 101 446 771 944 959 508 48;
  • 22) 0.870 153 119 101 446 771 944 959 508 48 × 2 = 1 + 0.740 306 238 202 893 543 889 919 016 96;
  • 23) 0.740 306 238 202 893 543 889 919 016 96 × 2 = 1 + 0.480 612 476 405 787 087 779 838 033 92;
  • 24) 0.480 612 476 405 787 087 779 838 033 92 × 2 = 0 + 0.961 224 952 811 574 175 559 676 067 84;
  • 25) 0.961 224 952 811 574 175 559 676 067 84 × 2 = 1 + 0.922 449 905 623 148 351 119 352 135 68;
  • 26) 0.922 449 905 623 148 351 119 352 135 68 × 2 = 1 + 0.844 899 811 246 296 702 238 704 271 36;
  • 27) 0.844 899 811 246 296 702 238 704 271 36 × 2 = 1 + 0.689 799 622 492 593 404 477 408 542 72;
  • 28) 0.689 799 622 492 593 404 477 408 542 72 × 2 = 1 + 0.379 599 244 985 186 808 954 817 085 44;
  • 29) 0.379 599 244 985 186 808 954 817 085 44 × 2 = 0 + 0.759 198 489 970 373 617 909 634 170 88;
  • 30) 0.759 198 489 970 373 617 909 634 170 88 × 2 = 1 + 0.518 396 979 940 747 235 819 268 341 76;
  • 31) 0.518 396 979 940 747 235 819 268 341 76 × 2 = 1 + 0.036 793 959 881 494 471 638 536 683 52;
  • 32) 0.036 793 959 881 494 471 638 536 683 52 × 2 = 0 + 0.073 587 919 762 988 943 277 073 367 04;
  • 33) 0.073 587 919 762 988 943 277 073 367 04 × 2 = 0 + 0.147 175 839 525 977 886 554 146 734 08;
  • 34) 0.147 175 839 525 977 886 554 146 734 08 × 2 = 0 + 0.294 351 679 051 955 773 108 293 468 16;
  • 35) 0.294 351 679 051 955 773 108 293 468 16 × 2 = 0 + 0.588 703 358 103 911 546 216 586 936 32;
  • 36) 0.588 703 358 103 911 546 216 586 936 32 × 2 = 1 + 0.177 406 716 207 823 092 433 173 872 64;
  • 37) 0.177 406 716 207 823 092 433 173 872 64 × 2 = 0 + 0.354 813 432 415 646 184 866 347 745 28;
  • 38) 0.354 813 432 415 646 184 866 347 745 28 × 2 = 0 + 0.709 626 864 831 292 369 732 695 490 56;
  • 39) 0.709 626 864 831 292 369 732 695 490 56 × 2 = 1 + 0.419 253 729 662 584 739 465 390 981 12;
  • 40) 0.419 253 729 662 584 739 465 390 981 12 × 2 = 0 + 0.838 507 459 325 169 478 930 781 962 24;
  • 41) 0.838 507 459 325 169 478 930 781 962 24 × 2 = 1 + 0.677 014 918 650 338 957 861 563 924 48;
  • 42) 0.677 014 918 650 338 957 861 563 924 48 × 2 = 1 + 0.354 029 837 300 677 915 723 127 848 96;
  • 43) 0.354 029 837 300 677 915 723 127 848 96 × 2 = 0 + 0.708 059 674 601 355 831 446 255 697 92;
  • 44) 0.708 059 674 601 355 831 446 255 697 92 × 2 = 1 + 0.416 119 349 202 711 662 892 511 395 84;
  • 45) 0.416 119 349 202 711 662 892 511 395 84 × 2 = 0 + 0.832 238 698 405 423 325 785 022 791 68;
  • 46) 0.832 238 698 405 423 325 785 022 791 68 × 2 = 1 + 0.664 477 396 810 846 651 570 045 583 36;
  • 47) 0.664 477 396 810 846 651 570 045 583 36 × 2 = 1 + 0.328 954 793 621 693 303 140 091 166 72;
  • 48) 0.328 954 793 621 693 303 140 091 166 72 × 2 = 0 + 0.657 909 587 243 386 606 280 182 333 44;
  • 49) 0.657 909 587 243 386 606 280 182 333 44 × 2 = 1 + 0.315 819 174 486 773 212 560 364 666 88;
  • 50) 0.315 819 174 486 773 212 560 364 666 88 × 2 = 0 + 0.631 638 348 973 546 425 120 729 333 76;
  • 51) 0.631 638 348 973 546 425 120 729 333 76 × 2 = 1 + 0.263 276 697 947 092 850 241 458 667 52;
  • 52) 0.263 276 697 947 092 850 241 458 667 52 × 2 = 0 + 0.526 553 395 894 185 700 482 917 335 04;
  • 53) 0.526 553 395 894 185 700 482 917 335 04 × 2 = 1 + 0.053 106 791 788 371 400 965 834 670 08;
  • 54) 0.053 106 791 788 371 400 965 834 670 08 × 2 = 0 + 0.106 213 583 576 742 801 931 669 340 16;
  • 55) 0.106 213 583 576 742 801 931 669 340 16 × 2 = 0 + 0.212 427 167 153 485 603 863 338 680 32;
  • 56) 0.212 427 167 153 485 603 863 338 680 32 × 2 = 0 + 0.424 854 334 306 971 207 726 677 360 64;
  • 57) 0.424 854 334 306 971 207 726 677 360 64 × 2 = 0 + 0.849 708 668 613 942 415 453 354 721 28;
  • 58) 0.849 708 668 613 942 415 453 354 721 28 × 2 = 1 + 0.699 417 337 227 884 830 906 709 442 56;

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:


0.019 932 208 134 231 138 919 244 739 99(10) =

0.0000 0101 0001 1010 0100 0110 1111 0110 0001 0010 1101 0110 1010 1000 01(2)

5. Positive number before normalization:

0.019 932 208 134 231 138 919 244 739 99(10) =

0.0000 0101 0001 1010 0100 0110 1111 0110 0001 0010 1101 0110 1010 1000 01(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 6 positions to the right, so that only one non zero digit remains to the left of it:


0.019 932 208 134 231 138 919 244 739 99(10) =

0.0000 0101 0001 1010 0100 0110 1111 0110 0001 0010 1101 0110 1010 1000 01(2) =

0.0000 0101 0001 1010 0100 0110 1111 0110 0001 0010 1101 0110 1010 1000 01(2) × 20 =

1.0100 0110 1001 0001 1011 1101 1000 0100 1011 0101 1010 1010 0001(2) × 2-6


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): -6

Mantissa (not normalized):
1.0100 0110 1001 0001 1011 1101 1000 0100 1011 0101 1010 1010 0001


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-6 + 2(11-1) - 1 =

(-6 + 1 023)(10) =

1 017(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;
  • 1 017 ÷ 2 = 508 + 1;
  • 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) =

1017(10) =

011 1111 1001(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. 0100 0110 1001 0001 1011 1101 1000 0100 1011 0101 1010 1010 0001 =

0100 0110 1001 0001 1011 1101 1000 0100 1011 0101 1010 1010 0001


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 1001

Mantissa (52 bits) =
0100 0110 1001 0001 1011 1101 1000 0100 1011 0101 1010 1010 0001


Decimal number 0.019 932 208 134 231 138 919 244 739 99 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1001 - 0100 0110 1001 0001 1011 1101 1000 0100 1011 0101 1010 1010 0001

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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