0.019 932 208 134 231 138 919 244 738 69 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.019 932 208 134 231 138 919 244 738 69(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
0.019 932 208 134 231 138 919 244 738 69(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: 0.019 932 208 134 231 138 919 244 738 69.

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) 0.019 932 208 134 231 138 919 244 738 69 × 2 = 0 + 0.039 864 416 268 462 277 838 489 477 38;
  • 2) 0.039 864 416 268 462 277 838 489 477 38 × 2 = 0 + 0.079 728 832 536 924 555 676 978 954 76;
  • 3) 0.079 728 832 536 924 555 676 978 954 76 × 2 = 0 + 0.159 457 665 073 849 111 353 957 909 52;
  • 4) 0.159 457 665 073 849 111 353 957 909 52 × 2 = 0 + 0.318 915 330 147 698 222 707 915 819 04;
  • 5) 0.318 915 330 147 698 222 707 915 819 04 × 2 = 0 + 0.637 830 660 295 396 445 415 831 638 08;
  • 6) 0.637 830 660 295 396 445 415 831 638 08 × 2 = 1 + 0.275 661 320 590 792 890 831 663 276 16;
  • 7) 0.275 661 320 590 792 890 831 663 276 16 × 2 = 0 + 0.551 322 641 181 585 781 663 326 552 32;
  • 8) 0.551 322 641 181 585 781 663 326 552 32 × 2 = 1 + 0.102 645 282 363 171 563 326 653 104 64;
  • 9) 0.102 645 282 363 171 563 326 653 104 64 × 2 = 0 + 0.205 290 564 726 343 126 653 306 209 28;
  • 10) 0.205 290 564 726 343 126 653 306 209 28 × 2 = 0 + 0.410 581 129 452 686 253 306 612 418 56;
  • 11) 0.410 581 129 452 686 253 306 612 418 56 × 2 = 0 + 0.821 162 258 905 372 506 613 224 837 12;
  • 12) 0.821 162 258 905 372 506 613 224 837 12 × 2 = 1 + 0.642 324 517 810 745 013 226 449 674 24;
  • 13) 0.642 324 517 810 745 013 226 449 674 24 × 2 = 1 + 0.284 649 035 621 490 026 452 899 348 48;
  • 14) 0.284 649 035 621 490 026 452 899 348 48 × 2 = 0 + 0.569 298 071 242 980 052 905 798 696 96;
  • 15) 0.569 298 071 242 980 052 905 798 696 96 × 2 = 1 + 0.138 596 142 485 960 105 811 597 393 92;
  • 16) 0.138 596 142 485 960 105 811 597 393 92 × 2 = 0 + 0.277 192 284 971 920 211 623 194 787 84;
  • 17) 0.277 192 284 971 920 211 623 194 787 84 × 2 = 0 + 0.554 384 569 943 840 423 246 389 575 68;
  • 18) 0.554 384 569 943 840 423 246 389 575 68 × 2 = 1 + 0.108 769 139 887 680 846 492 779 151 36;
  • 19) 0.108 769 139 887 680 846 492 779 151 36 × 2 = 0 + 0.217 538 279 775 361 692 985 558 302 72;
  • 20) 0.217 538 279 775 361 692 985 558 302 72 × 2 = 0 + 0.435 076 559 550 723 385 971 116 605 44;
  • 21) 0.435 076 559 550 723 385 971 116 605 44 × 2 = 0 + 0.870 153 119 101 446 771 942 233 210 88;
  • 22) 0.870 153 119 101 446 771 942 233 210 88 × 2 = 1 + 0.740 306 238 202 893 543 884 466 421 76;
  • 23) 0.740 306 238 202 893 543 884 466 421 76 × 2 = 1 + 0.480 612 476 405 787 087 768 932 843 52;
  • 24) 0.480 612 476 405 787 087 768 932 843 52 × 2 = 0 + 0.961 224 952 811 574 175 537 865 687 04;
  • 25) 0.961 224 952 811 574 175 537 865 687 04 × 2 = 1 + 0.922 449 905 623 148 351 075 731 374 08;
  • 26) 0.922 449 905 623 148 351 075 731 374 08 × 2 = 1 + 0.844 899 811 246 296 702 151 462 748 16;
  • 27) 0.844 899 811 246 296 702 151 462 748 16 × 2 = 1 + 0.689 799 622 492 593 404 302 925 496 32;
  • 28) 0.689 799 622 492 593 404 302 925 496 32 × 2 = 1 + 0.379 599 244 985 186 808 605 850 992 64;
  • 29) 0.379 599 244 985 186 808 605 850 992 64 × 2 = 0 + 0.759 198 489 970 373 617 211 701 985 28;
  • 30) 0.759 198 489 970 373 617 211 701 985 28 × 2 = 1 + 0.518 396 979 940 747 234 423 403 970 56;
  • 31) 0.518 396 979 940 747 234 423 403 970 56 × 2 = 1 + 0.036 793 959 881 494 468 846 807 941 12;
  • 32) 0.036 793 959 881 494 468 846 807 941 12 × 2 = 0 + 0.073 587 919 762 988 937 693 615 882 24;
  • 33) 0.073 587 919 762 988 937 693 615 882 24 × 2 = 0 + 0.147 175 839 525 977 875 387 231 764 48;
  • 34) 0.147 175 839 525 977 875 387 231 764 48 × 2 = 0 + 0.294 351 679 051 955 750 774 463 528 96;
  • 35) 0.294 351 679 051 955 750 774 463 528 96 × 2 = 0 + 0.588 703 358 103 911 501 548 927 057 92;
  • 36) 0.588 703 358 103 911 501 548 927 057 92 × 2 = 1 + 0.177 406 716 207 823 003 097 854 115 84;
  • 37) 0.177 406 716 207 823 003 097 854 115 84 × 2 = 0 + 0.354 813 432 415 646 006 195 708 231 68;
  • 38) 0.354 813 432 415 646 006 195 708 231 68 × 2 = 0 + 0.709 626 864 831 292 012 391 416 463 36;
  • 39) 0.709 626 864 831 292 012 391 416 463 36 × 2 = 1 + 0.419 253 729 662 584 024 782 832 926 72;
  • 40) 0.419 253 729 662 584 024 782 832 926 72 × 2 = 0 + 0.838 507 459 325 168 049 565 665 853 44;
  • 41) 0.838 507 459 325 168 049 565 665 853 44 × 2 = 1 + 0.677 014 918 650 336 099 131 331 706 88;
  • 42) 0.677 014 918 650 336 099 131 331 706 88 × 2 = 1 + 0.354 029 837 300 672 198 262 663 413 76;
  • 43) 0.354 029 837 300 672 198 262 663 413 76 × 2 = 0 + 0.708 059 674 601 344 396 525 326 827 52;
  • 44) 0.708 059 674 601 344 396 525 326 827 52 × 2 = 1 + 0.416 119 349 202 688 793 050 653 655 04;
  • 45) 0.416 119 349 202 688 793 050 653 655 04 × 2 = 0 + 0.832 238 698 405 377 586 101 307 310 08;
  • 46) 0.832 238 698 405 377 586 101 307 310 08 × 2 = 1 + 0.664 477 396 810 755 172 202 614 620 16;
  • 47) 0.664 477 396 810 755 172 202 614 620 16 × 2 = 1 + 0.328 954 793 621 510 344 405 229 240 32;
  • 48) 0.328 954 793 621 510 344 405 229 240 32 × 2 = 0 + 0.657 909 587 243 020 688 810 458 480 64;
  • 49) 0.657 909 587 243 020 688 810 458 480 64 × 2 = 1 + 0.315 819 174 486 041 377 620 916 961 28;
  • 50) 0.315 819 174 486 041 377 620 916 961 28 × 2 = 0 + 0.631 638 348 972 082 755 241 833 922 56;
  • 51) 0.631 638 348 972 082 755 241 833 922 56 × 2 = 1 + 0.263 276 697 944 165 510 483 667 845 12;
  • 52) 0.263 276 697 944 165 510 483 667 845 12 × 2 = 0 + 0.526 553 395 888 331 020 967 335 690 24;
  • 53) 0.526 553 395 888 331 020 967 335 690 24 × 2 = 1 + 0.053 106 791 776 662 041 934 671 380 48;
  • 54) 0.053 106 791 776 662 041 934 671 380 48 × 2 = 0 + 0.106 213 583 553 324 083 869 342 760 96;
  • 55) 0.106 213 583 553 324 083 869 342 760 96 × 2 = 0 + 0.212 427 167 106 648 167 738 685 521 92;
  • 56) 0.212 427 167 106 648 167 738 685 521 92 × 2 = 0 + 0.424 854 334 213 296 335 477 371 043 84;
  • 57) 0.424 854 334 213 296 335 477 371 043 84 × 2 = 0 + 0.849 708 668 426 592 670 954 742 087 68;
  • 58) 0.849 708 668 426 592 670 954 742 087 68 × 2 = 1 + 0.699 417 336 853 185 341 909 484 175 36;

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:


0.019 932 208 134 231 138 919 244 738 69(10) =

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

5. Positive number before normalization:

0.019 932 208 134 231 138 919 244 738 69(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:


0.019 932 208 134 231 138 919 244 738 69(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 0.019 932 208 134 231 138 919 244 738 69 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

Share

» Convert decimal 0.019 932 208 134 231 138 919 244 739 08 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 08, 2026 [August]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: August - by our visitors

» Improve your social skills: Reputation: Bridge Between Company's Actions & Market Value. Gemini Notebook YouTube Video of 6.55 minutes

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