0.019 932 208 134 231 138 919 244 740 22 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.019 932 208 134 231 138 919 244 740 22(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
0.019 932 208 134 231 138 919 244 740 22(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: 0.019 932 208 134 231 138 919 244 740 22.

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) 0.019 932 208 134 231 138 919 244 740 22 × 2 = 0 + 0.039 864 416 268 462 277 838 489 480 44;
  • 2) 0.039 864 416 268 462 277 838 489 480 44 × 2 = 0 + 0.079 728 832 536 924 555 676 978 960 88;
  • 3) 0.079 728 832 536 924 555 676 978 960 88 × 2 = 0 + 0.159 457 665 073 849 111 353 957 921 76;
  • 4) 0.159 457 665 073 849 111 353 957 921 76 × 2 = 0 + 0.318 915 330 147 698 222 707 915 843 52;
  • 5) 0.318 915 330 147 698 222 707 915 843 52 × 2 = 0 + 0.637 830 660 295 396 445 415 831 687 04;
  • 6) 0.637 830 660 295 396 445 415 831 687 04 × 2 = 1 + 0.275 661 320 590 792 890 831 663 374 08;
  • 7) 0.275 661 320 590 792 890 831 663 374 08 × 2 = 0 + 0.551 322 641 181 585 781 663 326 748 16;
  • 8) 0.551 322 641 181 585 781 663 326 748 16 × 2 = 1 + 0.102 645 282 363 171 563 326 653 496 32;
  • 9) 0.102 645 282 363 171 563 326 653 496 32 × 2 = 0 + 0.205 290 564 726 343 126 653 306 992 64;
  • 10) 0.205 290 564 726 343 126 653 306 992 64 × 2 = 0 + 0.410 581 129 452 686 253 306 613 985 28;
  • 11) 0.410 581 129 452 686 253 306 613 985 28 × 2 = 0 + 0.821 162 258 905 372 506 613 227 970 56;
  • 12) 0.821 162 258 905 372 506 613 227 970 56 × 2 = 1 + 0.642 324 517 810 745 013 226 455 941 12;
  • 13) 0.642 324 517 810 745 013 226 455 941 12 × 2 = 1 + 0.284 649 035 621 490 026 452 911 882 24;
  • 14) 0.284 649 035 621 490 026 452 911 882 24 × 2 = 0 + 0.569 298 071 242 980 052 905 823 764 48;
  • 15) 0.569 298 071 242 980 052 905 823 764 48 × 2 = 1 + 0.138 596 142 485 960 105 811 647 528 96;
  • 16) 0.138 596 142 485 960 105 811 647 528 96 × 2 = 0 + 0.277 192 284 971 920 211 623 295 057 92;
  • 17) 0.277 192 284 971 920 211 623 295 057 92 × 2 = 0 + 0.554 384 569 943 840 423 246 590 115 84;
  • 18) 0.554 384 569 943 840 423 246 590 115 84 × 2 = 1 + 0.108 769 139 887 680 846 493 180 231 68;
  • 19) 0.108 769 139 887 680 846 493 180 231 68 × 2 = 0 + 0.217 538 279 775 361 692 986 360 463 36;
  • 20) 0.217 538 279 775 361 692 986 360 463 36 × 2 = 0 + 0.435 076 559 550 723 385 972 720 926 72;
  • 21) 0.435 076 559 550 723 385 972 720 926 72 × 2 = 0 + 0.870 153 119 101 446 771 945 441 853 44;
  • 22) 0.870 153 119 101 446 771 945 441 853 44 × 2 = 1 + 0.740 306 238 202 893 543 890 883 706 88;
  • 23) 0.740 306 238 202 893 543 890 883 706 88 × 2 = 1 + 0.480 612 476 405 787 087 781 767 413 76;
  • 24) 0.480 612 476 405 787 087 781 767 413 76 × 2 = 0 + 0.961 224 952 811 574 175 563 534 827 52;
  • 25) 0.961 224 952 811 574 175 563 534 827 52 × 2 = 1 + 0.922 449 905 623 148 351 127 069 655 04;
  • 26) 0.922 449 905 623 148 351 127 069 655 04 × 2 = 1 + 0.844 899 811 246 296 702 254 139 310 08;
  • 27) 0.844 899 811 246 296 702 254 139 310 08 × 2 = 1 + 0.689 799 622 492 593 404 508 278 620 16;
  • 28) 0.689 799 622 492 593 404 508 278 620 16 × 2 = 1 + 0.379 599 244 985 186 809 016 557 240 32;
  • 29) 0.379 599 244 985 186 809 016 557 240 32 × 2 = 0 + 0.759 198 489 970 373 618 033 114 480 64;
  • 30) 0.759 198 489 970 373 618 033 114 480 64 × 2 = 1 + 0.518 396 979 940 747 236 066 228 961 28;
  • 31) 0.518 396 979 940 747 236 066 228 961 28 × 2 = 1 + 0.036 793 959 881 494 472 132 457 922 56;
  • 32) 0.036 793 959 881 494 472 132 457 922 56 × 2 = 0 + 0.073 587 919 762 988 944 264 915 845 12;
  • 33) 0.073 587 919 762 988 944 264 915 845 12 × 2 = 0 + 0.147 175 839 525 977 888 529 831 690 24;
  • 34) 0.147 175 839 525 977 888 529 831 690 24 × 2 = 0 + 0.294 351 679 051 955 777 059 663 380 48;
  • 35) 0.294 351 679 051 955 777 059 663 380 48 × 2 = 0 + 0.588 703 358 103 911 554 119 326 760 96;
  • 36) 0.588 703 358 103 911 554 119 326 760 96 × 2 = 1 + 0.177 406 716 207 823 108 238 653 521 92;
  • 37) 0.177 406 716 207 823 108 238 653 521 92 × 2 = 0 + 0.354 813 432 415 646 216 477 307 043 84;
  • 38) 0.354 813 432 415 646 216 477 307 043 84 × 2 = 0 + 0.709 626 864 831 292 432 954 614 087 68;
  • 39) 0.709 626 864 831 292 432 954 614 087 68 × 2 = 1 + 0.419 253 729 662 584 865 909 228 175 36;
  • 40) 0.419 253 729 662 584 865 909 228 175 36 × 2 = 0 + 0.838 507 459 325 169 731 818 456 350 72;
  • 41) 0.838 507 459 325 169 731 818 456 350 72 × 2 = 1 + 0.677 014 918 650 339 463 636 912 701 44;
  • 42) 0.677 014 918 650 339 463 636 912 701 44 × 2 = 1 + 0.354 029 837 300 678 927 273 825 402 88;
  • 43) 0.354 029 837 300 678 927 273 825 402 88 × 2 = 0 + 0.708 059 674 601 357 854 547 650 805 76;
  • 44) 0.708 059 674 601 357 854 547 650 805 76 × 2 = 1 + 0.416 119 349 202 715 709 095 301 611 52;
  • 45) 0.416 119 349 202 715 709 095 301 611 52 × 2 = 0 + 0.832 238 698 405 431 418 190 603 223 04;
  • 46) 0.832 238 698 405 431 418 190 603 223 04 × 2 = 1 + 0.664 477 396 810 862 836 381 206 446 08;
  • 47) 0.664 477 396 810 862 836 381 206 446 08 × 2 = 1 + 0.328 954 793 621 725 672 762 412 892 16;
  • 48) 0.328 954 793 621 725 672 762 412 892 16 × 2 = 0 + 0.657 909 587 243 451 345 524 825 784 32;
  • 49) 0.657 909 587 243 451 345 524 825 784 32 × 2 = 1 + 0.315 819 174 486 902 691 049 651 568 64;
  • 50) 0.315 819 174 486 902 691 049 651 568 64 × 2 = 0 + 0.631 638 348 973 805 382 099 303 137 28;
  • 51) 0.631 638 348 973 805 382 099 303 137 28 × 2 = 1 + 0.263 276 697 947 610 764 198 606 274 56;
  • 52) 0.263 276 697 947 610 764 198 606 274 56 × 2 = 0 + 0.526 553 395 895 221 528 397 212 549 12;
  • 53) 0.526 553 395 895 221 528 397 212 549 12 × 2 = 1 + 0.053 106 791 790 443 056 794 425 098 24;
  • 54) 0.053 106 791 790 443 056 794 425 098 24 × 2 = 0 + 0.106 213 583 580 886 113 588 850 196 48;
  • 55) 0.106 213 583 580 886 113 588 850 196 48 × 2 = 0 + 0.212 427 167 161 772 227 177 700 392 96;
  • 56) 0.212 427 167 161 772 227 177 700 392 96 × 2 = 0 + 0.424 854 334 323 544 454 355 400 785 92;
  • 57) 0.424 854 334 323 544 454 355 400 785 92 × 2 = 0 + 0.849 708 668 647 088 908 710 801 571 84;
  • 58) 0.849 708 668 647 088 908 710 801 571 84 × 2 = 1 + 0.699 417 337 294 177 817 421 603 143 68;

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:


0.019 932 208 134 231 138 919 244 740 22(10) =

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

5. Positive number before normalization:

0.019 932 208 134 231 138 919 244 740 22(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:


0.019 932 208 134 231 138 919 244 740 22(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 0.019 932 208 134 231 138 919 244 740 22 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