0.000 000 021 979 552 668 137 24 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 021 979 552 668 137 24(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
0.000 000 021 979 552 668 137 24(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: 0.000 000 021 979 552 668 137 24.

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) 0.000 000 021 979 552 668 137 24 × 2 = 0 + 0.000 000 043 959 105 336 274 48;
  • 2) 0.000 000 043 959 105 336 274 48 × 2 = 0 + 0.000 000 087 918 210 672 548 96;
  • 3) 0.000 000 087 918 210 672 548 96 × 2 = 0 + 0.000 000 175 836 421 345 097 92;
  • 4) 0.000 000 175 836 421 345 097 92 × 2 = 0 + 0.000 000 351 672 842 690 195 84;
  • 5) 0.000 000 351 672 842 690 195 84 × 2 = 0 + 0.000 000 703 345 685 380 391 68;
  • 6) 0.000 000 703 345 685 380 391 68 × 2 = 0 + 0.000 001 406 691 370 760 783 36;
  • 7) 0.000 001 406 691 370 760 783 36 × 2 = 0 + 0.000 002 813 382 741 521 566 72;
  • 8) 0.000 002 813 382 741 521 566 72 × 2 = 0 + 0.000 005 626 765 483 043 133 44;
  • 9) 0.000 005 626 765 483 043 133 44 × 2 = 0 + 0.000 011 253 530 966 086 266 88;
  • 10) 0.000 011 253 530 966 086 266 88 × 2 = 0 + 0.000 022 507 061 932 172 533 76;
  • 11) 0.000 022 507 061 932 172 533 76 × 2 = 0 + 0.000 045 014 123 864 345 067 52;
  • 12) 0.000 045 014 123 864 345 067 52 × 2 = 0 + 0.000 090 028 247 728 690 135 04;
  • 13) 0.000 090 028 247 728 690 135 04 × 2 = 0 + 0.000 180 056 495 457 380 270 08;
  • 14) 0.000 180 056 495 457 380 270 08 × 2 = 0 + 0.000 360 112 990 914 760 540 16;
  • 15) 0.000 360 112 990 914 760 540 16 × 2 = 0 + 0.000 720 225 981 829 521 080 32;
  • 16) 0.000 720 225 981 829 521 080 32 × 2 = 0 + 0.001 440 451 963 659 042 160 64;
  • 17) 0.001 440 451 963 659 042 160 64 × 2 = 0 + 0.002 880 903 927 318 084 321 28;
  • 18) 0.002 880 903 927 318 084 321 28 × 2 = 0 + 0.005 761 807 854 636 168 642 56;
  • 19) 0.005 761 807 854 636 168 642 56 × 2 = 0 + 0.011 523 615 709 272 337 285 12;
  • 20) 0.011 523 615 709 272 337 285 12 × 2 = 0 + 0.023 047 231 418 544 674 570 24;
  • 21) 0.023 047 231 418 544 674 570 24 × 2 = 0 + 0.046 094 462 837 089 349 140 48;
  • 22) 0.046 094 462 837 089 349 140 48 × 2 = 0 + 0.092 188 925 674 178 698 280 96;
  • 23) 0.092 188 925 674 178 698 280 96 × 2 = 0 + 0.184 377 851 348 357 396 561 92;
  • 24) 0.184 377 851 348 357 396 561 92 × 2 = 0 + 0.368 755 702 696 714 793 123 84;
  • 25) 0.368 755 702 696 714 793 123 84 × 2 = 0 + 0.737 511 405 393 429 586 247 68;
  • 26) 0.737 511 405 393 429 586 247 68 × 2 = 1 + 0.475 022 810 786 859 172 495 36;
  • 27) 0.475 022 810 786 859 172 495 36 × 2 = 0 + 0.950 045 621 573 718 344 990 72;
  • 28) 0.950 045 621 573 718 344 990 72 × 2 = 1 + 0.900 091 243 147 436 689 981 44;
  • 29) 0.900 091 243 147 436 689 981 44 × 2 = 1 + 0.800 182 486 294 873 379 962 88;
  • 30) 0.800 182 486 294 873 379 962 88 × 2 = 1 + 0.600 364 972 589 746 759 925 76;
  • 31) 0.600 364 972 589 746 759 925 76 × 2 = 1 + 0.200 729 945 179 493 519 851 52;
  • 32) 0.200 729 945 179 493 519 851 52 × 2 = 0 + 0.401 459 890 358 987 039 703 04;
  • 33) 0.401 459 890 358 987 039 703 04 × 2 = 0 + 0.802 919 780 717 974 079 406 08;
  • 34) 0.802 919 780 717 974 079 406 08 × 2 = 1 + 0.605 839 561 435 948 158 812 16;
  • 35) 0.605 839 561 435 948 158 812 16 × 2 = 1 + 0.211 679 122 871 896 317 624 32;
  • 36) 0.211 679 122 871 896 317 624 32 × 2 = 0 + 0.423 358 245 743 792 635 248 64;
  • 37) 0.423 358 245 743 792 635 248 64 × 2 = 0 + 0.846 716 491 487 585 270 497 28;
  • 38) 0.846 716 491 487 585 270 497 28 × 2 = 1 + 0.693 432 982 975 170 540 994 56;
  • 39) 0.693 432 982 975 170 540 994 56 × 2 = 1 + 0.386 865 965 950 341 081 989 12;
  • 40) 0.386 865 965 950 341 081 989 12 × 2 = 0 + 0.773 731 931 900 682 163 978 24;
  • 41) 0.773 731 931 900 682 163 978 24 × 2 = 1 + 0.547 463 863 801 364 327 956 48;
  • 42) 0.547 463 863 801 364 327 956 48 × 2 = 1 + 0.094 927 727 602 728 655 912 96;
  • 43) 0.094 927 727 602 728 655 912 96 × 2 = 0 + 0.189 855 455 205 457 311 825 92;
  • 44) 0.189 855 455 205 457 311 825 92 × 2 = 0 + 0.379 710 910 410 914 623 651 84;
  • 45) 0.379 710 910 410 914 623 651 84 × 2 = 0 + 0.759 421 820 821 829 247 303 68;
  • 46) 0.759 421 820 821 829 247 303 68 × 2 = 1 + 0.518 843 641 643 658 494 607 36;
  • 47) 0.518 843 641 643 658 494 607 36 × 2 = 1 + 0.037 687 283 287 316 989 214 72;
  • 48) 0.037 687 283 287 316 989 214 72 × 2 = 0 + 0.075 374 566 574 633 978 429 44;
  • 49) 0.075 374 566 574 633 978 429 44 × 2 = 0 + 0.150 749 133 149 267 956 858 88;
  • 50) 0.150 749 133 149 267 956 858 88 × 2 = 0 + 0.301 498 266 298 535 913 717 76;
  • 51) 0.301 498 266 298 535 913 717 76 × 2 = 0 + 0.602 996 532 597 071 827 435 52;
  • 52) 0.602 996 532 597 071 827 435 52 × 2 = 1 + 0.205 993 065 194 143 654 871 04;
  • 53) 0.205 993 065 194 143 654 871 04 × 2 = 0 + 0.411 986 130 388 287 309 742 08;
  • 54) 0.411 986 130 388 287 309 742 08 × 2 = 0 + 0.823 972 260 776 574 619 484 16;
  • 55) 0.823 972 260 776 574 619 484 16 × 2 = 1 + 0.647 944 521 553 149 238 968 32;
  • 56) 0.647 944 521 553 149 238 968 32 × 2 = 1 + 0.295 889 043 106 298 477 936 64;
  • 57) 0.295 889 043 106 298 477 936 64 × 2 = 0 + 0.591 778 086 212 596 955 873 28;
  • 58) 0.591 778 086 212 596 955 873 28 × 2 = 1 + 0.183 556 172 425 193 911 746 56;
  • 59) 0.183 556 172 425 193 911 746 56 × 2 = 0 + 0.367 112 344 850 387 823 493 12;
  • 60) 0.367 112 344 850 387 823 493 12 × 2 = 0 + 0.734 224 689 700 775 646 986 24;
  • 61) 0.734 224 689 700 775 646 986 24 × 2 = 1 + 0.468 449 379 401 551 293 972 48;
  • 62) 0.468 449 379 401 551 293 972 48 × 2 = 0 + 0.936 898 758 803 102 587 944 96;
  • 63) 0.936 898 758 803 102 587 944 96 × 2 = 1 + 0.873 797 517 606 205 175 889 92;
  • 64) 0.873 797 517 606 205 175 889 92 × 2 = 1 + 0.747 595 035 212 410 351 779 84;
  • 65) 0.747 595 035 212 410 351 779 84 × 2 = 1 + 0.495 190 070 424 820 703 559 68;
  • 66) 0.495 190 070 424 820 703 559 68 × 2 = 0 + 0.990 380 140 849 641 407 119 36;
  • 67) 0.990 380 140 849 641 407 119 36 × 2 = 1 + 0.980 760 281 699 282 814 238 72;
  • 68) 0.980 760 281 699 282 814 238 72 × 2 = 1 + 0.961 520 563 398 565 628 477 44;
  • 69) 0.961 520 563 398 565 628 477 44 × 2 = 1 + 0.923 041 126 797 131 256 954 88;
  • 70) 0.923 041 126 797 131 256 954 88 × 2 = 1 + 0.846 082 253 594 262 513 909 76;
  • 71) 0.846 082 253 594 262 513 909 76 × 2 = 1 + 0.692 164 507 188 525 027 819 52;
  • 72) 0.692 164 507 188 525 027 819 52 × 2 = 1 + 0.384 329 014 377 050 055 639 04;
  • 73) 0.384 329 014 377 050 055 639 04 × 2 = 0 + 0.768 658 028 754 100 111 278 08;
  • 74) 0.768 658 028 754 100 111 278 08 × 2 = 1 + 0.537 316 057 508 200 222 556 16;
  • 75) 0.537 316 057 508 200 222 556 16 × 2 = 1 + 0.074 632 115 016 400 445 112 32;
  • 76) 0.074 632 115 016 400 445 112 32 × 2 = 0 + 0.149 264 230 032 800 890 224 64;
  • 77) 0.149 264 230 032 800 890 224 64 × 2 = 0 + 0.298 528 460 065 601 780 449 28;
  • 78) 0.298 528 460 065 601 780 449 28 × 2 = 0 + 0.597 056 920 131 203 560 898 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:


0.000 000 021 979 552 668 137 24(10) =

0.0000 0000 0000 0000 0000 0000 0101 1110 0110 0110 1100 0110 0001 0011 0100 1011 1011 1111 0110 00(2)

5. Positive number before normalization:

0.000 000 021 979 552 668 137 24(10) =

0.0000 0000 0000 0000 0000 0000 0101 1110 0110 0110 1100 0110 0001 0011 0100 1011 1011 1111 0110 00(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 26 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 021 979 552 668 137 24(10) =

0.0000 0000 0000 0000 0000 0000 0101 1110 0110 0110 1100 0110 0001 0011 0100 1011 1011 1111 0110 00(2) =

0.0000 0000 0000 0000 0000 0000 0101 1110 0110 0110 1100 0110 0001 0011 0100 1011 1011 1111 0110 00(2) × 20 =

1.0111 1001 1001 1011 0001 1000 0100 1101 0010 1110 1111 1101 1000(2) × 2-26


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): -26

Mantissa (not normalized):
1.0111 1001 1001 1011 0001 1000 0100 1101 0010 1110 1111 1101 1000


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-26 + 2(11-1) - 1 =

(-26 + 1 023)(10) =

997(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;
  • 997 ÷ 2 = 498 + 1;
  • 498 ÷ 2 = 249 + 0;
  • 249 ÷ 2 = 124 + 1;
  • 124 ÷ 2 = 62 + 0;
  • 62 ÷ 2 = 31 + 0;
  • 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) =

997(10) =

011 1110 0101(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. 0111 1001 1001 1011 0001 1000 0100 1101 0010 1110 1111 1101 1000 =

0111 1001 1001 1011 0001 1000 0100 1101 0010 1110 1111 1101 1000


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 1110 0101

Mantissa (52 bits) =
0111 1001 1001 1011 0001 1000 0100 1101 0010 1110 1111 1101 1000


Decimal number 0.000 000 021 979 552 668 137 24 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1110 0101 - 0111 1001 1001 1011 0001 1000 0100 1101 0010 1110 1111 1101 1000

» Convert decimal 0.000 000 021 979 552 668 136 95 to 64 bit double precision IEEE 754 binary floating point representation standard

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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