0.000 000 021 979 29 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 021 979 29(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 29(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 29.

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 29 × 2 = 0 + 0.000 000 043 958 58;
  • 2) 0.000 000 043 958 58 × 2 = 0 + 0.000 000 087 917 16;
  • 3) 0.000 000 087 917 16 × 2 = 0 + 0.000 000 175 834 32;
  • 4) 0.000 000 175 834 32 × 2 = 0 + 0.000 000 351 668 64;
  • 5) 0.000 000 351 668 64 × 2 = 0 + 0.000 000 703 337 28;
  • 6) 0.000 000 703 337 28 × 2 = 0 + 0.000 001 406 674 56;
  • 7) 0.000 001 406 674 56 × 2 = 0 + 0.000 002 813 349 12;
  • 8) 0.000 002 813 349 12 × 2 = 0 + 0.000 005 626 698 24;
  • 9) 0.000 005 626 698 24 × 2 = 0 + 0.000 011 253 396 48;
  • 10) 0.000 011 253 396 48 × 2 = 0 + 0.000 022 506 792 96;
  • 11) 0.000 022 506 792 96 × 2 = 0 + 0.000 045 013 585 92;
  • 12) 0.000 045 013 585 92 × 2 = 0 + 0.000 090 027 171 84;
  • 13) 0.000 090 027 171 84 × 2 = 0 + 0.000 180 054 343 68;
  • 14) 0.000 180 054 343 68 × 2 = 0 + 0.000 360 108 687 36;
  • 15) 0.000 360 108 687 36 × 2 = 0 + 0.000 720 217 374 72;
  • 16) 0.000 720 217 374 72 × 2 = 0 + 0.001 440 434 749 44;
  • 17) 0.001 440 434 749 44 × 2 = 0 + 0.002 880 869 498 88;
  • 18) 0.002 880 869 498 88 × 2 = 0 + 0.005 761 738 997 76;
  • 19) 0.005 761 738 997 76 × 2 = 0 + 0.011 523 477 995 52;
  • 20) 0.011 523 477 995 52 × 2 = 0 + 0.023 046 955 991 04;
  • 21) 0.023 046 955 991 04 × 2 = 0 + 0.046 093 911 982 08;
  • 22) 0.046 093 911 982 08 × 2 = 0 + 0.092 187 823 964 16;
  • 23) 0.092 187 823 964 16 × 2 = 0 + 0.184 375 647 928 32;
  • 24) 0.184 375 647 928 32 × 2 = 0 + 0.368 751 295 856 64;
  • 25) 0.368 751 295 856 64 × 2 = 0 + 0.737 502 591 713 28;
  • 26) 0.737 502 591 713 28 × 2 = 1 + 0.475 005 183 426 56;
  • 27) 0.475 005 183 426 56 × 2 = 0 + 0.950 010 366 853 12;
  • 28) 0.950 010 366 853 12 × 2 = 1 + 0.900 020 733 706 24;
  • 29) 0.900 020 733 706 24 × 2 = 1 + 0.800 041 467 412 48;
  • 30) 0.800 041 467 412 48 × 2 = 1 + 0.600 082 934 824 96;
  • 31) 0.600 082 934 824 96 × 2 = 1 + 0.200 165 869 649 92;
  • 32) 0.200 165 869 649 92 × 2 = 0 + 0.400 331 739 299 84;
  • 33) 0.400 331 739 299 84 × 2 = 0 + 0.800 663 478 599 68;
  • 34) 0.800 663 478 599 68 × 2 = 1 + 0.601 326 957 199 36;
  • 35) 0.601 326 957 199 36 × 2 = 1 + 0.202 653 914 398 72;
  • 36) 0.202 653 914 398 72 × 2 = 0 + 0.405 307 828 797 44;
  • 37) 0.405 307 828 797 44 × 2 = 0 + 0.810 615 657 594 88;
  • 38) 0.810 615 657 594 88 × 2 = 1 + 0.621 231 315 189 76;
  • 39) 0.621 231 315 189 76 × 2 = 1 + 0.242 462 630 379 52;
  • 40) 0.242 462 630 379 52 × 2 = 0 + 0.484 925 260 759 04;
  • 41) 0.484 925 260 759 04 × 2 = 0 + 0.969 850 521 518 08;
  • 42) 0.969 850 521 518 08 × 2 = 1 + 0.939 701 043 036 16;
  • 43) 0.939 701 043 036 16 × 2 = 1 + 0.879 402 086 072 32;
  • 44) 0.879 402 086 072 32 × 2 = 1 + 0.758 804 172 144 64;
  • 45) 0.758 804 172 144 64 × 2 = 1 + 0.517 608 344 289 28;
  • 46) 0.517 608 344 289 28 × 2 = 1 + 0.035 216 688 578 56;
  • 47) 0.035 216 688 578 56 × 2 = 0 + 0.070 433 377 157 12;
  • 48) 0.070 433 377 157 12 × 2 = 0 + 0.140 866 754 314 24;
  • 49) 0.140 866 754 314 24 × 2 = 0 + 0.281 733 508 628 48;
  • 50) 0.281 733 508 628 48 × 2 = 0 + 0.563 467 017 256 96;
  • 51) 0.563 467 017 256 96 × 2 = 1 + 0.126 934 034 513 92;
  • 52) 0.126 934 034 513 92 × 2 = 0 + 0.253 868 069 027 84;
  • 53) 0.253 868 069 027 84 × 2 = 0 + 0.507 736 138 055 68;
  • 54) 0.507 736 138 055 68 × 2 = 1 + 0.015 472 276 111 36;
  • 55) 0.015 472 276 111 36 × 2 = 0 + 0.030 944 552 222 72;
  • 56) 0.030 944 552 222 72 × 2 = 0 + 0.061 889 104 445 44;
  • 57) 0.061 889 104 445 44 × 2 = 0 + 0.123 778 208 890 88;
  • 58) 0.123 778 208 890 88 × 2 = 0 + 0.247 556 417 781 76;
  • 59) 0.247 556 417 781 76 × 2 = 0 + 0.495 112 835 563 52;
  • 60) 0.495 112 835 563 52 × 2 = 0 + 0.990 225 671 127 04;
  • 61) 0.990 225 671 127 04 × 2 = 1 + 0.980 451 342 254 08;
  • 62) 0.980 451 342 254 08 × 2 = 1 + 0.960 902 684 508 16;
  • 63) 0.960 902 684 508 16 × 2 = 1 + 0.921 805 369 016 32;
  • 64) 0.921 805 369 016 32 × 2 = 1 + 0.843 610 738 032 64;
  • 65) 0.843 610 738 032 64 × 2 = 1 + 0.687 221 476 065 28;
  • 66) 0.687 221 476 065 28 × 2 = 1 + 0.374 442 952 130 56;
  • 67) 0.374 442 952 130 56 × 2 = 0 + 0.748 885 904 261 12;
  • 68) 0.748 885 904 261 12 × 2 = 1 + 0.497 771 808 522 24;
  • 69) 0.497 771 808 522 24 × 2 = 0 + 0.995 543 617 044 48;
  • 70) 0.995 543 617 044 48 × 2 = 1 + 0.991 087 234 088 96;
  • 71) 0.991 087 234 088 96 × 2 = 1 + 0.982 174 468 177 92;
  • 72) 0.982 174 468 177 92 × 2 = 1 + 0.964 348 936 355 84;
  • 73) 0.964 348 936 355 84 × 2 = 1 + 0.928 697 872 711 68;
  • 74) 0.928 697 872 711 68 × 2 = 1 + 0.857 395 745 423 36;
  • 75) 0.857 395 745 423 36 × 2 = 1 + 0.714 791 490 846 72;
  • 76) 0.714 791 490 846 72 × 2 = 1 + 0.429 582 981 693 44;
  • 77) 0.429 582 981 693 44 × 2 = 0 + 0.859 165 963 386 88;
  • 78) 0.859 165 963 386 88 × 2 = 1 + 0.718 331 926 773 76;

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 29(10) =

0.0000 0000 0000 0000 0000 0000 0101 1110 0110 0110 0111 1100 0010 0100 0000 1111 1101 0111 1111 01(2)

5. Positive number before normalization:

0.000 000 021 979 29(10) =

0.0000 0000 0000 0000 0000 0000 0101 1110 0110 0110 0111 1100 0010 0100 0000 1111 1101 0111 1111 01(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 29(10) =

0.0000 0000 0000 0000 0000 0000 0101 1110 0110 0110 0111 1100 0010 0100 0000 1111 1101 0111 1111 01(2) =

0.0000 0000 0000 0000 0000 0000 0101 1110 0110 0110 0111 1100 0010 0100 0000 1111 1101 0111 1111 01(2) × 20 =

1.0111 1001 1001 1001 1111 0000 1001 0000 0011 1111 0101 1111 1101(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 1001 1111 0000 1001 0000 0011 1111 0101 1111 1101


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 1001 1111 0000 1001 0000 0011 1111 0101 1111 1101 =

0111 1001 1001 1001 1111 0000 1001 0000 0011 1111 0101 1111 1101


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 1001 1111 0000 1001 0000 0011 1111 0101 1111 1101


Decimal number 0.000 000 021 979 29 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1110 0101 - 0111 1001 1001 1001 1111 0000 1001 0000 0011 1111 0101 1111 1101

» Convert decimal 0.000 000 021 978 3 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