0.000 000 021 979 35 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 021 979 35(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 35(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 35.

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 35 × 2 = 0 + 0.000 000 043 958 7;
  • 2) 0.000 000 043 958 7 × 2 = 0 + 0.000 000 087 917 4;
  • 3) 0.000 000 087 917 4 × 2 = 0 + 0.000 000 175 834 8;
  • 4) 0.000 000 175 834 8 × 2 = 0 + 0.000 000 351 669 6;
  • 5) 0.000 000 351 669 6 × 2 = 0 + 0.000 000 703 339 2;
  • 6) 0.000 000 703 339 2 × 2 = 0 + 0.000 001 406 678 4;
  • 7) 0.000 001 406 678 4 × 2 = 0 + 0.000 002 813 356 8;
  • 8) 0.000 002 813 356 8 × 2 = 0 + 0.000 005 626 713 6;
  • 9) 0.000 005 626 713 6 × 2 = 0 + 0.000 011 253 427 2;
  • 10) 0.000 011 253 427 2 × 2 = 0 + 0.000 022 506 854 4;
  • 11) 0.000 022 506 854 4 × 2 = 0 + 0.000 045 013 708 8;
  • 12) 0.000 045 013 708 8 × 2 = 0 + 0.000 090 027 417 6;
  • 13) 0.000 090 027 417 6 × 2 = 0 + 0.000 180 054 835 2;
  • 14) 0.000 180 054 835 2 × 2 = 0 + 0.000 360 109 670 4;
  • 15) 0.000 360 109 670 4 × 2 = 0 + 0.000 720 219 340 8;
  • 16) 0.000 720 219 340 8 × 2 = 0 + 0.001 440 438 681 6;
  • 17) 0.001 440 438 681 6 × 2 = 0 + 0.002 880 877 363 2;
  • 18) 0.002 880 877 363 2 × 2 = 0 + 0.005 761 754 726 4;
  • 19) 0.005 761 754 726 4 × 2 = 0 + 0.011 523 509 452 8;
  • 20) 0.011 523 509 452 8 × 2 = 0 + 0.023 047 018 905 6;
  • 21) 0.023 047 018 905 6 × 2 = 0 + 0.046 094 037 811 2;
  • 22) 0.046 094 037 811 2 × 2 = 0 + 0.092 188 075 622 4;
  • 23) 0.092 188 075 622 4 × 2 = 0 + 0.184 376 151 244 8;
  • 24) 0.184 376 151 244 8 × 2 = 0 + 0.368 752 302 489 6;
  • 25) 0.368 752 302 489 6 × 2 = 0 + 0.737 504 604 979 2;
  • 26) 0.737 504 604 979 2 × 2 = 1 + 0.475 009 209 958 4;
  • 27) 0.475 009 209 958 4 × 2 = 0 + 0.950 018 419 916 8;
  • 28) 0.950 018 419 916 8 × 2 = 1 + 0.900 036 839 833 6;
  • 29) 0.900 036 839 833 6 × 2 = 1 + 0.800 073 679 667 2;
  • 30) 0.800 073 679 667 2 × 2 = 1 + 0.600 147 359 334 4;
  • 31) 0.600 147 359 334 4 × 2 = 1 + 0.200 294 718 668 8;
  • 32) 0.200 294 718 668 8 × 2 = 0 + 0.400 589 437 337 6;
  • 33) 0.400 589 437 337 6 × 2 = 0 + 0.801 178 874 675 2;
  • 34) 0.801 178 874 675 2 × 2 = 1 + 0.602 357 749 350 4;
  • 35) 0.602 357 749 350 4 × 2 = 1 + 0.204 715 498 700 8;
  • 36) 0.204 715 498 700 8 × 2 = 0 + 0.409 430 997 401 6;
  • 37) 0.409 430 997 401 6 × 2 = 0 + 0.818 861 994 803 2;
  • 38) 0.818 861 994 803 2 × 2 = 1 + 0.637 723 989 606 4;
  • 39) 0.637 723 989 606 4 × 2 = 1 + 0.275 447 979 212 8;
  • 40) 0.275 447 979 212 8 × 2 = 0 + 0.550 895 958 425 6;
  • 41) 0.550 895 958 425 6 × 2 = 1 + 0.101 791 916 851 2;
  • 42) 0.101 791 916 851 2 × 2 = 0 + 0.203 583 833 702 4;
  • 43) 0.203 583 833 702 4 × 2 = 0 + 0.407 167 667 404 8;
  • 44) 0.407 167 667 404 8 × 2 = 0 + 0.814 335 334 809 6;
  • 45) 0.814 335 334 809 6 × 2 = 1 + 0.628 670 669 619 2;
  • 46) 0.628 670 669 619 2 × 2 = 1 + 0.257 341 339 238 4;
  • 47) 0.257 341 339 238 4 × 2 = 0 + 0.514 682 678 476 8;
  • 48) 0.514 682 678 476 8 × 2 = 1 + 0.029 365 356 953 6;
  • 49) 0.029 365 356 953 6 × 2 = 0 + 0.058 730 713 907 2;
  • 50) 0.058 730 713 907 2 × 2 = 0 + 0.117 461 427 814 4;
  • 51) 0.117 461 427 814 4 × 2 = 0 + 0.234 922 855 628 8;
  • 52) 0.234 922 855 628 8 × 2 = 0 + 0.469 845 711 257 6;
  • 53) 0.469 845 711 257 6 × 2 = 0 + 0.939 691 422 515 2;
  • 54) 0.939 691 422 515 2 × 2 = 1 + 0.879 382 845 030 4;
  • 55) 0.879 382 845 030 4 × 2 = 1 + 0.758 765 690 060 8;
  • 56) 0.758 765 690 060 8 × 2 = 1 + 0.517 531 380 121 6;
  • 57) 0.517 531 380 121 6 × 2 = 1 + 0.035 062 760 243 2;
  • 58) 0.035 062 760 243 2 × 2 = 0 + 0.070 125 520 486 4;
  • 59) 0.070 125 520 486 4 × 2 = 0 + 0.140 251 040 972 8;
  • 60) 0.140 251 040 972 8 × 2 = 0 + 0.280 502 081 945 6;
  • 61) 0.280 502 081 945 6 × 2 = 0 + 0.561 004 163 891 2;
  • 62) 0.561 004 163 891 2 × 2 = 1 + 0.122 008 327 782 4;
  • 63) 0.122 008 327 782 4 × 2 = 0 + 0.244 016 655 564 8;
  • 64) 0.244 016 655 564 8 × 2 = 0 + 0.488 033 311 129 6;
  • 65) 0.488 033 311 129 6 × 2 = 0 + 0.976 066 622 259 2;
  • 66) 0.976 066 622 259 2 × 2 = 1 + 0.952 133 244 518 4;
  • 67) 0.952 133 244 518 4 × 2 = 1 + 0.904 266 489 036 8;
  • 68) 0.904 266 489 036 8 × 2 = 1 + 0.808 532 978 073 6;
  • 69) 0.808 532 978 073 6 × 2 = 1 + 0.617 065 956 147 2;
  • 70) 0.617 065 956 147 2 × 2 = 1 + 0.234 131 912 294 4;
  • 71) 0.234 131 912 294 4 × 2 = 0 + 0.468 263 824 588 8;
  • 72) 0.468 263 824 588 8 × 2 = 0 + 0.936 527 649 177 6;
  • 73) 0.936 527 649 177 6 × 2 = 1 + 0.873 055 298 355 2;
  • 74) 0.873 055 298 355 2 × 2 = 1 + 0.746 110 596 710 4;
  • 75) 0.746 110 596 710 4 × 2 = 1 + 0.492 221 193 420 8;
  • 76) 0.492 221 193 420 8 × 2 = 0 + 0.984 442 386 841 6;
  • 77) 0.984 442 386 841 6 × 2 = 1 + 0.968 884 773 683 2;
  • 78) 0.968 884 773 683 2 × 2 = 1 + 0.937 769 547 366 4;

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 35(10) =

0.0000 0000 0000 0000 0000 0000 0101 1110 0110 0110 1000 1101 0000 0111 1000 0100 0111 1100 1110 11(2)

5. Positive number before normalization:

0.000 000 021 979 35(10) =

0.0000 0000 0000 0000 0000 0000 0101 1110 0110 0110 1000 1101 0000 0111 1000 0100 0111 1100 1110 11(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 35(10) =

0.0000 0000 0000 0000 0000 0000 0101 1110 0110 0110 1000 1101 0000 0111 1000 0100 0111 1100 1110 11(2) =

0.0000 0000 0000 0000 0000 0000 0101 1110 0110 0110 1000 1101 0000 0111 1000 0100 0111 1100 1110 11(2) × 20 =

1.0111 1001 1001 1010 0011 0100 0001 1110 0001 0001 1111 0011 1011(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 1010 0011 0100 0001 1110 0001 0001 1111 0011 1011


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 1010 0011 0100 0001 1110 0001 0001 1111 0011 1011 =

0111 1001 1001 1010 0011 0100 0001 1110 0001 0001 1111 0011 1011


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 1010 0011 0100 0001 1110 0001 0001 1111 0011 1011


Decimal number 0.000 000 021 979 35 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1110 0101 - 0111 1001 1001 1010 0011 0100 0001 1110 0001 0001 1111 0011 1011

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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