0.000 000 000 008 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 008 9(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
0.000 000 000 008 9(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: 0.000 000 000 008 9.

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) 0.000 000 000 008 9 × 2 = 0 + 0.000 000 000 017 8;
  • 2) 0.000 000 000 017 8 × 2 = 0 + 0.000 000 000 035 6;
  • 3) 0.000 000 000 035 6 × 2 = 0 + 0.000 000 000 071 2;
  • 4) 0.000 000 000 071 2 × 2 = 0 + 0.000 000 000 142 4;
  • 5) 0.000 000 000 142 4 × 2 = 0 + 0.000 000 000 284 8;
  • 6) 0.000 000 000 284 8 × 2 = 0 + 0.000 000 000 569 6;
  • 7) 0.000 000 000 569 6 × 2 = 0 + 0.000 000 001 139 2;
  • 8) 0.000 000 001 139 2 × 2 = 0 + 0.000 000 002 278 4;
  • 9) 0.000 000 002 278 4 × 2 = 0 + 0.000 000 004 556 8;
  • 10) 0.000 000 004 556 8 × 2 = 0 + 0.000 000 009 113 6;
  • 11) 0.000 000 009 113 6 × 2 = 0 + 0.000 000 018 227 2;
  • 12) 0.000 000 018 227 2 × 2 = 0 + 0.000 000 036 454 4;
  • 13) 0.000 000 036 454 4 × 2 = 0 + 0.000 000 072 908 8;
  • 14) 0.000 000 072 908 8 × 2 = 0 + 0.000 000 145 817 6;
  • 15) 0.000 000 145 817 6 × 2 = 0 + 0.000 000 291 635 2;
  • 16) 0.000 000 291 635 2 × 2 = 0 + 0.000 000 583 270 4;
  • 17) 0.000 000 583 270 4 × 2 = 0 + 0.000 001 166 540 8;
  • 18) 0.000 001 166 540 8 × 2 = 0 + 0.000 002 333 081 6;
  • 19) 0.000 002 333 081 6 × 2 = 0 + 0.000 004 666 163 2;
  • 20) 0.000 004 666 163 2 × 2 = 0 + 0.000 009 332 326 4;
  • 21) 0.000 009 332 326 4 × 2 = 0 + 0.000 018 664 652 8;
  • 22) 0.000 018 664 652 8 × 2 = 0 + 0.000 037 329 305 6;
  • 23) 0.000 037 329 305 6 × 2 = 0 + 0.000 074 658 611 2;
  • 24) 0.000 074 658 611 2 × 2 = 0 + 0.000 149 317 222 4;
  • 25) 0.000 149 317 222 4 × 2 = 0 + 0.000 298 634 444 8;
  • 26) 0.000 298 634 444 8 × 2 = 0 + 0.000 597 268 889 6;
  • 27) 0.000 597 268 889 6 × 2 = 0 + 0.001 194 537 779 2;
  • 28) 0.001 194 537 779 2 × 2 = 0 + 0.002 389 075 558 4;
  • 29) 0.002 389 075 558 4 × 2 = 0 + 0.004 778 151 116 8;
  • 30) 0.004 778 151 116 8 × 2 = 0 + 0.009 556 302 233 6;
  • 31) 0.009 556 302 233 6 × 2 = 0 + 0.019 112 604 467 2;
  • 32) 0.019 112 604 467 2 × 2 = 0 + 0.038 225 208 934 4;
  • 33) 0.038 225 208 934 4 × 2 = 0 + 0.076 450 417 868 8;
  • 34) 0.076 450 417 868 8 × 2 = 0 + 0.152 900 835 737 6;
  • 35) 0.152 900 835 737 6 × 2 = 0 + 0.305 801 671 475 2;
  • 36) 0.305 801 671 475 2 × 2 = 0 + 0.611 603 342 950 4;
  • 37) 0.611 603 342 950 4 × 2 = 1 + 0.223 206 685 900 8;
  • 38) 0.223 206 685 900 8 × 2 = 0 + 0.446 413 371 801 6;
  • 39) 0.446 413 371 801 6 × 2 = 0 + 0.892 826 743 603 2;
  • 40) 0.892 826 743 603 2 × 2 = 1 + 0.785 653 487 206 4;
  • 41) 0.785 653 487 206 4 × 2 = 1 + 0.571 306 974 412 8;
  • 42) 0.571 306 974 412 8 × 2 = 1 + 0.142 613 948 825 6;
  • 43) 0.142 613 948 825 6 × 2 = 0 + 0.285 227 897 651 2;
  • 44) 0.285 227 897 651 2 × 2 = 0 + 0.570 455 795 302 4;
  • 45) 0.570 455 795 302 4 × 2 = 1 + 0.140 911 590 604 8;
  • 46) 0.140 911 590 604 8 × 2 = 0 + 0.281 823 181 209 6;
  • 47) 0.281 823 181 209 6 × 2 = 0 + 0.563 646 362 419 2;
  • 48) 0.563 646 362 419 2 × 2 = 1 + 0.127 292 724 838 4;
  • 49) 0.127 292 724 838 4 × 2 = 0 + 0.254 585 449 676 8;
  • 50) 0.254 585 449 676 8 × 2 = 0 + 0.509 170 899 353 6;
  • 51) 0.509 170 899 353 6 × 2 = 1 + 0.018 341 798 707 2;
  • 52) 0.018 341 798 707 2 × 2 = 0 + 0.036 683 597 414 4;
  • 53) 0.036 683 597 414 4 × 2 = 0 + 0.073 367 194 828 8;
  • 54) 0.073 367 194 828 8 × 2 = 0 + 0.146 734 389 657 6;
  • 55) 0.146 734 389 657 6 × 2 = 0 + 0.293 468 779 315 2;
  • 56) 0.293 468 779 315 2 × 2 = 0 + 0.586 937 558 630 4;
  • 57) 0.586 937 558 630 4 × 2 = 1 + 0.173 875 117 260 8;
  • 58) 0.173 875 117 260 8 × 2 = 0 + 0.347 750 234 521 6;
  • 59) 0.347 750 234 521 6 × 2 = 0 + 0.695 500 469 043 2;
  • 60) 0.695 500 469 043 2 × 2 = 1 + 0.391 000 938 086 4;
  • 61) 0.391 000 938 086 4 × 2 = 0 + 0.782 001 876 172 8;
  • 62) 0.782 001 876 172 8 × 2 = 1 + 0.564 003 752 345 6;
  • 63) 0.564 003 752 345 6 × 2 = 1 + 0.128 007 504 691 2;
  • 64) 0.128 007 504 691 2 × 2 = 0 + 0.256 015 009 382 4;
  • 65) 0.256 015 009 382 4 × 2 = 0 + 0.512 030 018 764 8;
  • 66) 0.512 030 018 764 8 × 2 = 1 + 0.024 060 037 529 6;
  • 67) 0.024 060 037 529 6 × 2 = 0 + 0.048 120 075 059 2;
  • 68) 0.048 120 075 059 2 × 2 = 0 + 0.096 240 150 118 4;
  • 69) 0.096 240 150 118 4 × 2 = 0 + 0.192 480 300 236 8;
  • 70) 0.192 480 300 236 8 × 2 = 0 + 0.384 960 600 473 6;
  • 71) 0.384 960 600 473 6 × 2 = 0 + 0.769 921 200 947 2;
  • 72) 0.769 921 200 947 2 × 2 = 1 + 0.539 842 401 894 4;
  • 73) 0.539 842 401 894 4 × 2 = 1 + 0.079 684 803 788 8;
  • 74) 0.079 684 803 788 8 × 2 = 0 + 0.159 369 607 577 6;
  • 75) 0.159 369 607 577 6 × 2 = 0 + 0.318 739 215 155 2;
  • 76) 0.318 739 215 155 2 × 2 = 0 + 0.637 478 430 310 4;
  • 77) 0.637 478 430 310 4 × 2 = 1 + 0.274 956 860 620 8;
  • 78) 0.274 956 860 620 8 × 2 = 0 + 0.549 913 721 241 6;
  • 79) 0.549 913 721 241 6 × 2 = 1 + 0.099 827 442 483 2;
  • 80) 0.099 827 442 483 2 × 2 = 0 + 0.199 654 884 966 4;
  • 81) 0.199 654 884 966 4 × 2 = 0 + 0.399 309 769 932 8;
  • 82) 0.399 309 769 932 8 × 2 = 0 + 0.798 619 539 865 6;
  • 83) 0.798 619 539 865 6 × 2 = 1 + 0.597 239 079 731 2;
  • 84) 0.597 239 079 731 2 × 2 = 1 + 0.194 478 159 462 4;
  • 85) 0.194 478 159 462 4 × 2 = 0 + 0.388 956 318 924 8;
  • 86) 0.388 956 318 924 8 × 2 = 0 + 0.777 912 637 849 6;
  • 87) 0.777 912 637 849 6 × 2 = 1 + 0.555 825 275 699 2;
  • 88) 0.555 825 275 699 2 × 2 = 1 + 0.111 650 551 398 4;
  • 89) 0.111 650 551 398 4 × 2 = 0 + 0.223 301 102 796 8;

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:


0.000 000 000 008 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1001 1100 1001 0010 0000 1001 0110 0100 0001 1000 1010 0011 0011 0(2)

5. Positive number before normalization:

0.000 000 000 008 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1001 1100 1001 0010 0000 1001 0110 0100 0001 1000 1010 0011 0011 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 37 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 008 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1001 1100 1001 0010 0000 1001 0110 0100 0001 1000 1010 0011 0011 0(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1001 1100 1001 0010 0000 1001 0110 0100 0001 1000 1010 0011 0011 0(2) × 20 =

1.0011 1001 0010 0100 0001 0010 1100 1000 0011 0001 0100 0110 0110(2) × 2-37


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): -37

Mantissa (not normalized):
1.0011 1001 0010 0100 0001 0010 1100 1000 0011 0001 0100 0110 0110


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-37 + 2(11-1) - 1 =

(-37 + 1 023)(10) =

986(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;
  • 986 ÷ 2 = 493 + 0;
  • 493 ÷ 2 = 246 + 1;
  • 246 ÷ 2 = 123 + 0;
  • 123 ÷ 2 = 61 + 1;
  • 61 ÷ 2 = 30 + 1;
  • 30 ÷ 2 = 15 + 0;
  • 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) =

986(10) =

011 1101 1010(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. 0011 1001 0010 0100 0001 0010 1100 1000 0011 0001 0100 0110 0110 =

0011 1001 0010 0100 0001 0010 1100 1000 0011 0001 0100 0110 0110


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 1101 1010

Mantissa (52 bits) =
0011 1001 0010 0100 0001 0010 1100 1000 0011 0001 0100 0110 0110


Decimal number 0.000 000 000 008 9 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1101 1010 - 0011 1001 0010 0100 0001 0010 1100 1000 0011 0001 0100 0110 0110

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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