-0.000 035 666 873 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 035 666 873(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
-0.000 035 666 873(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. Start with the positive version of the number:

|-0.000 035 666 873| = 0.000 035 666 873


2. 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;

3. 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)


4. Convert to binary (base 2) the fractional part: 0.000 035 666 873.

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) 0.000 035 666 873 × 2 = 0 + 0.000 071 333 746;
  • 2) 0.000 071 333 746 × 2 = 0 + 0.000 142 667 492;
  • 3) 0.000 142 667 492 × 2 = 0 + 0.000 285 334 984;
  • 4) 0.000 285 334 984 × 2 = 0 + 0.000 570 669 968;
  • 5) 0.000 570 669 968 × 2 = 0 + 0.001 141 339 936;
  • 6) 0.001 141 339 936 × 2 = 0 + 0.002 282 679 872;
  • 7) 0.002 282 679 872 × 2 = 0 + 0.004 565 359 744;
  • 8) 0.004 565 359 744 × 2 = 0 + 0.009 130 719 488;
  • 9) 0.009 130 719 488 × 2 = 0 + 0.018 261 438 976;
  • 10) 0.018 261 438 976 × 2 = 0 + 0.036 522 877 952;
  • 11) 0.036 522 877 952 × 2 = 0 + 0.073 045 755 904;
  • 12) 0.073 045 755 904 × 2 = 0 + 0.146 091 511 808;
  • 13) 0.146 091 511 808 × 2 = 0 + 0.292 183 023 616;
  • 14) 0.292 183 023 616 × 2 = 0 + 0.584 366 047 232;
  • 15) 0.584 366 047 232 × 2 = 1 + 0.168 732 094 464;
  • 16) 0.168 732 094 464 × 2 = 0 + 0.337 464 188 928;
  • 17) 0.337 464 188 928 × 2 = 0 + 0.674 928 377 856;
  • 18) 0.674 928 377 856 × 2 = 1 + 0.349 856 755 712;
  • 19) 0.349 856 755 712 × 2 = 0 + 0.699 713 511 424;
  • 20) 0.699 713 511 424 × 2 = 1 + 0.399 427 022 848;
  • 21) 0.399 427 022 848 × 2 = 0 + 0.798 854 045 696;
  • 22) 0.798 854 045 696 × 2 = 1 + 0.597 708 091 392;
  • 23) 0.597 708 091 392 × 2 = 1 + 0.195 416 182 784;
  • 24) 0.195 416 182 784 × 2 = 0 + 0.390 832 365 568;
  • 25) 0.390 832 365 568 × 2 = 0 + 0.781 664 731 136;
  • 26) 0.781 664 731 136 × 2 = 1 + 0.563 329 462 272;
  • 27) 0.563 329 462 272 × 2 = 1 + 0.126 658 924 544;
  • 28) 0.126 658 924 544 × 2 = 0 + 0.253 317 849 088;
  • 29) 0.253 317 849 088 × 2 = 0 + 0.506 635 698 176;
  • 30) 0.506 635 698 176 × 2 = 1 + 0.013 271 396 352;
  • 31) 0.013 271 396 352 × 2 = 0 + 0.026 542 792 704;
  • 32) 0.026 542 792 704 × 2 = 0 + 0.053 085 585 408;
  • 33) 0.053 085 585 408 × 2 = 0 + 0.106 171 170 816;
  • 34) 0.106 171 170 816 × 2 = 0 + 0.212 342 341 632;
  • 35) 0.212 342 341 632 × 2 = 0 + 0.424 684 683 264;
  • 36) 0.424 684 683 264 × 2 = 0 + 0.849 369 366 528;
  • 37) 0.849 369 366 528 × 2 = 1 + 0.698 738 733 056;
  • 38) 0.698 738 733 056 × 2 = 1 + 0.397 477 466 112;
  • 39) 0.397 477 466 112 × 2 = 0 + 0.794 954 932 224;
  • 40) 0.794 954 932 224 × 2 = 1 + 0.589 909 864 448;
  • 41) 0.589 909 864 448 × 2 = 1 + 0.179 819 728 896;
  • 42) 0.179 819 728 896 × 2 = 0 + 0.359 639 457 792;
  • 43) 0.359 639 457 792 × 2 = 0 + 0.719 278 915 584;
  • 44) 0.719 278 915 584 × 2 = 1 + 0.438 557 831 168;
  • 45) 0.438 557 831 168 × 2 = 0 + 0.877 115 662 336;
  • 46) 0.877 115 662 336 × 2 = 1 + 0.754 231 324 672;
  • 47) 0.754 231 324 672 × 2 = 1 + 0.508 462 649 344;
  • 48) 0.508 462 649 344 × 2 = 1 + 0.016 925 298 688;
  • 49) 0.016 925 298 688 × 2 = 0 + 0.033 850 597 376;
  • 50) 0.033 850 597 376 × 2 = 0 + 0.067 701 194 752;
  • 51) 0.067 701 194 752 × 2 = 0 + 0.135 402 389 504;
  • 52) 0.135 402 389 504 × 2 = 0 + 0.270 804 779 008;
  • 53) 0.270 804 779 008 × 2 = 0 + 0.541 609 558 016;
  • 54) 0.541 609 558 016 × 2 = 1 + 0.083 219 116 032;
  • 55) 0.083 219 116 032 × 2 = 0 + 0.166 438 232 064;
  • 56) 0.166 438 232 064 × 2 = 0 + 0.332 876 464 128;
  • 57) 0.332 876 464 128 × 2 = 0 + 0.665 752 928 256;
  • 58) 0.665 752 928 256 × 2 = 1 + 0.331 505 856 512;
  • 59) 0.331 505 856 512 × 2 = 0 + 0.663 011 713 024;
  • 60) 0.663 011 713 024 × 2 = 1 + 0.326 023 426 048;
  • 61) 0.326 023 426 048 × 2 = 0 + 0.652 046 852 096;
  • 62) 0.652 046 852 096 × 2 = 1 + 0.304 093 704 192;
  • 63) 0.304 093 704 192 × 2 = 0 + 0.608 187 408 384;
  • 64) 0.608 187 408 384 × 2 = 1 + 0.216 374 816 768;
  • 65) 0.216 374 816 768 × 2 = 0 + 0.432 749 633 536;
  • 66) 0.432 749 633 536 × 2 = 0 + 0.865 499 267 072;
  • 67) 0.865 499 267 072 × 2 = 1 + 0.730 998 534 144;

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).


5. 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:


0.000 035 666 873(10) =

0.0000 0000 0000 0010 0101 0110 0110 0100 0000 1101 1001 0111 0000 0100 0101 0101 001(2)

6. Positive number before normalization:

0.000 035 666 873(10) =

0.0000 0000 0000 0010 0101 0110 0110 0100 0000 1101 1001 0111 0000 0100 0101 0101 001(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 15 positions to the right, so that only one non zero digit remains to the left of it:


0.000 035 666 873(10) =

0.0000 0000 0000 0010 0101 0110 0110 0100 0000 1101 1001 0111 0000 0100 0101 0101 001(2) =

0.0000 0000 0000 0010 0101 0110 0110 0100 0000 1101 1001 0111 0000 0100 0101 0101 001(2) × 20 =

1.0010 1011 0011 0010 0000 0110 1100 1011 1000 0010 0010 1010 1001(2) × 2-15


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): -15

Mantissa (not normalized):
1.0010 1011 0011 0010 0000 0110 1100 1011 1000 0010 0010 1010 1001


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-15 + 2(11-1) - 1 =

(-15 + 1 023)(10) =

1 008(10)


10. 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;
  • 1 008 ÷ 2 = 504 + 0;
  • 504 ÷ 2 = 252 + 0;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1008(10) =

011 1111 0000(2)


12. 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. 0010 1011 0011 0010 0000 0110 1100 1011 1000 0010 0010 1010 1001 =

0010 1011 0011 0010 0000 0110 1100 1011 1000 0010 0010 1010 1001


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
011 1111 0000

Mantissa (52 bits) =
0010 1011 0011 0010 0000 0110 1100 1011 1000 0010 0010 1010 1001


Decimal number -0.000 035 666 873 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0000 - 0010 1011 0011 0010 0000 0110 1100 1011 1000 0010 0010 1010 1001

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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