-0.000 035 666 906 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 035 666 906(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
-0.000 035 666 906(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:

|-0.000 035 666 906| = 0.000 035 666 906


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: 0.000 035 666 906.

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) 0.000 035 666 906 × 2 = 0 + 0.000 071 333 812;
  • 2) 0.000 071 333 812 × 2 = 0 + 0.000 142 667 624;
  • 3) 0.000 142 667 624 × 2 = 0 + 0.000 285 335 248;
  • 4) 0.000 285 335 248 × 2 = 0 + 0.000 570 670 496;
  • 5) 0.000 570 670 496 × 2 = 0 + 0.001 141 340 992;
  • 6) 0.001 141 340 992 × 2 = 0 + 0.002 282 681 984;
  • 7) 0.002 282 681 984 × 2 = 0 + 0.004 565 363 968;
  • 8) 0.004 565 363 968 × 2 = 0 + 0.009 130 727 936;
  • 9) 0.009 130 727 936 × 2 = 0 + 0.018 261 455 872;
  • 10) 0.018 261 455 872 × 2 = 0 + 0.036 522 911 744;
  • 11) 0.036 522 911 744 × 2 = 0 + 0.073 045 823 488;
  • 12) 0.073 045 823 488 × 2 = 0 + 0.146 091 646 976;
  • 13) 0.146 091 646 976 × 2 = 0 + 0.292 183 293 952;
  • 14) 0.292 183 293 952 × 2 = 0 + 0.584 366 587 904;
  • 15) 0.584 366 587 904 × 2 = 1 + 0.168 733 175 808;
  • 16) 0.168 733 175 808 × 2 = 0 + 0.337 466 351 616;
  • 17) 0.337 466 351 616 × 2 = 0 + 0.674 932 703 232;
  • 18) 0.674 932 703 232 × 2 = 1 + 0.349 865 406 464;
  • 19) 0.349 865 406 464 × 2 = 0 + 0.699 730 812 928;
  • 20) 0.699 730 812 928 × 2 = 1 + 0.399 461 625 856;
  • 21) 0.399 461 625 856 × 2 = 0 + 0.798 923 251 712;
  • 22) 0.798 923 251 712 × 2 = 1 + 0.597 846 503 424;
  • 23) 0.597 846 503 424 × 2 = 1 + 0.195 693 006 848;
  • 24) 0.195 693 006 848 × 2 = 0 + 0.391 386 013 696;
  • 25) 0.391 386 013 696 × 2 = 0 + 0.782 772 027 392;
  • 26) 0.782 772 027 392 × 2 = 1 + 0.565 544 054 784;
  • 27) 0.565 544 054 784 × 2 = 1 + 0.131 088 109 568;
  • 28) 0.131 088 109 568 × 2 = 0 + 0.262 176 219 136;
  • 29) 0.262 176 219 136 × 2 = 0 + 0.524 352 438 272;
  • 30) 0.524 352 438 272 × 2 = 1 + 0.048 704 876 544;
  • 31) 0.048 704 876 544 × 2 = 0 + 0.097 409 753 088;
  • 32) 0.097 409 753 088 × 2 = 0 + 0.194 819 506 176;
  • 33) 0.194 819 506 176 × 2 = 0 + 0.389 639 012 352;
  • 34) 0.389 639 012 352 × 2 = 0 + 0.779 278 024 704;
  • 35) 0.779 278 024 704 × 2 = 1 + 0.558 556 049 408;
  • 36) 0.558 556 049 408 × 2 = 1 + 0.117 112 098 816;
  • 37) 0.117 112 098 816 × 2 = 0 + 0.234 224 197 632;
  • 38) 0.234 224 197 632 × 2 = 0 + 0.468 448 395 264;
  • 39) 0.468 448 395 264 × 2 = 0 + 0.936 896 790 528;
  • 40) 0.936 896 790 528 × 2 = 1 + 0.873 793 581 056;
  • 41) 0.873 793 581 056 × 2 = 1 + 0.747 587 162 112;
  • 42) 0.747 587 162 112 × 2 = 1 + 0.495 174 324 224;
  • 43) 0.495 174 324 224 × 2 = 0 + 0.990 348 648 448;
  • 44) 0.990 348 648 448 × 2 = 1 + 0.980 697 296 896;
  • 45) 0.980 697 296 896 × 2 = 1 + 0.961 394 593 792;
  • 46) 0.961 394 593 792 × 2 = 1 + 0.922 789 187 584;
  • 47) 0.922 789 187 584 × 2 = 1 + 0.845 578 375 168;
  • 48) 0.845 578 375 168 × 2 = 1 + 0.691 156 750 336;
  • 49) 0.691 156 750 336 × 2 = 1 + 0.382 313 500 672;
  • 50) 0.382 313 500 672 × 2 = 0 + 0.764 627 001 344;
  • 51) 0.764 627 001 344 × 2 = 1 + 0.529 254 002 688;
  • 52) 0.529 254 002 688 × 2 = 1 + 0.058 508 005 376;
  • 53) 0.058 508 005 376 × 2 = 0 + 0.117 016 010 752;
  • 54) 0.117 016 010 752 × 2 = 0 + 0.234 032 021 504;
  • 55) 0.234 032 021 504 × 2 = 0 + 0.468 064 043 008;
  • 56) 0.468 064 043 008 × 2 = 0 + 0.936 128 086 016;
  • 57) 0.936 128 086 016 × 2 = 1 + 0.872 256 172 032;
  • 58) 0.872 256 172 032 × 2 = 1 + 0.744 512 344 064;
  • 59) 0.744 512 344 064 × 2 = 1 + 0.489 024 688 128;
  • 60) 0.489 024 688 128 × 2 = 0 + 0.978 049 376 256;
  • 61) 0.978 049 376 256 × 2 = 1 + 0.956 098 752 512;
  • 62) 0.956 098 752 512 × 2 = 1 + 0.912 197 505 024;
  • 63) 0.912 197 505 024 × 2 = 1 + 0.824 395 010 048;
  • 64) 0.824 395 010 048 × 2 = 1 + 0.648 790 020 096;
  • 65) 0.648 790 020 096 × 2 = 1 + 0.297 580 040 192;
  • 66) 0.297 580 040 192 × 2 = 0 + 0.595 160 080 384;
  • 67) 0.595 160 080 384 × 2 = 1 + 0.190 320 160 768;

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:


0.000 035 666 906(10) =

0.0000 0000 0000 0010 0101 0110 0110 0100 0011 0001 1101 1111 1011 0000 1110 1111 101(2)

6. Positive number before normalization:

0.000 035 666 906(10) =

0.0000 0000 0000 0010 0101 0110 0110 0100 0011 0001 1101 1111 1011 0000 1110 1111 101(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:


0.000 035 666 906(10) =

0.0000 0000 0000 0010 0101 0110 0110 0100 0011 0001 1101 1111 1011 0000 1110 1111 101(2) =

0.0000 0000 0000 0010 0101 0110 0110 0100 0011 0001 1101 1111 1011 0000 1110 1111 101(2) × 20 =

1.0010 1011 0011 0010 0001 1000 1110 1111 1101 1000 0111 0111 1101(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 0001 1000 1110 1111 1101 1000 0111 0111 1101


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 0001 1000 1110 1111 1101 1000 0111 0111 1101 =

0010 1011 0011 0010 0001 1000 1110 1111 1101 1000 0111 0111 1101


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 0001 1000 1110 1111 1101 1000 0111 0111 1101


Decimal number -0.000 035 666 906 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0000 - 0010 1011 0011 0010 0001 1000 1110 1111 1101 1000 0111 0111 1101

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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