-0.000 035 666 835 509 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 035 666 835 509(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
-0.000 035 666 835 509(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:

|-0.000 035 666 835 509| = 0.000 035 666 835 509


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: 0.000 035 666 835 509.

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) 0.000 035 666 835 509 × 2 = 0 + 0.000 071 333 671 018;
  • 2) 0.000 071 333 671 018 × 2 = 0 + 0.000 142 667 342 036;
  • 3) 0.000 142 667 342 036 × 2 = 0 + 0.000 285 334 684 072;
  • 4) 0.000 285 334 684 072 × 2 = 0 + 0.000 570 669 368 144;
  • 5) 0.000 570 669 368 144 × 2 = 0 + 0.001 141 338 736 288;
  • 6) 0.001 141 338 736 288 × 2 = 0 + 0.002 282 677 472 576;
  • 7) 0.002 282 677 472 576 × 2 = 0 + 0.004 565 354 945 152;
  • 8) 0.004 565 354 945 152 × 2 = 0 + 0.009 130 709 890 304;
  • 9) 0.009 130 709 890 304 × 2 = 0 + 0.018 261 419 780 608;
  • 10) 0.018 261 419 780 608 × 2 = 0 + 0.036 522 839 561 216;
  • 11) 0.036 522 839 561 216 × 2 = 0 + 0.073 045 679 122 432;
  • 12) 0.073 045 679 122 432 × 2 = 0 + 0.146 091 358 244 864;
  • 13) 0.146 091 358 244 864 × 2 = 0 + 0.292 182 716 489 728;
  • 14) 0.292 182 716 489 728 × 2 = 0 + 0.584 365 432 979 456;
  • 15) 0.584 365 432 979 456 × 2 = 1 + 0.168 730 865 958 912;
  • 16) 0.168 730 865 958 912 × 2 = 0 + 0.337 461 731 917 824;
  • 17) 0.337 461 731 917 824 × 2 = 0 + 0.674 923 463 835 648;
  • 18) 0.674 923 463 835 648 × 2 = 1 + 0.349 846 927 671 296;
  • 19) 0.349 846 927 671 296 × 2 = 0 + 0.699 693 855 342 592;
  • 20) 0.699 693 855 342 592 × 2 = 1 + 0.399 387 710 685 184;
  • 21) 0.399 387 710 685 184 × 2 = 0 + 0.798 775 421 370 368;
  • 22) 0.798 775 421 370 368 × 2 = 1 + 0.597 550 842 740 736;
  • 23) 0.597 550 842 740 736 × 2 = 1 + 0.195 101 685 481 472;
  • 24) 0.195 101 685 481 472 × 2 = 0 + 0.390 203 370 962 944;
  • 25) 0.390 203 370 962 944 × 2 = 0 + 0.780 406 741 925 888;
  • 26) 0.780 406 741 925 888 × 2 = 1 + 0.560 813 483 851 776;
  • 27) 0.560 813 483 851 776 × 2 = 1 + 0.121 626 967 703 552;
  • 28) 0.121 626 967 703 552 × 2 = 0 + 0.243 253 935 407 104;
  • 29) 0.243 253 935 407 104 × 2 = 0 + 0.486 507 870 814 208;
  • 30) 0.486 507 870 814 208 × 2 = 0 + 0.973 015 741 628 416;
  • 31) 0.973 015 741 628 416 × 2 = 1 + 0.946 031 483 256 832;
  • 32) 0.946 031 483 256 832 × 2 = 1 + 0.892 062 966 513 664;
  • 33) 0.892 062 966 513 664 × 2 = 1 + 0.784 125 933 027 328;
  • 34) 0.784 125 933 027 328 × 2 = 1 + 0.568 251 866 054 656;
  • 35) 0.568 251 866 054 656 × 2 = 1 + 0.136 503 732 109 312;
  • 36) 0.136 503 732 109 312 × 2 = 0 + 0.273 007 464 218 624;
  • 37) 0.273 007 464 218 624 × 2 = 0 + 0.546 014 928 437 248;
  • 38) 0.546 014 928 437 248 × 2 = 1 + 0.092 029 856 874 496;
  • 39) 0.092 029 856 874 496 × 2 = 0 + 0.184 059 713 748 992;
  • 40) 0.184 059 713 748 992 × 2 = 0 + 0.368 119 427 497 984;
  • 41) 0.368 119 427 497 984 × 2 = 0 + 0.736 238 854 995 968;
  • 42) 0.736 238 854 995 968 × 2 = 1 + 0.472 477 709 991 936;
  • 43) 0.472 477 709 991 936 × 2 = 0 + 0.944 955 419 983 872;
  • 44) 0.944 955 419 983 872 × 2 = 1 + 0.889 910 839 967 744;
  • 45) 0.889 910 839 967 744 × 2 = 1 + 0.779 821 679 935 488;
  • 46) 0.779 821 679 935 488 × 2 = 1 + 0.559 643 359 870 976;
  • 47) 0.559 643 359 870 976 × 2 = 1 + 0.119 286 719 741 952;
  • 48) 0.119 286 719 741 952 × 2 = 0 + 0.238 573 439 483 904;
  • 49) 0.238 573 439 483 904 × 2 = 0 + 0.477 146 878 967 808;
  • 50) 0.477 146 878 967 808 × 2 = 0 + 0.954 293 757 935 616;
  • 51) 0.954 293 757 935 616 × 2 = 1 + 0.908 587 515 871 232;
  • 52) 0.908 587 515 871 232 × 2 = 1 + 0.817 175 031 742 464;
  • 53) 0.817 175 031 742 464 × 2 = 1 + 0.634 350 063 484 928;
  • 54) 0.634 350 063 484 928 × 2 = 1 + 0.268 700 126 969 856;
  • 55) 0.268 700 126 969 856 × 2 = 0 + 0.537 400 253 939 712;
  • 56) 0.537 400 253 939 712 × 2 = 1 + 0.074 800 507 879 424;
  • 57) 0.074 800 507 879 424 × 2 = 0 + 0.149 601 015 758 848;
  • 58) 0.149 601 015 758 848 × 2 = 0 + 0.299 202 031 517 696;
  • 59) 0.299 202 031 517 696 × 2 = 0 + 0.598 404 063 035 392;
  • 60) 0.598 404 063 035 392 × 2 = 1 + 0.196 808 126 070 784;
  • 61) 0.196 808 126 070 784 × 2 = 0 + 0.393 616 252 141 568;
  • 62) 0.393 616 252 141 568 × 2 = 0 + 0.787 232 504 283 136;
  • 63) 0.787 232 504 283 136 × 2 = 1 + 0.574 465 008 566 272;
  • 64) 0.574 465 008 566 272 × 2 = 1 + 0.148 930 017 132 544;
  • 65) 0.148 930 017 132 544 × 2 = 0 + 0.297 860 034 265 088;
  • 66) 0.297 860 034 265 088 × 2 = 0 + 0.595 720 068 530 176;
  • 67) 0.595 720 068 530 176 × 2 = 1 + 0.191 440 137 060 352;

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:


0.000 035 666 835 509(10) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1110 0100 0101 1110 0011 1101 0001 0011 001(2)

6. Positive number before normalization:

0.000 035 666 835 509(10) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1110 0100 0101 1110 0011 1101 0001 0011 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:


0.000 035 666 835 509(10) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1110 0100 0101 1110 0011 1101 0001 0011 001(2) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1110 0100 0101 1110 0011 1101 0001 0011 001(2) × 20 =

1.0010 1011 0011 0001 1111 0010 0010 1111 0001 1110 1000 1001 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 0001 1111 0010 0010 1111 0001 1110 1000 1001 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 0001 1111 0010 0010 1111 0001 1110 1000 1001 1001 =

0010 1011 0011 0001 1111 0010 0010 1111 0001 1110 1000 1001 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 0001 1111 0010 0010 1111 0001 1110 1000 1001 1001


Decimal number -0.000 035 666 835 509 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0000 - 0010 1011 0011 0001 1111 0010 0010 1111 0001 1110 1000 1001 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