-0.000 035 666 829 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 035 666 829(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
-0.000 035 666 829(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:

|-0.000 035 666 829| = 0.000 035 666 829


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: 0.000 035 666 829.

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) 0.000 035 666 829 × 2 = 0 + 0.000 071 333 658;
  • 2) 0.000 071 333 658 × 2 = 0 + 0.000 142 667 316;
  • 3) 0.000 142 667 316 × 2 = 0 + 0.000 285 334 632;
  • 4) 0.000 285 334 632 × 2 = 0 + 0.000 570 669 264;
  • 5) 0.000 570 669 264 × 2 = 0 + 0.001 141 338 528;
  • 6) 0.001 141 338 528 × 2 = 0 + 0.002 282 677 056;
  • 7) 0.002 282 677 056 × 2 = 0 + 0.004 565 354 112;
  • 8) 0.004 565 354 112 × 2 = 0 + 0.009 130 708 224;
  • 9) 0.009 130 708 224 × 2 = 0 + 0.018 261 416 448;
  • 10) 0.018 261 416 448 × 2 = 0 + 0.036 522 832 896;
  • 11) 0.036 522 832 896 × 2 = 0 + 0.073 045 665 792;
  • 12) 0.073 045 665 792 × 2 = 0 + 0.146 091 331 584;
  • 13) 0.146 091 331 584 × 2 = 0 + 0.292 182 663 168;
  • 14) 0.292 182 663 168 × 2 = 0 + 0.584 365 326 336;
  • 15) 0.584 365 326 336 × 2 = 1 + 0.168 730 652 672;
  • 16) 0.168 730 652 672 × 2 = 0 + 0.337 461 305 344;
  • 17) 0.337 461 305 344 × 2 = 0 + 0.674 922 610 688;
  • 18) 0.674 922 610 688 × 2 = 1 + 0.349 845 221 376;
  • 19) 0.349 845 221 376 × 2 = 0 + 0.699 690 442 752;
  • 20) 0.699 690 442 752 × 2 = 1 + 0.399 380 885 504;
  • 21) 0.399 380 885 504 × 2 = 0 + 0.798 761 771 008;
  • 22) 0.798 761 771 008 × 2 = 1 + 0.597 523 542 016;
  • 23) 0.597 523 542 016 × 2 = 1 + 0.195 047 084 032;
  • 24) 0.195 047 084 032 × 2 = 0 + 0.390 094 168 064;
  • 25) 0.390 094 168 064 × 2 = 0 + 0.780 188 336 128;
  • 26) 0.780 188 336 128 × 2 = 1 + 0.560 376 672 256;
  • 27) 0.560 376 672 256 × 2 = 1 + 0.120 753 344 512;
  • 28) 0.120 753 344 512 × 2 = 0 + 0.241 506 689 024;
  • 29) 0.241 506 689 024 × 2 = 0 + 0.483 013 378 048;
  • 30) 0.483 013 378 048 × 2 = 0 + 0.966 026 756 096;
  • 31) 0.966 026 756 096 × 2 = 1 + 0.932 053 512 192;
  • 32) 0.932 053 512 192 × 2 = 1 + 0.864 107 024 384;
  • 33) 0.864 107 024 384 × 2 = 1 + 0.728 214 048 768;
  • 34) 0.728 214 048 768 × 2 = 1 + 0.456 428 097 536;
  • 35) 0.456 428 097 536 × 2 = 0 + 0.912 856 195 072;
  • 36) 0.912 856 195 072 × 2 = 1 + 0.825 712 390 144;
  • 37) 0.825 712 390 144 × 2 = 1 + 0.651 424 780 288;
  • 38) 0.651 424 780 288 × 2 = 1 + 0.302 849 560 576;
  • 39) 0.302 849 560 576 × 2 = 0 + 0.605 699 121 152;
  • 40) 0.605 699 121 152 × 2 = 1 + 0.211 398 242 304;
  • 41) 0.211 398 242 304 × 2 = 0 + 0.422 796 484 608;
  • 42) 0.422 796 484 608 × 2 = 0 + 0.845 592 969 216;
  • 43) 0.845 592 969 216 × 2 = 1 + 0.691 185 938 432;
  • 44) 0.691 185 938 432 × 2 = 1 + 0.382 371 876 864;
  • 45) 0.382 371 876 864 × 2 = 0 + 0.764 743 753 728;
  • 46) 0.764 743 753 728 × 2 = 1 + 0.529 487 507 456;
  • 47) 0.529 487 507 456 × 2 = 1 + 0.058 975 014 912;
  • 48) 0.058 975 014 912 × 2 = 0 + 0.117 950 029 824;
  • 49) 0.117 950 029 824 × 2 = 0 + 0.235 900 059 648;
  • 50) 0.235 900 059 648 × 2 = 0 + 0.471 800 119 296;
  • 51) 0.471 800 119 296 × 2 = 0 + 0.943 600 238 592;
  • 52) 0.943 600 238 592 × 2 = 1 + 0.887 200 477 184;
  • 53) 0.887 200 477 184 × 2 = 1 + 0.774 400 954 368;
  • 54) 0.774 400 954 368 × 2 = 1 + 0.548 801 908 736;
  • 55) 0.548 801 908 736 × 2 = 1 + 0.097 603 817 472;
  • 56) 0.097 603 817 472 × 2 = 0 + 0.195 207 634 944;
  • 57) 0.195 207 634 944 × 2 = 0 + 0.390 415 269 888;
  • 58) 0.390 415 269 888 × 2 = 0 + 0.780 830 539 776;
  • 59) 0.780 830 539 776 × 2 = 1 + 0.561 661 079 552;
  • 60) 0.561 661 079 552 × 2 = 1 + 0.123 322 159 104;
  • 61) 0.123 322 159 104 × 2 = 0 + 0.246 644 318 208;
  • 62) 0.246 644 318 208 × 2 = 0 + 0.493 288 636 416;
  • 63) 0.493 288 636 416 × 2 = 0 + 0.986 577 272 832;
  • 64) 0.986 577 272 832 × 2 = 1 + 0.973 154 545 664;
  • 65) 0.973 154 545 664 × 2 = 1 + 0.946 309 091 328;
  • 66) 0.946 309 091 328 × 2 = 1 + 0.892 618 182 656;
  • 67) 0.892 618 182 656 × 2 = 1 + 0.785 236 365 312;

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:


0.000 035 666 829(10) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1101 1101 0011 0110 0001 1110 0011 0001 111(2)

6. Positive number before normalization:

0.000 035 666 829(10) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1101 1101 0011 0110 0001 1110 0011 0001 111(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:


0.000 035 666 829(10) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1101 1101 0011 0110 0001 1110 0011 0001 111(2) =

0.0000 0000 0000 0010 0101 0110 0110 0011 1101 1101 0011 0110 0001 1110 0011 0001 111(2) × 20 =

1.0010 1011 0011 0001 1110 1110 1001 1011 0000 1111 0001 1000 1111(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 1110 1110 1001 1011 0000 1111 0001 1000 1111


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 1110 1110 1001 1011 0000 1111 0001 1000 1111 =

0010 1011 0011 0001 1110 1110 1001 1011 0000 1111 0001 1000 1111


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 1110 1110 1001 1011 0000 1111 0001 1000 1111


Decimal number -0.000 035 666 829 converted to 64 bit double precision IEEE 754 binary floating point representation:

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

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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