-0.000 323 037 664 299 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 323 037 664 299(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
-0.000 323 037 664 299(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:

|-0.000 323 037 664 299| = 0.000 323 037 664 299


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: 0.000 323 037 664 299.

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) 0.000 323 037 664 299 × 2 = 0 + 0.000 646 075 328 598;
  • 2) 0.000 646 075 328 598 × 2 = 0 + 0.001 292 150 657 196;
  • 3) 0.001 292 150 657 196 × 2 = 0 + 0.002 584 301 314 392;
  • 4) 0.002 584 301 314 392 × 2 = 0 + 0.005 168 602 628 784;
  • 5) 0.005 168 602 628 784 × 2 = 0 + 0.010 337 205 257 568;
  • 6) 0.010 337 205 257 568 × 2 = 0 + 0.020 674 410 515 136;
  • 7) 0.020 674 410 515 136 × 2 = 0 + 0.041 348 821 030 272;
  • 8) 0.041 348 821 030 272 × 2 = 0 + 0.082 697 642 060 544;
  • 9) 0.082 697 642 060 544 × 2 = 0 + 0.165 395 284 121 088;
  • 10) 0.165 395 284 121 088 × 2 = 0 + 0.330 790 568 242 176;
  • 11) 0.330 790 568 242 176 × 2 = 0 + 0.661 581 136 484 352;
  • 12) 0.661 581 136 484 352 × 2 = 1 + 0.323 162 272 968 704;
  • 13) 0.323 162 272 968 704 × 2 = 0 + 0.646 324 545 937 408;
  • 14) 0.646 324 545 937 408 × 2 = 1 + 0.292 649 091 874 816;
  • 15) 0.292 649 091 874 816 × 2 = 0 + 0.585 298 183 749 632;
  • 16) 0.585 298 183 749 632 × 2 = 1 + 0.170 596 367 499 264;
  • 17) 0.170 596 367 499 264 × 2 = 0 + 0.341 192 734 998 528;
  • 18) 0.341 192 734 998 528 × 2 = 0 + 0.682 385 469 997 056;
  • 19) 0.682 385 469 997 056 × 2 = 1 + 0.364 770 939 994 112;
  • 20) 0.364 770 939 994 112 × 2 = 0 + 0.729 541 879 988 224;
  • 21) 0.729 541 879 988 224 × 2 = 1 + 0.459 083 759 976 448;
  • 22) 0.459 083 759 976 448 × 2 = 0 + 0.918 167 519 952 896;
  • 23) 0.918 167 519 952 896 × 2 = 1 + 0.836 335 039 905 792;
  • 24) 0.836 335 039 905 792 × 2 = 1 + 0.672 670 079 811 584;
  • 25) 0.672 670 079 811 584 × 2 = 1 + 0.345 340 159 623 168;
  • 26) 0.345 340 159 623 168 × 2 = 0 + 0.690 680 319 246 336;
  • 27) 0.690 680 319 246 336 × 2 = 1 + 0.381 360 638 492 672;
  • 28) 0.381 360 638 492 672 × 2 = 0 + 0.762 721 276 985 344;
  • 29) 0.762 721 276 985 344 × 2 = 1 + 0.525 442 553 970 688;
  • 30) 0.525 442 553 970 688 × 2 = 1 + 0.050 885 107 941 376;
  • 31) 0.050 885 107 941 376 × 2 = 0 + 0.101 770 215 882 752;
  • 32) 0.101 770 215 882 752 × 2 = 0 + 0.203 540 431 765 504;
  • 33) 0.203 540 431 765 504 × 2 = 0 + 0.407 080 863 531 008;
  • 34) 0.407 080 863 531 008 × 2 = 0 + 0.814 161 727 062 016;
  • 35) 0.814 161 727 062 016 × 2 = 1 + 0.628 323 454 124 032;
  • 36) 0.628 323 454 124 032 × 2 = 1 + 0.256 646 908 248 064;
  • 37) 0.256 646 908 248 064 × 2 = 0 + 0.513 293 816 496 128;
  • 38) 0.513 293 816 496 128 × 2 = 1 + 0.026 587 632 992 256;
  • 39) 0.026 587 632 992 256 × 2 = 0 + 0.053 175 265 984 512;
  • 40) 0.053 175 265 984 512 × 2 = 0 + 0.106 350 531 969 024;
  • 41) 0.106 350 531 969 024 × 2 = 0 + 0.212 701 063 938 048;
  • 42) 0.212 701 063 938 048 × 2 = 0 + 0.425 402 127 876 096;
  • 43) 0.425 402 127 876 096 × 2 = 0 + 0.850 804 255 752 192;
  • 44) 0.850 804 255 752 192 × 2 = 1 + 0.701 608 511 504 384;
  • 45) 0.701 608 511 504 384 × 2 = 1 + 0.403 217 023 008 768;
  • 46) 0.403 217 023 008 768 × 2 = 0 + 0.806 434 046 017 536;
  • 47) 0.806 434 046 017 536 × 2 = 1 + 0.612 868 092 035 072;
  • 48) 0.612 868 092 035 072 × 2 = 1 + 0.225 736 184 070 144;
  • 49) 0.225 736 184 070 144 × 2 = 0 + 0.451 472 368 140 288;
  • 50) 0.451 472 368 140 288 × 2 = 0 + 0.902 944 736 280 576;
  • 51) 0.902 944 736 280 576 × 2 = 1 + 0.805 889 472 561 152;
  • 52) 0.805 889 472 561 152 × 2 = 1 + 0.611 778 945 122 304;
  • 53) 0.611 778 945 122 304 × 2 = 1 + 0.223 557 890 244 608;
  • 54) 0.223 557 890 244 608 × 2 = 0 + 0.447 115 780 489 216;
  • 55) 0.447 115 780 489 216 × 2 = 0 + 0.894 231 560 978 432;
  • 56) 0.894 231 560 978 432 × 2 = 1 + 0.788 463 121 956 864;
  • 57) 0.788 463 121 956 864 × 2 = 1 + 0.576 926 243 913 728;
  • 58) 0.576 926 243 913 728 × 2 = 1 + 0.153 852 487 827 456;
  • 59) 0.153 852 487 827 456 × 2 = 0 + 0.307 704 975 654 912;
  • 60) 0.307 704 975 654 912 × 2 = 0 + 0.615 409 951 309 824;
  • 61) 0.615 409 951 309 824 × 2 = 1 + 0.230 819 902 619 648;
  • 62) 0.230 819 902 619 648 × 2 = 0 + 0.461 639 805 239 296;
  • 63) 0.461 639 805 239 296 × 2 = 0 + 0.923 279 610 478 592;
  • 64) 0.923 279 610 478 592 × 2 = 1 + 0.846 559 220 957 184;

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:


0.000 323 037 664 299(10) =

0.0000 0000 0001 0101 0010 1011 1010 1100 0011 0100 0001 1011 0011 1001 1100 1001(2)

6. Positive number before normalization:

0.000 323 037 664 299(10) =

0.0000 0000 0001 0101 0010 1011 1010 1100 0011 0100 0001 1011 0011 1001 1100 1001(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 12 positions to the right, so that only one non zero digit remains to the left of it:


0.000 323 037 664 299(10) =

0.0000 0000 0001 0101 0010 1011 1010 1100 0011 0100 0001 1011 0011 1001 1100 1001(2) =

0.0000 0000 0001 0101 0010 1011 1010 1100 0011 0100 0001 1011 0011 1001 1100 1001(2) × 20 =

1.0101 0010 1011 1010 1100 0011 0100 0001 1011 0011 1001 1100 1001(2) × 2-12


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): -12

Mantissa (not normalized):
1.0101 0010 1011 1010 1100 0011 0100 0001 1011 0011 1001 1100 1001


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-12 + 2(11-1) - 1 =

(-12 + 1 023)(10) =

1 011(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;
  • 1 011 ÷ 2 = 505 + 1;
  • 505 ÷ 2 = 252 + 1;
  • 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) =

1011(10) =

011 1111 0011(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. 0101 0010 1011 1010 1100 0011 0100 0001 1011 0011 1001 1100 1001 =

0101 0010 1011 1010 1100 0011 0100 0001 1011 0011 1001 1100 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 0011

Mantissa (52 bits) =
0101 0010 1011 1010 1100 0011 0100 0001 1011 0011 1001 1100 1001


Decimal number -0.000 323 037 664 299 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0101 0010 1011 1010 1100 0011 0100 0001 1011 0011 1001 1100 1001

» Convert decimal -0.000 323 037 664 224 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 11, 2026 [November]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: November - by our visitors

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