-0.000 282 005 664 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 664(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
-0.000 282 005 664(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:

|-0.000 282 005 664| = 0.000 282 005 664


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: 0.000 282 005 664.

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) 0.000 282 005 664 × 2 = 0 + 0.000 564 011 328;
  • 2) 0.000 564 011 328 × 2 = 0 + 0.001 128 022 656;
  • 3) 0.001 128 022 656 × 2 = 0 + 0.002 256 045 312;
  • 4) 0.002 256 045 312 × 2 = 0 + 0.004 512 090 624;
  • 5) 0.004 512 090 624 × 2 = 0 + 0.009 024 181 248;
  • 6) 0.009 024 181 248 × 2 = 0 + 0.018 048 362 496;
  • 7) 0.018 048 362 496 × 2 = 0 + 0.036 096 724 992;
  • 8) 0.036 096 724 992 × 2 = 0 + 0.072 193 449 984;
  • 9) 0.072 193 449 984 × 2 = 0 + 0.144 386 899 968;
  • 10) 0.144 386 899 968 × 2 = 0 + 0.288 773 799 936;
  • 11) 0.288 773 799 936 × 2 = 0 + 0.577 547 599 872;
  • 12) 0.577 547 599 872 × 2 = 1 + 0.155 095 199 744;
  • 13) 0.155 095 199 744 × 2 = 0 + 0.310 190 399 488;
  • 14) 0.310 190 399 488 × 2 = 0 + 0.620 380 798 976;
  • 15) 0.620 380 798 976 × 2 = 1 + 0.240 761 597 952;
  • 16) 0.240 761 597 952 × 2 = 0 + 0.481 523 195 904;
  • 17) 0.481 523 195 904 × 2 = 0 + 0.963 046 391 808;
  • 18) 0.963 046 391 808 × 2 = 1 + 0.926 092 783 616;
  • 19) 0.926 092 783 616 × 2 = 1 + 0.852 185 567 232;
  • 20) 0.852 185 567 232 × 2 = 1 + 0.704 371 134 464;
  • 21) 0.704 371 134 464 × 2 = 1 + 0.408 742 268 928;
  • 22) 0.408 742 268 928 × 2 = 0 + 0.817 484 537 856;
  • 23) 0.817 484 537 856 × 2 = 1 + 0.634 969 075 712;
  • 24) 0.634 969 075 712 × 2 = 1 + 0.269 938 151 424;
  • 25) 0.269 938 151 424 × 2 = 0 + 0.539 876 302 848;
  • 26) 0.539 876 302 848 × 2 = 1 + 0.079 752 605 696;
  • 27) 0.079 752 605 696 × 2 = 0 + 0.159 505 211 392;
  • 28) 0.159 505 211 392 × 2 = 0 + 0.319 010 422 784;
  • 29) 0.319 010 422 784 × 2 = 0 + 0.638 020 845 568;
  • 30) 0.638 020 845 568 × 2 = 1 + 0.276 041 691 136;
  • 31) 0.276 041 691 136 × 2 = 0 + 0.552 083 382 272;
  • 32) 0.552 083 382 272 × 2 = 1 + 0.104 166 764 544;
  • 33) 0.104 166 764 544 × 2 = 0 + 0.208 333 529 088;
  • 34) 0.208 333 529 088 × 2 = 0 + 0.416 667 058 176;
  • 35) 0.416 667 058 176 × 2 = 0 + 0.833 334 116 352;
  • 36) 0.833 334 116 352 × 2 = 1 + 0.666 668 232 704;
  • 37) 0.666 668 232 704 × 2 = 1 + 0.333 336 465 408;
  • 38) 0.333 336 465 408 × 2 = 0 + 0.666 672 930 816;
  • 39) 0.666 672 930 816 × 2 = 1 + 0.333 345 861 632;
  • 40) 0.333 345 861 632 × 2 = 0 + 0.666 691 723 264;
  • 41) 0.666 691 723 264 × 2 = 1 + 0.333 383 446 528;
  • 42) 0.333 383 446 528 × 2 = 0 + 0.666 766 893 056;
  • 43) 0.666 766 893 056 × 2 = 1 + 0.333 533 786 112;
  • 44) 0.333 533 786 112 × 2 = 0 + 0.667 067 572 224;
  • 45) 0.667 067 572 224 × 2 = 1 + 0.334 135 144 448;
  • 46) 0.334 135 144 448 × 2 = 0 + 0.668 270 288 896;
  • 47) 0.668 270 288 896 × 2 = 1 + 0.336 540 577 792;
  • 48) 0.336 540 577 792 × 2 = 0 + 0.673 081 155 584;
  • 49) 0.673 081 155 584 × 2 = 1 + 0.346 162 311 168;
  • 50) 0.346 162 311 168 × 2 = 0 + 0.692 324 622 336;
  • 51) 0.692 324 622 336 × 2 = 1 + 0.384 649 244 672;
  • 52) 0.384 649 244 672 × 2 = 0 + 0.769 298 489 344;
  • 53) 0.769 298 489 344 × 2 = 1 + 0.538 596 978 688;
  • 54) 0.538 596 978 688 × 2 = 1 + 0.077 193 957 376;
  • 55) 0.077 193 957 376 × 2 = 0 + 0.154 387 914 752;
  • 56) 0.154 387 914 752 × 2 = 0 + 0.308 775 829 504;
  • 57) 0.308 775 829 504 × 2 = 0 + 0.617 551 659 008;
  • 58) 0.617 551 659 008 × 2 = 1 + 0.235 103 318 016;
  • 59) 0.235 103 318 016 × 2 = 0 + 0.470 206 636 032;
  • 60) 0.470 206 636 032 × 2 = 0 + 0.940 413 272 064;
  • 61) 0.940 413 272 064 × 2 = 1 + 0.880 826 544 128;
  • 62) 0.880 826 544 128 × 2 = 1 + 0.761 653 088 256;
  • 63) 0.761 653 088 256 × 2 = 1 + 0.523 306 176 512;
  • 64) 0.523 306 176 512 × 2 = 1 + 0.046 612 353 024;

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:


0.000 282 005 664(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0001 1010 1010 1010 1010 1100 0100 1111(2)

6. Positive number before normalization:

0.000 282 005 664(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0001 1010 1010 1010 1010 1100 0100 1111(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:


0.000 282 005 664(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0001 1010 1010 1010 1010 1100 0100 1111(2) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0001 1010 1010 1010 1010 1100 0100 1111(2) × 20 =

1.0010 0111 1011 0100 0101 0001 1010 1010 1010 1010 1100 0100 1111(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):
1.0010 0111 1011 0100 0101 0001 1010 1010 1010 1010 1100 0100 1111


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. 0010 0111 1011 0100 0101 0001 1010 1010 1010 1010 1100 0100 1111 =

0010 0111 1011 0100 0101 0001 1010 1010 1010 1010 1100 0100 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 0011

Mantissa (52 bits) =
0010 0111 1011 0100 0101 0001 1010 1010 1010 1010 1100 0100 1111


Decimal number -0.000 282 005 664 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0101 0001 1010 1010 1010 1010 1100 0100 1111

Share

» Convert decimal -0.000 282 005 667 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 05, 2026 [May]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: The Hidden Requirement in Every Single Job Description. NotebookLM YouTube Video of 8.15 minutes

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