-0.000 282 008 33 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 008 33(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
-0.000 282 008 33(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:

|-0.000 282 008 33| = 0.000 282 008 33


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: 0.000 282 008 33.

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) 0.000 282 008 33 × 2 = 0 + 0.000 564 016 66;
  • 2) 0.000 564 016 66 × 2 = 0 + 0.001 128 033 32;
  • 3) 0.001 128 033 32 × 2 = 0 + 0.002 256 066 64;
  • 4) 0.002 256 066 64 × 2 = 0 + 0.004 512 133 28;
  • 5) 0.004 512 133 28 × 2 = 0 + 0.009 024 266 56;
  • 6) 0.009 024 266 56 × 2 = 0 + 0.018 048 533 12;
  • 7) 0.018 048 533 12 × 2 = 0 + 0.036 097 066 24;
  • 8) 0.036 097 066 24 × 2 = 0 + 0.072 194 132 48;
  • 9) 0.072 194 132 48 × 2 = 0 + 0.144 388 264 96;
  • 10) 0.144 388 264 96 × 2 = 0 + 0.288 776 529 92;
  • 11) 0.288 776 529 92 × 2 = 0 + 0.577 553 059 84;
  • 12) 0.577 553 059 84 × 2 = 1 + 0.155 106 119 68;
  • 13) 0.155 106 119 68 × 2 = 0 + 0.310 212 239 36;
  • 14) 0.310 212 239 36 × 2 = 0 + 0.620 424 478 72;
  • 15) 0.620 424 478 72 × 2 = 1 + 0.240 848 957 44;
  • 16) 0.240 848 957 44 × 2 = 0 + 0.481 697 914 88;
  • 17) 0.481 697 914 88 × 2 = 0 + 0.963 395 829 76;
  • 18) 0.963 395 829 76 × 2 = 1 + 0.926 791 659 52;
  • 19) 0.926 791 659 52 × 2 = 1 + 0.853 583 319 04;
  • 20) 0.853 583 319 04 × 2 = 1 + 0.707 166 638 08;
  • 21) 0.707 166 638 08 × 2 = 1 + 0.414 333 276 16;
  • 22) 0.414 333 276 16 × 2 = 0 + 0.828 666 552 32;
  • 23) 0.828 666 552 32 × 2 = 1 + 0.657 333 104 64;
  • 24) 0.657 333 104 64 × 2 = 1 + 0.314 666 209 28;
  • 25) 0.314 666 209 28 × 2 = 0 + 0.629 332 418 56;
  • 26) 0.629 332 418 56 × 2 = 1 + 0.258 664 837 12;
  • 27) 0.258 664 837 12 × 2 = 0 + 0.517 329 674 24;
  • 28) 0.517 329 674 24 × 2 = 1 + 0.034 659 348 48;
  • 29) 0.034 659 348 48 × 2 = 0 + 0.069 318 696 96;
  • 30) 0.069 318 696 96 × 2 = 0 + 0.138 637 393 92;
  • 31) 0.138 637 393 92 × 2 = 0 + 0.277 274 787 84;
  • 32) 0.277 274 787 84 × 2 = 0 + 0.554 549 575 68;
  • 33) 0.554 549 575 68 × 2 = 1 + 0.109 099 151 36;
  • 34) 0.109 099 151 36 × 2 = 0 + 0.218 198 302 72;
  • 35) 0.218 198 302 72 × 2 = 0 + 0.436 396 605 44;
  • 36) 0.436 396 605 44 × 2 = 0 + 0.872 793 210 88;
  • 37) 0.872 793 210 88 × 2 = 1 + 0.745 586 421 76;
  • 38) 0.745 586 421 76 × 2 = 1 + 0.491 172 843 52;
  • 39) 0.491 172 843 52 × 2 = 0 + 0.982 345 687 04;
  • 40) 0.982 345 687 04 × 2 = 1 + 0.964 691 374 08;
  • 41) 0.964 691 374 08 × 2 = 1 + 0.929 382 748 16;
  • 42) 0.929 382 748 16 × 2 = 1 + 0.858 765 496 32;
  • 43) 0.858 765 496 32 × 2 = 1 + 0.717 530 992 64;
  • 44) 0.717 530 992 64 × 2 = 1 + 0.435 061 985 28;
  • 45) 0.435 061 985 28 × 2 = 0 + 0.870 123 970 56;
  • 46) 0.870 123 970 56 × 2 = 1 + 0.740 247 941 12;
  • 47) 0.740 247 941 12 × 2 = 1 + 0.480 495 882 24;
  • 48) 0.480 495 882 24 × 2 = 0 + 0.960 991 764 48;
  • 49) 0.960 991 764 48 × 2 = 1 + 0.921 983 528 96;
  • 50) 0.921 983 528 96 × 2 = 1 + 0.843 967 057 92;
  • 51) 0.843 967 057 92 × 2 = 1 + 0.687 934 115 84;
  • 52) 0.687 934 115 84 × 2 = 1 + 0.375 868 231 68;
  • 53) 0.375 868 231 68 × 2 = 0 + 0.751 736 463 36;
  • 54) 0.751 736 463 36 × 2 = 1 + 0.503 472 926 72;
  • 55) 0.503 472 926 72 × 2 = 1 + 0.006 945 853 44;
  • 56) 0.006 945 853 44 × 2 = 0 + 0.013 891 706 88;
  • 57) 0.013 891 706 88 × 2 = 0 + 0.027 783 413 76;
  • 58) 0.027 783 413 76 × 2 = 0 + 0.055 566 827 52;
  • 59) 0.055 566 827 52 × 2 = 0 + 0.111 133 655 04;
  • 60) 0.111 133 655 04 × 2 = 0 + 0.222 267 310 08;
  • 61) 0.222 267 310 08 × 2 = 0 + 0.444 534 620 16;
  • 62) 0.444 534 620 16 × 2 = 0 + 0.889 069 240 32;
  • 63) 0.889 069 240 32 × 2 = 1 + 0.778 138 480 64;
  • 64) 0.778 138 480 64 × 2 = 1 + 0.556 276 961 28;

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:


0.000 282 008 33(10) =

0.0000 0000 0001 0010 0111 1011 0101 0000 1000 1101 1111 0110 1111 0110 0000 0011(2)

6. Positive number before normalization:

0.000 282 008 33(10) =

0.0000 0000 0001 0010 0111 1011 0101 0000 1000 1101 1111 0110 1111 0110 0000 0011(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:


0.000 282 008 33(10) =

0.0000 0000 0001 0010 0111 1011 0101 0000 1000 1101 1111 0110 1111 0110 0000 0011(2) =

0.0000 0000 0001 0010 0111 1011 0101 0000 1000 1101 1111 0110 1111 0110 0000 0011(2) × 20 =

1.0010 0111 1011 0101 0000 1000 1101 1111 0110 1111 0110 0000 0011(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 0101 0000 1000 1101 1111 0110 1111 0110 0000 0011


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 0101 0000 1000 1101 1111 0110 1111 0110 0000 0011 =

0010 0111 1011 0101 0000 1000 1101 1111 0110 1111 0110 0000 0011


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 0101 0000 1000 1101 1111 0110 1111 0110 0000 0011


Decimal number -0.000 282 008 33 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0101 0000 1000 1101 1111 0110 1111 0110 0000 0011

Share

» Convert decimal -0.000 282 008 63 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: Your greatest rival, your most powerful ally. NotebookLM YouTube Video of 7.54 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