-0.000 282 109 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 109(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
-0.000 282 109(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:

|-0.000 282 109| = 0.000 282 109


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: 0.000 282 109.

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) 0.000 282 109 × 2 = 0 + 0.000 564 218;
  • 2) 0.000 564 218 × 2 = 0 + 0.001 128 436;
  • 3) 0.001 128 436 × 2 = 0 + 0.002 256 872;
  • 4) 0.002 256 872 × 2 = 0 + 0.004 513 744;
  • 5) 0.004 513 744 × 2 = 0 + 0.009 027 488;
  • 6) 0.009 027 488 × 2 = 0 + 0.018 054 976;
  • 7) 0.018 054 976 × 2 = 0 + 0.036 109 952;
  • 8) 0.036 109 952 × 2 = 0 + 0.072 219 904;
  • 9) 0.072 219 904 × 2 = 0 + 0.144 439 808;
  • 10) 0.144 439 808 × 2 = 0 + 0.288 879 616;
  • 11) 0.288 879 616 × 2 = 0 + 0.577 759 232;
  • 12) 0.577 759 232 × 2 = 1 + 0.155 518 464;
  • 13) 0.155 518 464 × 2 = 0 + 0.311 036 928;
  • 14) 0.311 036 928 × 2 = 0 + 0.622 073 856;
  • 15) 0.622 073 856 × 2 = 1 + 0.244 147 712;
  • 16) 0.244 147 712 × 2 = 0 + 0.488 295 424;
  • 17) 0.488 295 424 × 2 = 0 + 0.976 590 848;
  • 18) 0.976 590 848 × 2 = 1 + 0.953 181 696;
  • 19) 0.953 181 696 × 2 = 1 + 0.906 363 392;
  • 20) 0.906 363 392 × 2 = 1 + 0.812 726 784;
  • 21) 0.812 726 784 × 2 = 1 + 0.625 453 568;
  • 22) 0.625 453 568 × 2 = 1 + 0.250 907 136;
  • 23) 0.250 907 136 × 2 = 0 + 0.501 814 272;
  • 24) 0.501 814 272 × 2 = 1 + 0.003 628 544;
  • 25) 0.003 628 544 × 2 = 0 + 0.007 257 088;
  • 26) 0.007 257 088 × 2 = 0 + 0.014 514 176;
  • 27) 0.014 514 176 × 2 = 0 + 0.029 028 352;
  • 28) 0.029 028 352 × 2 = 0 + 0.058 056 704;
  • 29) 0.058 056 704 × 2 = 0 + 0.116 113 408;
  • 30) 0.116 113 408 × 2 = 0 + 0.232 226 816;
  • 31) 0.232 226 816 × 2 = 0 + 0.464 453 632;
  • 32) 0.464 453 632 × 2 = 0 + 0.928 907 264;
  • 33) 0.928 907 264 × 2 = 1 + 0.857 814 528;
  • 34) 0.857 814 528 × 2 = 1 + 0.715 629 056;
  • 35) 0.715 629 056 × 2 = 1 + 0.431 258 112;
  • 36) 0.431 258 112 × 2 = 0 + 0.862 516 224;
  • 37) 0.862 516 224 × 2 = 1 + 0.725 032 448;
  • 38) 0.725 032 448 × 2 = 1 + 0.450 064 896;
  • 39) 0.450 064 896 × 2 = 0 + 0.900 129 792;
  • 40) 0.900 129 792 × 2 = 1 + 0.800 259 584;
  • 41) 0.800 259 584 × 2 = 1 + 0.600 519 168;
  • 42) 0.600 519 168 × 2 = 1 + 0.201 038 336;
  • 43) 0.201 038 336 × 2 = 0 + 0.402 076 672;
  • 44) 0.402 076 672 × 2 = 0 + 0.804 153 344;
  • 45) 0.804 153 344 × 2 = 1 + 0.608 306 688;
  • 46) 0.608 306 688 × 2 = 1 + 0.216 613 376;
  • 47) 0.216 613 376 × 2 = 0 + 0.433 226 752;
  • 48) 0.433 226 752 × 2 = 0 + 0.866 453 504;
  • 49) 0.866 453 504 × 2 = 1 + 0.732 907 008;
  • 50) 0.732 907 008 × 2 = 1 + 0.465 814 016;
  • 51) 0.465 814 016 × 2 = 0 + 0.931 628 032;
  • 52) 0.931 628 032 × 2 = 1 + 0.863 256 064;
  • 53) 0.863 256 064 × 2 = 1 + 0.726 512 128;
  • 54) 0.726 512 128 × 2 = 1 + 0.453 024 256;
  • 55) 0.453 024 256 × 2 = 0 + 0.906 048 512;
  • 56) 0.906 048 512 × 2 = 1 + 0.812 097 024;
  • 57) 0.812 097 024 × 2 = 1 + 0.624 194 048;
  • 58) 0.624 194 048 × 2 = 1 + 0.248 388 096;
  • 59) 0.248 388 096 × 2 = 0 + 0.496 776 192;
  • 60) 0.496 776 192 × 2 = 0 + 0.993 552 384;
  • 61) 0.993 552 384 × 2 = 1 + 0.987 104 768;
  • 62) 0.987 104 768 × 2 = 1 + 0.974 209 536;
  • 63) 0.974 209 536 × 2 = 1 + 0.948 419 072;
  • 64) 0.948 419 072 × 2 = 1 + 0.896 838 144;

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:


0.000 282 109(10) =

0.0000 0000 0001 0010 0111 1101 0000 0000 1110 1101 1100 1100 1101 1101 1100 1111(2)

6. Positive number before normalization:

0.000 282 109(10) =

0.0000 0000 0001 0010 0111 1101 0000 0000 1110 1101 1100 1100 1101 1101 1100 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:


0.000 282 109(10) =

0.0000 0000 0001 0010 0111 1101 0000 0000 1110 1101 1100 1100 1101 1101 1100 1111(2) =

0.0000 0000 0001 0010 0111 1101 0000 0000 1110 1101 1100 1100 1101 1101 1100 1111(2) × 20 =

1.0010 0111 1101 0000 0000 1110 1101 1100 1100 1101 1101 1100 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 1101 0000 0000 1110 1101 1100 1100 1101 1101 1100 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 1101 0000 0000 1110 1101 1100 1100 1101 1101 1100 1111 =

0010 0111 1101 0000 0000 1110 1101 1100 1100 1101 1101 1100 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 1101 0000 0000 1110 1101 1100 1100 1101 1101 1100 1111


Decimal number -0.000 282 109 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1101 0000 0000 1110 1101 1100 1100 1101 1101 1100 1111

Share

» Convert decimal -0.000 282 069 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 Attention Economy - The Battlefield For Your Focus. NotebookLM YouTube Video of 7.50 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