-0.000 282 010 08 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 010 08(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
-0.000 282 010 08(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:

|-0.000 282 010 08| = 0.000 282 010 08


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: 0.000 282 010 08.

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) 0.000 282 010 08 × 2 = 0 + 0.000 564 020 16;
  • 2) 0.000 564 020 16 × 2 = 0 + 0.001 128 040 32;
  • 3) 0.001 128 040 32 × 2 = 0 + 0.002 256 080 64;
  • 4) 0.002 256 080 64 × 2 = 0 + 0.004 512 161 28;
  • 5) 0.004 512 161 28 × 2 = 0 + 0.009 024 322 56;
  • 6) 0.009 024 322 56 × 2 = 0 + 0.018 048 645 12;
  • 7) 0.018 048 645 12 × 2 = 0 + 0.036 097 290 24;
  • 8) 0.036 097 290 24 × 2 = 0 + 0.072 194 580 48;
  • 9) 0.072 194 580 48 × 2 = 0 + 0.144 389 160 96;
  • 10) 0.144 389 160 96 × 2 = 0 + 0.288 778 321 92;
  • 11) 0.288 778 321 92 × 2 = 0 + 0.577 556 643 84;
  • 12) 0.577 556 643 84 × 2 = 1 + 0.155 113 287 68;
  • 13) 0.155 113 287 68 × 2 = 0 + 0.310 226 575 36;
  • 14) 0.310 226 575 36 × 2 = 0 + 0.620 453 150 72;
  • 15) 0.620 453 150 72 × 2 = 1 + 0.240 906 301 44;
  • 16) 0.240 906 301 44 × 2 = 0 + 0.481 812 602 88;
  • 17) 0.481 812 602 88 × 2 = 0 + 0.963 625 205 76;
  • 18) 0.963 625 205 76 × 2 = 1 + 0.927 250 411 52;
  • 19) 0.927 250 411 52 × 2 = 1 + 0.854 500 823 04;
  • 20) 0.854 500 823 04 × 2 = 1 + 0.709 001 646 08;
  • 21) 0.709 001 646 08 × 2 = 1 + 0.418 003 292 16;
  • 22) 0.418 003 292 16 × 2 = 0 + 0.836 006 584 32;
  • 23) 0.836 006 584 32 × 2 = 1 + 0.672 013 168 64;
  • 24) 0.672 013 168 64 × 2 = 1 + 0.344 026 337 28;
  • 25) 0.344 026 337 28 × 2 = 0 + 0.688 052 674 56;
  • 26) 0.688 052 674 56 × 2 = 1 + 0.376 105 349 12;
  • 27) 0.376 105 349 12 × 2 = 0 + 0.752 210 698 24;
  • 28) 0.752 210 698 24 × 2 = 1 + 0.504 421 396 48;
  • 29) 0.504 421 396 48 × 2 = 1 + 0.008 842 792 96;
  • 30) 0.008 842 792 96 × 2 = 0 + 0.017 685 585 92;
  • 31) 0.017 685 585 92 × 2 = 0 + 0.035 371 171 84;
  • 32) 0.035 371 171 84 × 2 = 0 + 0.070 742 343 68;
  • 33) 0.070 742 343 68 × 2 = 0 + 0.141 484 687 36;
  • 34) 0.141 484 687 36 × 2 = 0 + 0.282 969 374 72;
  • 35) 0.282 969 374 72 × 2 = 0 + 0.565 938 749 44;
  • 36) 0.565 938 749 44 × 2 = 1 + 0.131 877 498 88;
  • 37) 0.131 877 498 88 × 2 = 0 + 0.263 754 997 76;
  • 38) 0.263 754 997 76 × 2 = 0 + 0.527 509 995 52;
  • 39) 0.527 509 995 52 × 2 = 1 + 0.055 019 991 04;
  • 40) 0.055 019 991 04 × 2 = 0 + 0.110 039 982 08;
  • 41) 0.110 039 982 08 × 2 = 0 + 0.220 079 964 16;
  • 42) 0.220 079 964 16 × 2 = 0 + 0.440 159 928 32;
  • 43) 0.440 159 928 32 × 2 = 0 + 0.880 319 856 64;
  • 44) 0.880 319 856 64 × 2 = 1 + 0.760 639 713 28;
  • 45) 0.760 639 713 28 × 2 = 1 + 0.521 279 426 56;
  • 46) 0.521 279 426 56 × 2 = 1 + 0.042 558 853 12;
  • 47) 0.042 558 853 12 × 2 = 0 + 0.085 117 706 24;
  • 48) 0.085 117 706 24 × 2 = 0 + 0.170 235 412 48;
  • 49) 0.170 235 412 48 × 2 = 0 + 0.340 470 824 96;
  • 50) 0.340 470 824 96 × 2 = 0 + 0.680 941 649 92;
  • 51) 0.680 941 649 92 × 2 = 1 + 0.361 883 299 84;
  • 52) 0.361 883 299 84 × 2 = 0 + 0.723 766 599 68;
  • 53) 0.723 766 599 68 × 2 = 1 + 0.447 533 199 36;
  • 54) 0.447 533 199 36 × 2 = 0 + 0.895 066 398 72;
  • 55) 0.895 066 398 72 × 2 = 1 + 0.790 132 797 44;
  • 56) 0.790 132 797 44 × 2 = 1 + 0.580 265 594 88;
  • 57) 0.580 265 594 88 × 2 = 1 + 0.160 531 189 76;
  • 58) 0.160 531 189 76 × 2 = 0 + 0.321 062 379 52;
  • 59) 0.321 062 379 52 × 2 = 0 + 0.642 124 759 04;
  • 60) 0.642 124 759 04 × 2 = 1 + 0.284 249 518 08;
  • 61) 0.284 249 518 08 × 2 = 0 + 0.568 499 036 16;
  • 62) 0.568 499 036 16 × 2 = 1 + 0.136 998 072 32;
  • 63) 0.136 998 072 32 × 2 = 0 + 0.273 996 144 64;
  • 64) 0.273 996 144 64 × 2 = 0 + 0.547 992 289 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:


0.000 282 010 08(10) =

0.0000 0000 0001 0010 0111 1011 0101 1000 0001 0010 0001 1100 0010 1011 1001 0100(2)

6. Positive number before normalization:

0.000 282 010 08(10) =

0.0000 0000 0001 0010 0111 1011 0101 1000 0001 0010 0001 1100 0010 1011 1001 0100(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:


0.000 282 010 08(10) =

0.0000 0000 0001 0010 0111 1011 0101 1000 0001 0010 0001 1100 0010 1011 1001 0100(2) =

0.0000 0000 0001 0010 0111 1011 0101 1000 0001 0010 0001 1100 0010 1011 1001 0100(2) × 20 =

1.0010 0111 1011 0101 1000 0001 0010 0001 1100 0010 1011 1001 0100(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 1000 0001 0010 0001 1100 0010 1011 1001 0100


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 1000 0001 0010 0001 1100 0010 1011 1001 0100 =

0010 0111 1011 0101 1000 0001 0010 0001 1100 0010 1011 1001 0100


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 1000 0001 0010 0001 1100 0010 1011 1001 0100


Decimal number -0.000 282 010 08 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0101 1000 0001 0010 0001 1100 0010 1011 1001 0100

Share

» Convert decimal -0.000 282 010 75 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: Are you using communication by design, or by default? NotebookLM YouTube Video of 6.33 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