-0.000 282 009 49 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 009 49(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
-0.000 282 009 49(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:

|-0.000 282 009 49| = 0.000 282 009 49


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: 0.000 282 009 49.

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) 0.000 282 009 49 × 2 = 0 + 0.000 564 018 98;
  • 2) 0.000 564 018 98 × 2 = 0 + 0.001 128 037 96;
  • 3) 0.001 128 037 96 × 2 = 0 + 0.002 256 075 92;
  • 4) 0.002 256 075 92 × 2 = 0 + 0.004 512 151 84;
  • 5) 0.004 512 151 84 × 2 = 0 + 0.009 024 303 68;
  • 6) 0.009 024 303 68 × 2 = 0 + 0.018 048 607 36;
  • 7) 0.018 048 607 36 × 2 = 0 + 0.036 097 214 72;
  • 8) 0.036 097 214 72 × 2 = 0 + 0.072 194 429 44;
  • 9) 0.072 194 429 44 × 2 = 0 + 0.144 388 858 88;
  • 10) 0.144 388 858 88 × 2 = 0 + 0.288 777 717 76;
  • 11) 0.288 777 717 76 × 2 = 0 + 0.577 555 435 52;
  • 12) 0.577 555 435 52 × 2 = 1 + 0.155 110 871 04;
  • 13) 0.155 110 871 04 × 2 = 0 + 0.310 221 742 08;
  • 14) 0.310 221 742 08 × 2 = 0 + 0.620 443 484 16;
  • 15) 0.620 443 484 16 × 2 = 1 + 0.240 886 968 32;
  • 16) 0.240 886 968 32 × 2 = 0 + 0.481 773 936 64;
  • 17) 0.481 773 936 64 × 2 = 0 + 0.963 547 873 28;
  • 18) 0.963 547 873 28 × 2 = 1 + 0.927 095 746 56;
  • 19) 0.927 095 746 56 × 2 = 1 + 0.854 191 493 12;
  • 20) 0.854 191 493 12 × 2 = 1 + 0.708 382 986 24;
  • 21) 0.708 382 986 24 × 2 = 1 + 0.416 765 972 48;
  • 22) 0.416 765 972 48 × 2 = 0 + 0.833 531 944 96;
  • 23) 0.833 531 944 96 × 2 = 1 + 0.667 063 889 92;
  • 24) 0.667 063 889 92 × 2 = 1 + 0.334 127 779 84;
  • 25) 0.334 127 779 84 × 2 = 0 + 0.668 255 559 68;
  • 26) 0.668 255 559 68 × 2 = 1 + 0.336 511 119 36;
  • 27) 0.336 511 119 36 × 2 = 0 + 0.673 022 238 72;
  • 28) 0.673 022 238 72 × 2 = 1 + 0.346 044 477 44;
  • 29) 0.346 044 477 44 × 2 = 0 + 0.692 088 954 88;
  • 30) 0.692 088 954 88 × 2 = 1 + 0.384 177 909 76;
  • 31) 0.384 177 909 76 × 2 = 0 + 0.768 355 819 52;
  • 32) 0.768 355 819 52 × 2 = 1 + 0.536 711 639 04;
  • 33) 0.536 711 639 04 × 2 = 1 + 0.073 423 278 08;
  • 34) 0.073 423 278 08 × 2 = 0 + 0.146 846 556 16;
  • 35) 0.146 846 556 16 × 2 = 0 + 0.293 693 112 32;
  • 36) 0.293 693 112 32 × 2 = 0 + 0.587 386 224 64;
  • 37) 0.587 386 224 64 × 2 = 1 + 0.174 772 449 28;
  • 38) 0.174 772 449 28 × 2 = 0 + 0.349 544 898 56;
  • 39) 0.349 544 898 56 × 2 = 0 + 0.699 089 797 12;
  • 40) 0.699 089 797 12 × 2 = 1 + 0.398 179 594 24;
  • 41) 0.398 179 594 24 × 2 = 0 + 0.796 359 188 48;
  • 42) 0.796 359 188 48 × 2 = 1 + 0.592 718 376 96;
  • 43) 0.592 718 376 96 × 2 = 1 + 0.185 436 753 92;
  • 44) 0.185 436 753 92 × 2 = 0 + 0.370 873 507 84;
  • 45) 0.370 873 507 84 × 2 = 0 + 0.741 747 015 68;
  • 46) 0.741 747 015 68 × 2 = 1 + 0.483 494 031 36;
  • 47) 0.483 494 031 36 × 2 = 0 + 0.966 988 062 72;
  • 48) 0.966 988 062 72 × 2 = 1 + 0.933 976 125 44;
  • 49) 0.933 976 125 44 × 2 = 1 + 0.867 952 250 88;
  • 50) 0.867 952 250 88 × 2 = 1 + 0.735 904 501 76;
  • 51) 0.735 904 501 76 × 2 = 1 + 0.471 809 003 52;
  • 52) 0.471 809 003 52 × 2 = 0 + 0.943 618 007 04;
  • 53) 0.943 618 007 04 × 2 = 1 + 0.887 236 014 08;
  • 54) 0.887 236 014 08 × 2 = 1 + 0.774 472 028 16;
  • 55) 0.774 472 028 16 × 2 = 1 + 0.548 944 056 32;
  • 56) 0.548 944 056 32 × 2 = 1 + 0.097 888 112 64;
  • 57) 0.097 888 112 64 × 2 = 0 + 0.195 776 225 28;
  • 58) 0.195 776 225 28 × 2 = 0 + 0.391 552 450 56;
  • 59) 0.391 552 450 56 × 2 = 0 + 0.783 104 901 12;
  • 60) 0.783 104 901 12 × 2 = 1 + 0.566 209 802 24;
  • 61) 0.566 209 802 24 × 2 = 1 + 0.132 419 604 48;
  • 62) 0.132 419 604 48 × 2 = 0 + 0.264 839 208 96;
  • 63) 0.264 839 208 96 × 2 = 0 + 0.529 678 417 92;
  • 64) 0.529 678 417 92 × 2 = 1 + 0.059 356 835 84;

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:


0.000 282 009 49(10) =

0.0000 0000 0001 0010 0111 1011 0101 0101 1000 1001 0110 0101 1110 1111 0001 1001(2)

6. Positive number before normalization:

0.000 282 009 49(10) =

0.0000 0000 0001 0010 0111 1011 0101 0101 1000 1001 0110 0101 1110 1111 0001 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:


0.000 282 009 49(10) =

0.0000 0000 0001 0010 0111 1011 0101 0101 1000 1001 0110 0101 1110 1111 0001 1001(2) =

0.0000 0000 0001 0010 0111 1011 0101 0101 1000 1001 0110 0101 1110 1111 0001 1001(2) × 20 =

1.0010 0111 1011 0101 0101 1000 1001 0110 0101 1110 1111 0001 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):
1.0010 0111 1011 0101 0101 1000 1001 0110 0101 1110 1111 0001 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. 0010 0111 1011 0101 0101 1000 1001 0110 0101 1110 1111 0001 1001 =

0010 0111 1011 0101 0101 1000 1001 0110 0101 1110 1111 0001 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) =
0010 0111 1011 0101 0101 1000 1001 0110 0101 1110 1111 0001 1001


Decimal number -0.000 282 009 49 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0101 0101 1000 1001 0110 0101 1110 1111 0001 1001

Share

» Convert decimal -0.000 282 010 08 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