-0.000 282 004 12 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 004 12(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
-0.000 282 004 12(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:

|-0.000 282 004 12| = 0.000 282 004 12


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: 0.000 282 004 12.

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) 0.000 282 004 12 × 2 = 0 + 0.000 564 008 24;
  • 2) 0.000 564 008 24 × 2 = 0 + 0.001 128 016 48;
  • 3) 0.001 128 016 48 × 2 = 0 + 0.002 256 032 96;
  • 4) 0.002 256 032 96 × 2 = 0 + 0.004 512 065 92;
  • 5) 0.004 512 065 92 × 2 = 0 + 0.009 024 131 84;
  • 6) 0.009 024 131 84 × 2 = 0 + 0.018 048 263 68;
  • 7) 0.018 048 263 68 × 2 = 0 + 0.036 096 527 36;
  • 8) 0.036 096 527 36 × 2 = 0 + 0.072 193 054 72;
  • 9) 0.072 193 054 72 × 2 = 0 + 0.144 386 109 44;
  • 10) 0.144 386 109 44 × 2 = 0 + 0.288 772 218 88;
  • 11) 0.288 772 218 88 × 2 = 0 + 0.577 544 437 76;
  • 12) 0.577 544 437 76 × 2 = 1 + 0.155 088 875 52;
  • 13) 0.155 088 875 52 × 2 = 0 + 0.310 177 751 04;
  • 14) 0.310 177 751 04 × 2 = 0 + 0.620 355 502 08;
  • 15) 0.620 355 502 08 × 2 = 1 + 0.240 711 004 16;
  • 16) 0.240 711 004 16 × 2 = 0 + 0.481 422 008 32;
  • 17) 0.481 422 008 32 × 2 = 0 + 0.962 844 016 64;
  • 18) 0.962 844 016 64 × 2 = 1 + 0.925 688 033 28;
  • 19) 0.925 688 033 28 × 2 = 1 + 0.851 376 066 56;
  • 20) 0.851 376 066 56 × 2 = 1 + 0.702 752 133 12;
  • 21) 0.702 752 133 12 × 2 = 1 + 0.405 504 266 24;
  • 22) 0.405 504 266 24 × 2 = 0 + 0.811 008 532 48;
  • 23) 0.811 008 532 48 × 2 = 1 + 0.622 017 064 96;
  • 24) 0.622 017 064 96 × 2 = 1 + 0.244 034 129 92;
  • 25) 0.244 034 129 92 × 2 = 0 + 0.488 068 259 84;
  • 26) 0.488 068 259 84 × 2 = 0 + 0.976 136 519 68;
  • 27) 0.976 136 519 68 × 2 = 1 + 0.952 273 039 36;
  • 28) 0.952 273 039 36 × 2 = 1 + 0.904 546 078 72;
  • 29) 0.904 546 078 72 × 2 = 1 + 0.809 092 157 44;
  • 30) 0.809 092 157 44 × 2 = 1 + 0.618 184 314 88;
  • 31) 0.618 184 314 88 × 2 = 1 + 0.236 368 629 76;
  • 32) 0.236 368 629 76 × 2 = 0 + 0.472 737 259 52;
  • 33) 0.472 737 259 52 × 2 = 0 + 0.945 474 519 04;
  • 34) 0.945 474 519 04 × 2 = 1 + 0.890 949 038 08;
  • 35) 0.890 949 038 08 × 2 = 1 + 0.781 898 076 16;
  • 36) 0.781 898 076 16 × 2 = 1 + 0.563 796 152 32;
  • 37) 0.563 796 152 32 × 2 = 1 + 0.127 592 304 64;
  • 38) 0.127 592 304 64 × 2 = 0 + 0.255 184 609 28;
  • 39) 0.255 184 609 28 × 2 = 0 + 0.510 369 218 56;
  • 40) 0.510 369 218 56 × 2 = 1 + 0.020 738 437 12;
  • 41) 0.020 738 437 12 × 2 = 0 + 0.041 476 874 24;
  • 42) 0.041 476 874 24 × 2 = 0 + 0.082 953 748 48;
  • 43) 0.082 953 748 48 × 2 = 0 + 0.165 907 496 96;
  • 44) 0.165 907 496 96 × 2 = 0 + 0.331 814 993 92;
  • 45) 0.331 814 993 92 × 2 = 0 + 0.663 629 987 84;
  • 46) 0.663 629 987 84 × 2 = 1 + 0.327 259 975 68;
  • 47) 0.327 259 975 68 × 2 = 0 + 0.654 519 951 36;
  • 48) 0.654 519 951 36 × 2 = 1 + 0.309 039 902 72;
  • 49) 0.309 039 902 72 × 2 = 0 + 0.618 079 805 44;
  • 50) 0.618 079 805 44 × 2 = 1 + 0.236 159 610 88;
  • 51) 0.236 159 610 88 × 2 = 0 + 0.472 319 221 76;
  • 52) 0.472 319 221 76 × 2 = 0 + 0.944 638 443 52;
  • 53) 0.944 638 443 52 × 2 = 1 + 0.889 276 887 04;
  • 54) 0.889 276 887 04 × 2 = 1 + 0.778 553 774 08;
  • 55) 0.778 553 774 08 × 2 = 1 + 0.557 107 548 16;
  • 56) 0.557 107 548 16 × 2 = 1 + 0.114 215 096 32;
  • 57) 0.114 215 096 32 × 2 = 0 + 0.228 430 192 64;
  • 58) 0.228 430 192 64 × 2 = 0 + 0.456 860 385 28;
  • 59) 0.456 860 385 28 × 2 = 0 + 0.913 720 770 56;
  • 60) 0.913 720 770 56 × 2 = 1 + 0.827 441 541 12;
  • 61) 0.827 441 541 12 × 2 = 1 + 0.654 883 082 24;
  • 62) 0.654 883 082 24 × 2 = 1 + 0.309 766 164 48;
  • 63) 0.309 766 164 48 × 2 = 0 + 0.619 532 328 96;
  • 64) 0.619 532 328 96 × 2 = 1 + 0.239 064 657 92;

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:


0.000 282 004 12(10) =

0.0000 0000 0001 0010 0111 1011 0011 1110 0111 1001 0000 0101 0100 1111 0001 1101(2)

6. Positive number before normalization:

0.000 282 004 12(10) =

0.0000 0000 0001 0010 0111 1011 0011 1110 0111 1001 0000 0101 0100 1111 0001 1101(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:


0.000 282 004 12(10) =

0.0000 0000 0001 0010 0111 1011 0011 1110 0111 1001 0000 0101 0100 1111 0001 1101(2) =

0.0000 0000 0001 0010 0111 1011 0011 1110 0111 1001 0000 0101 0100 1111 0001 1101(2) × 20 =

1.0010 0111 1011 0011 1110 0111 1001 0000 0101 0100 1111 0001 1101(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 0011 1110 0111 1001 0000 0101 0100 1111 0001 1101


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 0011 1110 0111 1001 0000 0101 0100 1111 0001 1101 =

0010 0111 1011 0011 1110 0111 1001 0000 0101 0100 1111 0001 1101


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 0011 1110 0111 1001 0000 0101 0100 1111 0001 1101


Decimal number -0.000 282 004 12 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0011 1110 0111 1001 0000 0101 0100 1111 0001 1101

Share

» Convert decimal -0.000 282 004 19 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 Art of Strategic Ambiguity and Concealed Intentions. NotebookLM YouTube Video of 6.59 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