-0.000 282 005 813 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 813(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
-0.000 282 005 813(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:

|-0.000 282 005 813| = 0.000 282 005 813


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: 0.000 282 005 813.

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) 0.000 282 005 813 × 2 = 0 + 0.000 564 011 626;
  • 2) 0.000 564 011 626 × 2 = 0 + 0.001 128 023 252;
  • 3) 0.001 128 023 252 × 2 = 0 + 0.002 256 046 504;
  • 4) 0.002 256 046 504 × 2 = 0 + 0.004 512 093 008;
  • 5) 0.004 512 093 008 × 2 = 0 + 0.009 024 186 016;
  • 6) 0.009 024 186 016 × 2 = 0 + 0.018 048 372 032;
  • 7) 0.018 048 372 032 × 2 = 0 + 0.036 096 744 064;
  • 8) 0.036 096 744 064 × 2 = 0 + 0.072 193 488 128;
  • 9) 0.072 193 488 128 × 2 = 0 + 0.144 386 976 256;
  • 10) 0.144 386 976 256 × 2 = 0 + 0.288 773 952 512;
  • 11) 0.288 773 952 512 × 2 = 0 + 0.577 547 905 024;
  • 12) 0.577 547 905 024 × 2 = 1 + 0.155 095 810 048;
  • 13) 0.155 095 810 048 × 2 = 0 + 0.310 191 620 096;
  • 14) 0.310 191 620 096 × 2 = 0 + 0.620 383 240 192;
  • 15) 0.620 383 240 192 × 2 = 1 + 0.240 766 480 384;
  • 16) 0.240 766 480 384 × 2 = 0 + 0.481 532 960 768;
  • 17) 0.481 532 960 768 × 2 = 0 + 0.963 065 921 536;
  • 18) 0.963 065 921 536 × 2 = 1 + 0.926 131 843 072;
  • 19) 0.926 131 843 072 × 2 = 1 + 0.852 263 686 144;
  • 20) 0.852 263 686 144 × 2 = 1 + 0.704 527 372 288;
  • 21) 0.704 527 372 288 × 2 = 1 + 0.409 054 744 576;
  • 22) 0.409 054 744 576 × 2 = 0 + 0.818 109 489 152;
  • 23) 0.818 109 489 152 × 2 = 1 + 0.636 218 978 304;
  • 24) 0.636 218 978 304 × 2 = 1 + 0.272 437 956 608;
  • 25) 0.272 437 956 608 × 2 = 0 + 0.544 875 913 216;
  • 26) 0.544 875 913 216 × 2 = 1 + 0.089 751 826 432;
  • 27) 0.089 751 826 432 × 2 = 0 + 0.179 503 652 864;
  • 28) 0.179 503 652 864 × 2 = 0 + 0.359 007 305 728;
  • 29) 0.359 007 305 728 × 2 = 0 + 0.718 014 611 456;
  • 30) 0.718 014 611 456 × 2 = 1 + 0.436 029 222 912;
  • 31) 0.436 029 222 912 × 2 = 0 + 0.872 058 445 824;
  • 32) 0.872 058 445 824 × 2 = 1 + 0.744 116 891 648;
  • 33) 0.744 116 891 648 × 2 = 1 + 0.488 233 783 296;
  • 34) 0.488 233 783 296 × 2 = 0 + 0.976 467 566 592;
  • 35) 0.976 467 566 592 × 2 = 1 + 0.952 935 133 184;
  • 36) 0.952 935 133 184 × 2 = 1 + 0.905 870 266 368;
  • 37) 0.905 870 266 368 × 2 = 1 + 0.811 740 532 736;
  • 38) 0.811 740 532 736 × 2 = 1 + 0.623 481 065 472;
  • 39) 0.623 481 065 472 × 2 = 1 + 0.246 962 130 944;
  • 40) 0.246 962 130 944 × 2 = 0 + 0.493 924 261 888;
  • 41) 0.493 924 261 888 × 2 = 0 + 0.987 848 523 776;
  • 42) 0.987 848 523 776 × 2 = 1 + 0.975 697 047 552;
  • 43) 0.975 697 047 552 × 2 = 1 + 0.951 394 095 104;
  • 44) 0.951 394 095 104 × 2 = 1 + 0.902 788 190 208;
  • 45) 0.902 788 190 208 × 2 = 1 + 0.805 576 380 416;
  • 46) 0.805 576 380 416 × 2 = 1 + 0.611 152 760 832;
  • 47) 0.611 152 760 832 × 2 = 1 + 0.222 305 521 664;
  • 48) 0.222 305 521 664 × 2 = 0 + 0.444 611 043 328;
  • 49) 0.444 611 043 328 × 2 = 0 + 0.889 222 086 656;
  • 50) 0.889 222 086 656 × 2 = 1 + 0.778 444 173 312;
  • 51) 0.778 444 173 312 × 2 = 1 + 0.556 888 346 624;
  • 52) 0.556 888 346 624 × 2 = 1 + 0.113 776 693 248;
  • 53) 0.113 776 693 248 × 2 = 0 + 0.227 553 386 496;
  • 54) 0.227 553 386 496 × 2 = 0 + 0.455 106 772 992;
  • 55) 0.455 106 772 992 × 2 = 0 + 0.910 213 545 984;
  • 56) 0.910 213 545 984 × 2 = 1 + 0.820 427 091 968;
  • 57) 0.820 427 091 968 × 2 = 1 + 0.640 854 183 936;
  • 58) 0.640 854 183 936 × 2 = 1 + 0.281 708 367 872;
  • 59) 0.281 708 367 872 × 2 = 0 + 0.563 416 735 744;
  • 60) 0.563 416 735 744 × 2 = 1 + 0.126 833 471 488;
  • 61) 0.126 833 471 488 × 2 = 0 + 0.253 666 942 976;
  • 62) 0.253 666 942 976 × 2 = 0 + 0.507 333 885 952;
  • 63) 0.507 333 885 952 × 2 = 1 + 0.014 667 771 904;
  • 64) 0.014 667 771 904 × 2 = 0 + 0.029 335 543 808;

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:


0.000 282 005 813(10) =

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

6. Positive number before normalization:

0.000 282 005 813(10) =

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


0.000 282 005 813(10) =

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

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

1.0010 0111 1011 0100 0101 1011 1110 0111 1110 0111 0001 1101 0010(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 0100 0101 1011 1110 0111 1110 0111 0001 1101 0010


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 0100 0101 1011 1110 0111 1110 0111 0001 1101 0010 =

0010 0111 1011 0100 0101 1011 1110 0111 1110 0111 0001 1101 0010


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 0100 0101 1011 1110 0111 1110 0111 0001 1101 0010


Decimal number -0.000 282 005 813 converted to 64 bit double precision IEEE 754 binary floating point representation:

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

Share

» Convert decimal -0.000 282 005 744 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