-0.000 282 294 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 294(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
-0.000 282 294(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:

|-0.000 282 294| = 0.000 282 294


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: 0.000 282 294.

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) 0.000 282 294 × 2 = 0 + 0.000 564 588;
  • 2) 0.000 564 588 × 2 = 0 + 0.001 129 176;
  • 3) 0.001 129 176 × 2 = 0 + 0.002 258 352;
  • 4) 0.002 258 352 × 2 = 0 + 0.004 516 704;
  • 5) 0.004 516 704 × 2 = 0 + 0.009 033 408;
  • 6) 0.009 033 408 × 2 = 0 + 0.018 066 816;
  • 7) 0.018 066 816 × 2 = 0 + 0.036 133 632;
  • 8) 0.036 133 632 × 2 = 0 + 0.072 267 264;
  • 9) 0.072 267 264 × 2 = 0 + 0.144 534 528;
  • 10) 0.144 534 528 × 2 = 0 + 0.289 069 056;
  • 11) 0.289 069 056 × 2 = 0 + 0.578 138 112;
  • 12) 0.578 138 112 × 2 = 1 + 0.156 276 224;
  • 13) 0.156 276 224 × 2 = 0 + 0.312 552 448;
  • 14) 0.312 552 448 × 2 = 0 + 0.625 104 896;
  • 15) 0.625 104 896 × 2 = 1 + 0.250 209 792;
  • 16) 0.250 209 792 × 2 = 0 + 0.500 419 584;
  • 17) 0.500 419 584 × 2 = 1 + 0.000 839 168;
  • 18) 0.000 839 168 × 2 = 0 + 0.001 678 336;
  • 19) 0.001 678 336 × 2 = 0 + 0.003 356 672;
  • 20) 0.003 356 672 × 2 = 0 + 0.006 713 344;
  • 21) 0.006 713 344 × 2 = 0 + 0.013 426 688;
  • 22) 0.013 426 688 × 2 = 0 + 0.026 853 376;
  • 23) 0.026 853 376 × 2 = 0 + 0.053 706 752;
  • 24) 0.053 706 752 × 2 = 0 + 0.107 413 504;
  • 25) 0.107 413 504 × 2 = 0 + 0.214 827 008;
  • 26) 0.214 827 008 × 2 = 0 + 0.429 654 016;
  • 27) 0.429 654 016 × 2 = 0 + 0.859 308 032;
  • 28) 0.859 308 032 × 2 = 1 + 0.718 616 064;
  • 29) 0.718 616 064 × 2 = 1 + 0.437 232 128;
  • 30) 0.437 232 128 × 2 = 0 + 0.874 464 256;
  • 31) 0.874 464 256 × 2 = 1 + 0.748 928 512;
  • 32) 0.748 928 512 × 2 = 1 + 0.497 857 024;
  • 33) 0.497 857 024 × 2 = 0 + 0.995 714 048;
  • 34) 0.995 714 048 × 2 = 1 + 0.991 428 096;
  • 35) 0.991 428 096 × 2 = 1 + 0.982 856 192;
  • 36) 0.982 856 192 × 2 = 1 + 0.965 712 384;
  • 37) 0.965 712 384 × 2 = 1 + 0.931 424 768;
  • 38) 0.931 424 768 × 2 = 1 + 0.862 849 536;
  • 39) 0.862 849 536 × 2 = 1 + 0.725 699 072;
  • 40) 0.725 699 072 × 2 = 1 + 0.451 398 144;
  • 41) 0.451 398 144 × 2 = 0 + 0.902 796 288;
  • 42) 0.902 796 288 × 2 = 1 + 0.805 592 576;
  • 43) 0.805 592 576 × 2 = 1 + 0.611 185 152;
  • 44) 0.611 185 152 × 2 = 1 + 0.222 370 304;
  • 45) 0.222 370 304 × 2 = 0 + 0.444 740 608;
  • 46) 0.444 740 608 × 2 = 0 + 0.889 481 216;
  • 47) 0.889 481 216 × 2 = 1 + 0.778 962 432;
  • 48) 0.778 962 432 × 2 = 1 + 0.557 924 864;
  • 49) 0.557 924 864 × 2 = 1 + 0.115 849 728;
  • 50) 0.115 849 728 × 2 = 0 + 0.231 699 456;
  • 51) 0.231 699 456 × 2 = 0 + 0.463 398 912;
  • 52) 0.463 398 912 × 2 = 0 + 0.926 797 824;
  • 53) 0.926 797 824 × 2 = 1 + 0.853 595 648;
  • 54) 0.853 595 648 × 2 = 1 + 0.707 191 296;
  • 55) 0.707 191 296 × 2 = 1 + 0.414 382 592;
  • 56) 0.414 382 592 × 2 = 0 + 0.828 765 184;
  • 57) 0.828 765 184 × 2 = 1 + 0.657 530 368;
  • 58) 0.657 530 368 × 2 = 1 + 0.315 060 736;
  • 59) 0.315 060 736 × 2 = 0 + 0.630 121 472;
  • 60) 0.630 121 472 × 2 = 1 + 0.260 242 944;
  • 61) 0.260 242 944 × 2 = 0 + 0.520 485 888;
  • 62) 0.520 485 888 × 2 = 1 + 0.040 971 776;
  • 63) 0.040 971 776 × 2 = 0 + 0.081 943 552;
  • 64) 0.081 943 552 × 2 = 0 + 0.163 887 104;

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:


0.000 282 294(10) =

0.0000 0000 0001 0010 1000 0000 0001 1011 0111 1111 0111 0011 1000 1110 1101 0100(2)

6. Positive number before normalization:

0.000 282 294(10) =

0.0000 0000 0001 0010 1000 0000 0001 1011 0111 1111 0111 0011 1000 1110 1101 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:


0.000 282 294(10) =

0.0000 0000 0001 0010 1000 0000 0001 1011 0111 1111 0111 0011 1000 1110 1101 0100(2) =

0.0000 0000 0001 0010 1000 0000 0001 1011 0111 1111 0111 0011 1000 1110 1101 0100(2) × 20 =

1.0010 1000 0000 0001 1011 0111 1111 0111 0011 1000 1110 1101 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 1000 0000 0001 1011 0111 1111 0111 0011 1000 1110 1101 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 1000 0000 0001 1011 0111 1111 0111 0011 1000 1110 1101 0100 =

0010 1000 0000 0001 1011 0111 1111 0111 0011 1000 1110 1101 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 1000 0000 0001 1011 0111 1111 0111 0011 1000 1110 1101 0100


Decimal number -0.000 282 294 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1000 0000 0001 1011 0111 1111 0111 0011 1000 1110 1101 0100

Share

» Convert decimal -0.000 282 372 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