-0.000 282 018 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 018 3(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
-0.000 282 018 3(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:

|-0.000 282 018 3| = 0.000 282 018 3


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: 0.000 282 018 3.

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) 0.000 282 018 3 × 2 = 0 + 0.000 564 036 6;
  • 2) 0.000 564 036 6 × 2 = 0 + 0.001 128 073 2;
  • 3) 0.001 128 073 2 × 2 = 0 + 0.002 256 146 4;
  • 4) 0.002 256 146 4 × 2 = 0 + 0.004 512 292 8;
  • 5) 0.004 512 292 8 × 2 = 0 + 0.009 024 585 6;
  • 6) 0.009 024 585 6 × 2 = 0 + 0.018 049 171 2;
  • 7) 0.018 049 171 2 × 2 = 0 + 0.036 098 342 4;
  • 8) 0.036 098 342 4 × 2 = 0 + 0.072 196 684 8;
  • 9) 0.072 196 684 8 × 2 = 0 + 0.144 393 369 6;
  • 10) 0.144 393 369 6 × 2 = 0 + 0.288 786 739 2;
  • 11) 0.288 786 739 2 × 2 = 0 + 0.577 573 478 4;
  • 12) 0.577 573 478 4 × 2 = 1 + 0.155 146 956 8;
  • 13) 0.155 146 956 8 × 2 = 0 + 0.310 293 913 6;
  • 14) 0.310 293 913 6 × 2 = 0 + 0.620 587 827 2;
  • 15) 0.620 587 827 2 × 2 = 1 + 0.241 175 654 4;
  • 16) 0.241 175 654 4 × 2 = 0 + 0.482 351 308 8;
  • 17) 0.482 351 308 8 × 2 = 0 + 0.964 702 617 6;
  • 18) 0.964 702 617 6 × 2 = 1 + 0.929 405 235 2;
  • 19) 0.929 405 235 2 × 2 = 1 + 0.858 810 470 4;
  • 20) 0.858 810 470 4 × 2 = 1 + 0.717 620 940 8;
  • 21) 0.717 620 940 8 × 2 = 1 + 0.435 241 881 6;
  • 22) 0.435 241 881 6 × 2 = 0 + 0.870 483 763 2;
  • 23) 0.870 483 763 2 × 2 = 1 + 0.740 967 526 4;
  • 24) 0.740 967 526 4 × 2 = 1 + 0.481 935 052 8;
  • 25) 0.481 935 052 8 × 2 = 0 + 0.963 870 105 6;
  • 26) 0.963 870 105 6 × 2 = 1 + 0.927 740 211 2;
  • 27) 0.927 740 211 2 × 2 = 1 + 0.855 480 422 4;
  • 28) 0.855 480 422 4 × 2 = 1 + 0.710 960 844 8;
  • 29) 0.710 960 844 8 × 2 = 1 + 0.421 921 689 6;
  • 30) 0.421 921 689 6 × 2 = 0 + 0.843 843 379 2;
  • 31) 0.843 843 379 2 × 2 = 1 + 0.687 686 758 4;
  • 32) 0.687 686 758 4 × 2 = 1 + 0.375 373 516 8;
  • 33) 0.375 373 516 8 × 2 = 0 + 0.750 747 033 6;
  • 34) 0.750 747 033 6 × 2 = 1 + 0.501 494 067 2;
  • 35) 0.501 494 067 2 × 2 = 1 + 0.002 988 134 4;
  • 36) 0.002 988 134 4 × 2 = 0 + 0.005 976 268 8;
  • 37) 0.005 976 268 8 × 2 = 0 + 0.011 952 537 6;
  • 38) 0.011 952 537 6 × 2 = 0 + 0.023 905 075 2;
  • 39) 0.023 905 075 2 × 2 = 0 + 0.047 810 150 4;
  • 40) 0.047 810 150 4 × 2 = 0 + 0.095 620 300 8;
  • 41) 0.095 620 300 8 × 2 = 0 + 0.191 240 601 6;
  • 42) 0.191 240 601 6 × 2 = 0 + 0.382 481 203 2;
  • 43) 0.382 481 203 2 × 2 = 0 + 0.764 962 406 4;
  • 44) 0.764 962 406 4 × 2 = 1 + 0.529 924 812 8;
  • 45) 0.529 924 812 8 × 2 = 1 + 0.059 849 625 6;
  • 46) 0.059 849 625 6 × 2 = 0 + 0.119 699 251 2;
  • 47) 0.119 699 251 2 × 2 = 0 + 0.239 398 502 4;
  • 48) 0.239 398 502 4 × 2 = 0 + 0.478 797 004 8;
  • 49) 0.478 797 004 8 × 2 = 0 + 0.957 594 009 6;
  • 50) 0.957 594 009 6 × 2 = 1 + 0.915 188 019 2;
  • 51) 0.915 188 019 2 × 2 = 1 + 0.830 376 038 4;
  • 52) 0.830 376 038 4 × 2 = 1 + 0.660 752 076 8;
  • 53) 0.660 752 076 8 × 2 = 1 + 0.321 504 153 6;
  • 54) 0.321 504 153 6 × 2 = 0 + 0.643 008 307 2;
  • 55) 0.643 008 307 2 × 2 = 1 + 0.286 016 614 4;
  • 56) 0.286 016 614 4 × 2 = 0 + 0.572 033 228 8;
  • 57) 0.572 033 228 8 × 2 = 1 + 0.144 066 457 6;
  • 58) 0.144 066 457 6 × 2 = 0 + 0.288 132 915 2;
  • 59) 0.288 132 915 2 × 2 = 0 + 0.576 265 830 4;
  • 60) 0.576 265 830 4 × 2 = 1 + 0.152 531 660 8;
  • 61) 0.152 531 660 8 × 2 = 0 + 0.305 063 321 6;
  • 62) 0.305 063 321 6 × 2 = 0 + 0.610 126 643 2;
  • 63) 0.610 126 643 2 × 2 = 1 + 0.220 253 286 4;
  • 64) 0.220 253 286 4 × 2 = 0 + 0.440 506 572 8;

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:


0.000 282 018 3(10) =

0.0000 0000 0001 0010 0111 1011 0111 1011 0110 0000 0001 1000 0111 1010 1001 0010(2)

6. Positive number before normalization:

0.000 282 018 3(10) =

0.0000 0000 0001 0010 0111 1011 0111 1011 0110 0000 0001 1000 0111 1010 1001 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:


0.000 282 018 3(10) =

0.0000 0000 0001 0010 0111 1011 0111 1011 0110 0000 0001 1000 0111 1010 1001 0010(2) =

0.0000 0000 0001 0010 0111 1011 0111 1011 0110 0000 0001 1000 0111 1010 1001 0010(2) × 20 =

1.0010 0111 1011 0111 1011 0110 0000 0001 1000 0111 1010 1001 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 0111 1011 0110 0000 0001 1000 0111 1010 1001 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 0111 1011 0110 0000 0001 1000 0111 1010 1001 0010 =

0010 0111 1011 0111 1011 0110 0000 0001 1000 0111 1010 1001 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 0111 1011 0110 0000 0001 1000 0111 1010 1001 0010


Decimal number -0.000 282 018 3 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0111 1011 0110 0000 0001 1000 0111 1010 1001 0010

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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