-0.000 282 005 866 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 866 2(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
-0.000 282 005 866 2(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:

|-0.000 282 005 866 2| = 0.000 282 005 866 2


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: 0.000 282 005 866 2.

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) 0.000 282 005 866 2 × 2 = 0 + 0.000 564 011 732 4;
  • 2) 0.000 564 011 732 4 × 2 = 0 + 0.001 128 023 464 8;
  • 3) 0.001 128 023 464 8 × 2 = 0 + 0.002 256 046 929 6;
  • 4) 0.002 256 046 929 6 × 2 = 0 + 0.004 512 093 859 2;
  • 5) 0.004 512 093 859 2 × 2 = 0 + 0.009 024 187 718 4;
  • 6) 0.009 024 187 718 4 × 2 = 0 + 0.018 048 375 436 8;
  • 7) 0.018 048 375 436 8 × 2 = 0 + 0.036 096 750 873 6;
  • 8) 0.036 096 750 873 6 × 2 = 0 + 0.072 193 501 747 2;
  • 9) 0.072 193 501 747 2 × 2 = 0 + 0.144 387 003 494 4;
  • 10) 0.144 387 003 494 4 × 2 = 0 + 0.288 774 006 988 8;
  • 11) 0.288 774 006 988 8 × 2 = 0 + 0.577 548 013 977 6;
  • 12) 0.577 548 013 977 6 × 2 = 1 + 0.155 096 027 955 2;
  • 13) 0.155 096 027 955 2 × 2 = 0 + 0.310 192 055 910 4;
  • 14) 0.310 192 055 910 4 × 2 = 0 + 0.620 384 111 820 8;
  • 15) 0.620 384 111 820 8 × 2 = 1 + 0.240 768 223 641 6;
  • 16) 0.240 768 223 641 6 × 2 = 0 + 0.481 536 447 283 2;
  • 17) 0.481 536 447 283 2 × 2 = 0 + 0.963 072 894 566 4;
  • 18) 0.963 072 894 566 4 × 2 = 1 + 0.926 145 789 132 8;
  • 19) 0.926 145 789 132 8 × 2 = 1 + 0.852 291 578 265 6;
  • 20) 0.852 291 578 265 6 × 2 = 1 + 0.704 583 156 531 2;
  • 21) 0.704 583 156 531 2 × 2 = 1 + 0.409 166 313 062 4;
  • 22) 0.409 166 313 062 4 × 2 = 0 + 0.818 332 626 124 8;
  • 23) 0.818 332 626 124 8 × 2 = 1 + 0.636 665 252 249 6;
  • 24) 0.636 665 252 249 6 × 2 = 1 + 0.273 330 504 499 2;
  • 25) 0.273 330 504 499 2 × 2 = 0 + 0.546 661 008 998 4;
  • 26) 0.546 661 008 998 4 × 2 = 1 + 0.093 322 017 996 8;
  • 27) 0.093 322 017 996 8 × 2 = 0 + 0.186 644 035 993 6;
  • 28) 0.186 644 035 993 6 × 2 = 0 + 0.373 288 071 987 2;
  • 29) 0.373 288 071 987 2 × 2 = 0 + 0.746 576 143 974 4;
  • 30) 0.746 576 143 974 4 × 2 = 1 + 0.493 152 287 948 8;
  • 31) 0.493 152 287 948 8 × 2 = 0 + 0.986 304 575 897 6;
  • 32) 0.986 304 575 897 6 × 2 = 1 + 0.972 609 151 795 2;
  • 33) 0.972 609 151 795 2 × 2 = 1 + 0.945 218 303 590 4;
  • 34) 0.945 218 303 590 4 × 2 = 1 + 0.890 436 607 180 8;
  • 35) 0.890 436 607 180 8 × 2 = 1 + 0.780 873 214 361 6;
  • 36) 0.780 873 214 361 6 × 2 = 1 + 0.561 746 428 723 2;
  • 37) 0.561 746 428 723 2 × 2 = 1 + 0.123 492 857 446 4;
  • 38) 0.123 492 857 446 4 × 2 = 0 + 0.246 985 714 892 8;
  • 39) 0.246 985 714 892 8 × 2 = 0 + 0.493 971 429 785 6;
  • 40) 0.493 971 429 785 6 × 2 = 0 + 0.987 942 859 571 2;
  • 41) 0.987 942 859 571 2 × 2 = 1 + 0.975 885 719 142 4;
  • 42) 0.975 885 719 142 4 × 2 = 1 + 0.951 771 438 284 8;
  • 43) 0.951 771 438 284 8 × 2 = 1 + 0.903 542 876 569 6;
  • 44) 0.903 542 876 569 6 × 2 = 1 + 0.807 085 753 139 2;
  • 45) 0.807 085 753 139 2 × 2 = 1 + 0.614 171 506 278 4;
  • 46) 0.614 171 506 278 4 × 2 = 1 + 0.228 343 012 556 8;
  • 47) 0.228 343 012 556 8 × 2 = 0 + 0.456 686 025 113 6;
  • 48) 0.456 686 025 113 6 × 2 = 0 + 0.913 372 050 227 2;
  • 49) 0.913 372 050 227 2 × 2 = 1 + 0.826 744 100 454 4;
  • 50) 0.826 744 100 454 4 × 2 = 1 + 0.653 488 200 908 8;
  • 51) 0.653 488 200 908 8 × 2 = 1 + 0.306 976 401 817 6;
  • 52) 0.306 976 401 817 6 × 2 = 0 + 0.613 952 803 635 2;
  • 53) 0.613 952 803 635 2 × 2 = 1 + 0.227 905 607 270 4;
  • 54) 0.227 905 607 270 4 × 2 = 0 + 0.455 811 214 540 8;
  • 55) 0.455 811 214 540 8 × 2 = 0 + 0.911 622 429 081 6;
  • 56) 0.911 622 429 081 6 × 2 = 1 + 0.823 244 858 163 2;
  • 57) 0.823 244 858 163 2 × 2 = 1 + 0.646 489 716 326 4;
  • 58) 0.646 489 716 326 4 × 2 = 1 + 0.292 979 432 652 8;
  • 59) 0.292 979 432 652 8 × 2 = 0 + 0.585 958 865 305 6;
  • 60) 0.585 958 865 305 6 × 2 = 1 + 0.171 917 730 611 2;
  • 61) 0.171 917 730 611 2 × 2 = 0 + 0.343 835 461 222 4;
  • 62) 0.343 835 461 222 4 × 2 = 0 + 0.687 670 922 444 8;
  • 63) 0.687 670 922 444 8 × 2 = 1 + 0.375 341 844 889 6;
  • 64) 0.375 341 844 889 6 × 2 = 0 + 0.750 683 689 779 2;

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:


0.000 282 005 866 2(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1111 1000 1111 1100 1110 1001 1101 0010(2)

6. Positive number before normalization:

0.000 282 005 866 2(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1111 1000 1111 1100 1110 1001 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:


0.000 282 005 866 2(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1111 1000 1111 1100 1110 1001 1101 0010(2) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1111 1000 1111 1100 1110 1001 1101 0010(2) × 20 =

1.0010 0111 1011 0100 0101 1111 1000 1111 1100 1110 1001 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 1111 1000 1111 1100 1110 1001 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 1111 1000 1111 1100 1110 1001 1101 0010 =

0010 0111 1011 0100 0101 1111 1000 1111 1100 1110 1001 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 1111 1000 1111 1100 1110 1001 1101 0010


Decimal number -0.000 282 005 866 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0101 1111 1000 1111 1100 1110 1001 1101 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