-0.000 282 010 91 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 010 91(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
-0.000 282 010 91(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:

|-0.000 282 010 91| = 0.000 282 010 91


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: 0.000 282 010 91.

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) 0.000 282 010 91 × 2 = 0 + 0.000 564 021 82;
  • 2) 0.000 564 021 82 × 2 = 0 + 0.001 128 043 64;
  • 3) 0.001 128 043 64 × 2 = 0 + 0.002 256 087 28;
  • 4) 0.002 256 087 28 × 2 = 0 + 0.004 512 174 56;
  • 5) 0.004 512 174 56 × 2 = 0 + 0.009 024 349 12;
  • 6) 0.009 024 349 12 × 2 = 0 + 0.018 048 698 24;
  • 7) 0.018 048 698 24 × 2 = 0 + 0.036 097 396 48;
  • 8) 0.036 097 396 48 × 2 = 0 + 0.072 194 792 96;
  • 9) 0.072 194 792 96 × 2 = 0 + 0.144 389 585 92;
  • 10) 0.144 389 585 92 × 2 = 0 + 0.288 779 171 84;
  • 11) 0.288 779 171 84 × 2 = 0 + 0.577 558 343 68;
  • 12) 0.577 558 343 68 × 2 = 1 + 0.155 116 687 36;
  • 13) 0.155 116 687 36 × 2 = 0 + 0.310 233 374 72;
  • 14) 0.310 233 374 72 × 2 = 0 + 0.620 466 749 44;
  • 15) 0.620 466 749 44 × 2 = 1 + 0.240 933 498 88;
  • 16) 0.240 933 498 88 × 2 = 0 + 0.481 866 997 76;
  • 17) 0.481 866 997 76 × 2 = 0 + 0.963 733 995 52;
  • 18) 0.963 733 995 52 × 2 = 1 + 0.927 467 991 04;
  • 19) 0.927 467 991 04 × 2 = 1 + 0.854 935 982 08;
  • 20) 0.854 935 982 08 × 2 = 1 + 0.709 871 964 16;
  • 21) 0.709 871 964 16 × 2 = 1 + 0.419 743 928 32;
  • 22) 0.419 743 928 32 × 2 = 0 + 0.839 487 856 64;
  • 23) 0.839 487 856 64 × 2 = 1 + 0.678 975 713 28;
  • 24) 0.678 975 713 28 × 2 = 1 + 0.357 951 426 56;
  • 25) 0.357 951 426 56 × 2 = 0 + 0.715 902 853 12;
  • 26) 0.715 902 853 12 × 2 = 1 + 0.431 805 706 24;
  • 27) 0.431 805 706 24 × 2 = 0 + 0.863 611 412 48;
  • 28) 0.863 611 412 48 × 2 = 1 + 0.727 222 824 96;
  • 29) 0.727 222 824 96 × 2 = 1 + 0.454 445 649 92;
  • 30) 0.454 445 649 92 × 2 = 0 + 0.908 891 299 84;
  • 31) 0.908 891 299 84 × 2 = 1 + 0.817 782 599 68;
  • 32) 0.817 782 599 68 × 2 = 1 + 0.635 565 199 36;
  • 33) 0.635 565 199 36 × 2 = 1 + 0.271 130 398 72;
  • 34) 0.271 130 398 72 × 2 = 0 + 0.542 260 797 44;
  • 35) 0.542 260 797 44 × 2 = 1 + 0.084 521 594 88;
  • 36) 0.084 521 594 88 × 2 = 0 + 0.169 043 189 76;
  • 37) 0.169 043 189 76 × 2 = 0 + 0.338 086 379 52;
  • 38) 0.338 086 379 52 × 2 = 0 + 0.676 172 759 04;
  • 39) 0.676 172 759 04 × 2 = 1 + 0.352 345 518 08;
  • 40) 0.352 345 518 08 × 2 = 0 + 0.704 691 036 16;
  • 41) 0.704 691 036 16 × 2 = 1 + 0.409 382 072 32;
  • 42) 0.409 382 072 32 × 2 = 0 + 0.818 764 144 64;
  • 43) 0.818 764 144 64 × 2 = 1 + 0.637 528 289 28;
  • 44) 0.637 528 289 28 × 2 = 1 + 0.275 056 578 56;
  • 45) 0.275 056 578 56 × 2 = 0 + 0.550 113 157 12;
  • 46) 0.550 113 157 12 × 2 = 1 + 0.100 226 314 24;
  • 47) 0.100 226 314 24 × 2 = 0 + 0.200 452 628 48;
  • 48) 0.200 452 628 48 × 2 = 0 + 0.400 905 256 96;
  • 49) 0.400 905 256 96 × 2 = 0 + 0.801 810 513 92;
  • 50) 0.801 810 513 92 × 2 = 1 + 0.603 621 027 84;
  • 51) 0.603 621 027 84 × 2 = 1 + 0.207 242 055 68;
  • 52) 0.207 242 055 68 × 2 = 0 + 0.414 484 111 36;
  • 53) 0.414 484 111 36 × 2 = 0 + 0.828 968 222 72;
  • 54) 0.828 968 222 72 × 2 = 1 + 0.657 936 445 44;
  • 55) 0.657 936 445 44 × 2 = 1 + 0.315 872 890 88;
  • 56) 0.315 872 890 88 × 2 = 0 + 0.631 745 781 76;
  • 57) 0.631 745 781 76 × 2 = 1 + 0.263 491 563 52;
  • 58) 0.263 491 563 52 × 2 = 0 + 0.526 983 127 04;
  • 59) 0.526 983 127 04 × 2 = 1 + 0.053 966 254 08;
  • 60) 0.053 966 254 08 × 2 = 0 + 0.107 932 508 16;
  • 61) 0.107 932 508 16 × 2 = 0 + 0.215 865 016 32;
  • 62) 0.215 865 016 32 × 2 = 0 + 0.431 730 032 64;
  • 63) 0.431 730 032 64 × 2 = 0 + 0.863 460 065 28;
  • 64) 0.863 460 065 28 × 2 = 1 + 0.726 920 130 56;

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:


0.000 282 010 91(10) =

0.0000 0000 0001 0010 0111 1011 0101 1011 1010 0010 1011 0100 0110 0110 1010 0001(2)

6. Positive number before normalization:

0.000 282 010 91(10) =

0.0000 0000 0001 0010 0111 1011 0101 1011 1010 0010 1011 0100 0110 0110 1010 0001(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:


0.000 282 010 91(10) =

0.0000 0000 0001 0010 0111 1011 0101 1011 1010 0010 1011 0100 0110 0110 1010 0001(2) =

0.0000 0000 0001 0010 0111 1011 0101 1011 1010 0010 1011 0100 0110 0110 1010 0001(2) × 20 =

1.0010 0111 1011 0101 1011 1010 0010 1011 0100 0110 0110 1010 0001(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 0101 1011 1010 0010 1011 0100 0110 0110 1010 0001


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 0101 1011 1010 0010 1011 0100 0110 0110 1010 0001 =

0010 0111 1011 0101 1011 1010 0010 1011 0100 0110 0110 1010 0001


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 0101 1011 1010 0010 1011 0100 0110 0110 1010 0001


Decimal number -0.000 282 010 91 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0101 1011 1010 0010 1011 0100 0110 0110 1010 0001

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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