-0.000 282 007 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 007 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
-0.000 282 007 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:

|-0.000 282 007 3| = 0.000 282 007 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: 0.000 282 007 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) 0.000 282 007 3 × 2 = 0 + 0.000 564 014 6;
  • 2) 0.000 564 014 6 × 2 = 0 + 0.001 128 029 2;
  • 3) 0.001 128 029 2 × 2 = 0 + 0.002 256 058 4;
  • 4) 0.002 256 058 4 × 2 = 0 + 0.004 512 116 8;
  • 5) 0.004 512 116 8 × 2 = 0 + 0.009 024 233 6;
  • 6) 0.009 024 233 6 × 2 = 0 + 0.018 048 467 2;
  • 7) 0.018 048 467 2 × 2 = 0 + 0.036 096 934 4;
  • 8) 0.036 096 934 4 × 2 = 0 + 0.072 193 868 8;
  • 9) 0.072 193 868 8 × 2 = 0 + 0.144 387 737 6;
  • 10) 0.144 387 737 6 × 2 = 0 + 0.288 775 475 2;
  • 11) 0.288 775 475 2 × 2 = 0 + 0.577 550 950 4;
  • 12) 0.577 550 950 4 × 2 = 1 + 0.155 101 900 8;
  • 13) 0.155 101 900 8 × 2 = 0 + 0.310 203 801 6;
  • 14) 0.310 203 801 6 × 2 = 0 + 0.620 407 603 2;
  • 15) 0.620 407 603 2 × 2 = 1 + 0.240 815 206 4;
  • 16) 0.240 815 206 4 × 2 = 0 + 0.481 630 412 8;
  • 17) 0.481 630 412 8 × 2 = 0 + 0.963 260 825 6;
  • 18) 0.963 260 825 6 × 2 = 1 + 0.926 521 651 2;
  • 19) 0.926 521 651 2 × 2 = 1 + 0.853 043 302 4;
  • 20) 0.853 043 302 4 × 2 = 1 + 0.706 086 604 8;
  • 21) 0.706 086 604 8 × 2 = 1 + 0.412 173 209 6;
  • 22) 0.412 173 209 6 × 2 = 0 + 0.824 346 419 2;
  • 23) 0.824 346 419 2 × 2 = 1 + 0.648 692 838 4;
  • 24) 0.648 692 838 4 × 2 = 1 + 0.297 385 676 8;
  • 25) 0.297 385 676 8 × 2 = 0 + 0.594 771 353 6;
  • 26) 0.594 771 353 6 × 2 = 1 + 0.189 542 707 2;
  • 27) 0.189 542 707 2 × 2 = 0 + 0.379 085 414 4;
  • 28) 0.379 085 414 4 × 2 = 0 + 0.758 170 828 8;
  • 29) 0.758 170 828 8 × 2 = 1 + 0.516 341 657 6;
  • 30) 0.516 341 657 6 × 2 = 1 + 0.032 683 315 2;
  • 31) 0.032 683 315 2 × 2 = 0 + 0.065 366 630 4;
  • 32) 0.065 366 630 4 × 2 = 0 + 0.130 733 260 8;
  • 33) 0.130 733 260 8 × 2 = 0 + 0.261 466 521 6;
  • 34) 0.261 466 521 6 × 2 = 0 + 0.522 933 043 2;
  • 35) 0.522 933 043 2 × 2 = 1 + 0.045 866 086 4;
  • 36) 0.045 866 086 4 × 2 = 0 + 0.091 732 172 8;
  • 37) 0.091 732 172 8 × 2 = 0 + 0.183 464 345 6;
  • 38) 0.183 464 345 6 × 2 = 0 + 0.366 928 691 2;
  • 39) 0.366 928 691 2 × 2 = 0 + 0.733 857 382 4;
  • 40) 0.733 857 382 4 × 2 = 1 + 0.467 714 764 8;
  • 41) 0.467 714 764 8 × 2 = 0 + 0.935 429 529 6;
  • 42) 0.935 429 529 6 × 2 = 1 + 0.870 859 059 2;
  • 43) 0.870 859 059 2 × 2 = 1 + 0.741 718 118 4;
  • 44) 0.741 718 118 4 × 2 = 1 + 0.483 436 236 8;
  • 45) 0.483 436 236 8 × 2 = 0 + 0.966 872 473 6;
  • 46) 0.966 872 473 6 × 2 = 1 + 0.933 744 947 2;
  • 47) 0.933 744 947 2 × 2 = 1 + 0.867 489 894 4;
  • 48) 0.867 489 894 4 × 2 = 1 + 0.734 979 788 8;
  • 49) 0.734 979 788 8 × 2 = 1 + 0.469 959 577 6;
  • 50) 0.469 959 577 6 × 2 = 0 + 0.939 919 155 2;
  • 51) 0.939 919 155 2 × 2 = 1 + 0.879 838 310 4;
  • 52) 0.879 838 310 4 × 2 = 1 + 0.759 676 620 8;
  • 53) 0.759 676 620 8 × 2 = 1 + 0.519 353 241 6;
  • 54) 0.519 353 241 6 × 2 = 1 + 0.038 706 483 2;
  • 55) 0.038 706 483 2 × 2 = 0 + 0.077 412 966 4;
  • 56) 0.077 412 966 4 × 2 = 0 + 0.154 825 932 8;
  • 57) 0.154 825 932 8 × 2 = 0 + 0.309 651 865 6;
  • 58) 0.309 651 865 6 × 2 = 0 + 0.619 303 731 2;
  • 59) 0.619 303 731 2 × 2 = 1 + 0.238 607 462 4;
  • 60) 0.238 607 462 4 × 2 = 0 + 0.477 214 924 8;
  • 61) 0.477 214 924 8 × 2 = 0 + 0.954 429 849 6;
  • 62) 0.954 429 849 6 × 2 = 1 + 0.908 859 699 2;
  • 63) 0.908 859 699 2 × 2 = 1 + 0.817 719 398 4;
  • 64) 0.817 719 398 4 × 2 = 1 + 0.635 438 796 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:


0.000 282 007 3(10) =

0.0000 0000 0001 0010 0111 1011 0100 1100 0010 0001 0111 0111 1011 1100 0010 0111(2)

6. Positive number before normalization:

0.000 282 007 3(10) =

0.0000 0000 0001 0010 0111 1011 0100 1100 0010 0001 0111 0111 1011 1100 0010 0111(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:


0.000 282 007 3(10) =

0.0000 0000 0001 0010 0111 1011 0100 1100 0010 0001 0111 0111 1011 1100 0010 0111(2) =

0.0000 0000 0001 0010 0111 1011 0100 1100 0010 0001 0111 0111 1011 1100 0010 0111(2) × 20 =

1.0010 0111 1011 0100 1100 0010 0001 0111 0111 1011 1100 0010 0111(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 1100 0010 0001 0111 0111 1011 1100 0010 0111


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 1100 0010 0001 0111 0111 1011 1100 0010 0111 =

0010 0111 1011 0100 1100 0010 0001 0111 0111 1011 1100 0010 0111


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 1100 0010 0001 0111 0111 1011 1100 0010 0111


Decimal number -0.000 282 007 3 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 1100 0010 0001 0111 0111 1011 1100 0010 0111

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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