-0.000 282 005 912 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 912 9(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
-0.000 282 005 912 9(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:

|-0.000 282 005 912 9| = 0.000 282 005 912 9


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: 0.000 282 005 912 9.

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) 0.000 282 005 912 9 × 2 = 0 + 0.000 564 011 825 8;
  • 2) 0.000 564 011 825 8 × 2 = 0 + 0.001 128 023 651 6;
  • 3) 0.001 128 023 651 6 × 2 = 0 + 0.002 256 047 303 2;
  • 4) 0.002 256 047 303 2 × 2 = 0 + 0.004 512 094 606 4;
  • 5) 0.004 512 094 606 4 × 2 = 0 + 0.009 024 189 212 8;
  • 6) 0.009 024 189 212 8 × 2 = 0 + 0.018 048 378 425 6;
  • 7) 0.018 048 378 425 6 × 2 = 0 + 0.036 096 756 851 2;
  • 8) 0.036 096 756 851 2 × 2 = 0 + 0.072 193 513 702 4;
  • 9) 0.072 193 513 702 4 × 2 = 0 + 0.144 387 027 404 8;
  • 10) 0.144 387 027 404 8 × 2 = 0 + 0.288 774 054 809 6;
  • 11) 0.288 774 054 809 6 × 2 = 0 + 0.577 548 109 619 2;
  • 12) 0.577 548 109 619 2 × 2 = 1 + 0.155 096 219 238 4;
  • 13) 0.155 096 219 238 4 × 2 = 0 + 0.310 192 438 476 8;
  • 14) 0.310 192 438 476 8 × 2 = 0 + 0.620 384 876 953 6;
  • 15) 0.620 384 876 953 6 × 2 = 1 + 0.240 769 753 907 2;
  • 16) 0.240 769 753 907 2 × 2 = 0 + 0.481 539 507 814 4;
  • 17) 0.481 539 507 814 4 × 2 = 0 + 0.963 079 015 628 8;
  • 18) 0.963 079 015 628 8 × 2 = 1 + 0.926 158 031 257 6;
  • 19) 0.926 158 031 257 6 × 2 = 1 + 0.852 316 062 515 2;
  • 20) 0.852 316 062 515 2 × 2 = 1 + 0.704 632 125 030 4;
  • 21) 0.704 632 125 030 4 × 2 = 1 + 0.409 264 250 060 8;
  • 22) 0.409 264 250 060 8 × 2 = 0 + 0.818 528 500 121 6;
  • 23) 0.818 528 500 121 6 × 2 = 1 + 0.637 057 000 243 2;
  • 24) 0.637 057 000 243 2 × 2 = 1 + 0.274 114 000 486 4;
  • 25) 0.274 114 000 486 4 × 2 = 0 + 0.548 228 000 972 8;
  • 26) 0.548 228 000 972 8 × 2 = 1 + 0.096 456 001 945 6;
  • 27) 0.096 456 001 945 6 × 2 = 0 + 0.192 912 003 891 2;
  • 28) 0.192 912 003 891 2 × 2 = 0 + 0.385 824 007 782 4;
  • 29) 0.385 824 007 782 4 × 2 = 0 + 0.771 648 015 564 8;
  • 30) 0.771 648 015 564 8 × 2 = 1 + 0.543 296 031 129 6;
  • 31) 0.543 296 031 129 6 × 2 = 1 + 0.086 592 062 259 2;
  • 32) 0.086 592 062 259 2 × 2 = 0 + 0.173 184 124 518 4;
  • 33) 0.173 184 124 518 4 × 2 = 0 + 0.346 368 249 036 8;
  • 34) 0.346 368 249 036 8 × 2 = 0 + 0.692 736 498 073 6;
  • 35) 0.692 736 498 073 6 × 2 = 1 + 0.385 472 996 147 2;
  • 36) 0.385 472 996 147 2 × 2 = 0 + 0.770 945 992 294 4;
  • 37) 0.770 945 992 294 4 × 2 = 1 + 0.541 891 984 588 8;
  • 38) 0.541 891 984 588 8 × 2 = 1 + 0.083 783 969 177 6;
  • 39) 0.083 783 969 177 6 × 2 = 0 + 0.167 567 938 355 2;
  • 40) 0.167 567 938 355 2 × 2 = 0 + 0.335 135 876 710 4;
  • 41) 0.335 135 876 710 4 × 2 = 0 + 0.670 271 753 420 8;
  • 42) 0.670 271 753 420 8 × 2 = 1 + 0.340 543 506 841 6;
  • 43) 0.340 543 506 841 6 × 2 = 0 + 0.681 087 013 683 2;
  • 44) 0.681 087 013 683 2 × 2 = 1 + 0.362 174 027 366 4;
  • 45) 0.362 174 027 366 4 × 2 = 0 + 0.724 348 054 732 8;
  • 46) 0.724 348 054 732 8 × 2 = 1 + 0.448 696 109 465 6;
  • 47) 0.448 696 109 465 6 × 2 = 0 + 0.897 392 218 931 2;
  • 48) 0.897 392 218 931 2 × 2 = 1 + 0.794 784 437 862 4;
  • 49) 0.794 784 437 862 4 × 2 = 1 + 0.589 568 875 724 8;
  • 50) 0.589 568 875 724 8 × 2 = 1 + 0.179 137 751 449 6;
  • 51) 0.179 137 751 449 6 × 2 = 0 + 0.358 275 502 899 2;
  • 52) 0.358 275 502 899 2 × 2 = 0 + 0.716 551 005 798 4;
  • 53) 0.716 551 005 798 4 × 2 = 1 + 0.433 102 011 596 8;
  • 54) 0.433 102 011 596 8 × 2 = 0 + 0.866 204 023 193 6;
  • 55) 0.866 204 023 193 6 × 2 = 1 + 0.732 408 046 387 2;
  • 56) 0.732 408 046 387 2 × 2 = 1 + 0.464 816 092 774 4;
  • 57) 0.464 816 092 774 4 × 2 = 0 + 0.929 632 185 548 8;
  • 58) 0.929 632 185 548 8 × 2 = 1 + 0.859 264 371 097 6;
  • 59) 0.859 264 371 097 6 × 2 = 1 + 0.718 528 742 195 2;
  • 60) 0.718 528 742 195 2 × 2 = 1 + 0.437 057 484 390 4;
  • 61) 0.437 057 484 390 4 × 2 = 0 + 0.874 114 968 780 8;
  • 62) 0.874 114 968 780 8 × 2 = 1 + 0.748 229 937 561 6;
  • 63) 0.748 229 937 561 6 × 2 = 1 + 0.496 459 875 123 2;
  • 64) 0.496 459 875 123 2 × 2 = 0 + 0.992 919 750 246 4;

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:


0.000 282 005 912 9(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 0101 0101 1100 1011 0111 0110(2)

6. Positive number before normalization:

0.000 282 005 912 9(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 0101 0101 1100 1011 0111 0110(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:


0.000 282 005 912 9(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 0101 0101 1100 1011 0111 0110(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 0101 0101 1100 1011 0111 0110(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1100 0101 0101 1100 1011 0111 0110(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 0110 0010 1100 0101 0101 1100 1011 0111 0110


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 0110 0010 1100 0101 0101 1100 1011 0111 0110 =

0010 0111 1011 0100 0110 0010 1100 0101 0101 1100 1011 0111 0110


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 0110 0010 1100 0101 0101 1100 1011 0111 0110


Decimal number -0.000 282 005 912 9 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1100 0101 0101 1100 1011 0111 0110

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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