-0.000 282 005 913 64 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 913 64(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
-0.000 282 005 913 64(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:

|-0.000 282 005 913 64| = 0.000 282 005 913 64


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: 0.000 282 005 913 64.

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) 0.000 282 005 913 64 × 2 = 0 + 0.000 564 011 827 28;
  • 2) 0.000 564 011 827 28 × 2 = 0 + 0.001 128 023 654 56;
  • 3) 0.001 128 023 654 56 × 2 = 0 + 0.002 256 047 309 12;
  • 4) 0.002 256 047 309 12 × 2 = 0 + 0.004 512 094 618 24;
  • 5) 0.004 512 094 618 24 × 2 = 0 + 0.009 024 189 236 48;
  • 6) 0.009 024 189 236 48 × 2 = 0 + 0.018 048 378 472 96;
  • 7) 0.018 048 378 472 96 × 2 = 0 + 0.036 096 756 945 92;
  • 8) 0.036 096 756 945 92 × 2 = 0 + 0.072 193 513 891 84;
  • 9) 0.072 193 513 891 84 × 2 = 0 + 0.144 387 027 783 68;
  • 10) 0.144 387 027 783 68 × 2 = 0 + 0.288 774 055 567 36;
  • 11) 0.288 774 055 567 36 × 2 = 0 + 0.577 548 111 134 72;
  • 12) 0.577 548 111 134 72 × 2 = 1 + 0.155 096 222 269 44;
  • 13) 0.155 096 222 269 44 × 2 = 0 + 0.310 192 444 538 88;
  • 14) 0.310 192 444 538 88 × 2 = 0 + 0.620 384 889 077 76;
  • 15) 0.620 384 889 077 76 × 2 = 1 + 0.240 769 778 155 52;
  • 16) 0.240 769 778 155 52 × 2 = 0 + 0.481 539 556 311 04;
  • 17) 0.481 539 556 311 04 × 2 = 0 + 0.963 079 112 622 08;
  • 18) 0.963 079 112 622 08 × 2 = 1 + 0.926 158 225 244 16;
  • 19) 0.926 158 225 244 16 × 2 = 1 + 0.852 316 450 488 32;
  • 20) 0.852 316 450 488 32 × 2 = 1 + 0.704 632 900 976 64;
  • 21) 0.704 632 900 976 64 × 2 = 1 + 0.409 265 801 953 28;
  • 22) 0.409 265 801 953 28 × 2 = 0 + 0.818 531 603 906 56;
  • 23) 0.818 531 603 906 56 × 2 = 1 + 0.637 063 207 813 12;
  • 24) 0.637 063 207 813 12 × 2 = 1 + 0.274 126 415 626 24;
  • 25) 0.274 126 415 626 24 × 2 = 0 + 0.548 252 831 252 48;
  • 26) 0.548 252 831 252 48 × 2 = 1 + 0.096 505 662 504 96;
  • 27) 0.096 505 662 504 96 × 2 = 0 + 0.193 011 325 009 92;
  • 28) 0.193 011 325 009 92 × 2 = 0 + 0.386 022 650 019 84;
  • 29) 0.386 022 650 019 84 × 2 = 0 + 0.772 045 300 039 68;
  • 30) 0.772 045 300 039 68 × 2 = 1 + 0.544 090 600 079 36;
  • 31) 0.544 090 600 079 36 × 2 = 1 + 0.088 181 200 158 72;
  • 32) 0.088 181 200 158 72 × 2 = 0 + 0.176 362 400 317 44;
  • 33) 0.176 362 400 317 44 × 2 = 0 + 0.352 724 800 634 88;
  • 34) 0.352 724 800 634 88 × 2 = 0 + 0.705 449 601 269 76;
  • 35) 0.705 449 601 269 76 × 2 = 1 + 0.410 899 202 539 52;
  • 36) 0.410 899 202 539 52 × 2 = 0 + 0.821 798 405 079 04;
  • 37) 0.821 798 405 079 04 × 2 = 1 + 0.643 596 810 158 08;
  • 38) 0.643 596 810 158 08 × 2 = 1 + 0.287 193 620 316 16;
  • 39) 0.287 193 620 316 16 × 2 = 0 + 0.574 387 240 632 32;
  • 40) 0.574 387 240 632 32 × 2 = 1 + 0.148 774 481 264 64;
  • 41) 0.148 774 481 264 64 × 2 = 0 + 0.297 548 962 529 28;
  • 42) 0.297 548 962 529 28 × 2 = 0 + 0.595 097 925 058 56;
  • 43) 0.595 097 925 058 56 × 2 = 1 + 0.190 195 850 117 12;
  • 44) 0.190 195 850 117 12 × 2 = 0 + 0.380 391 700 234 24;
  • 45) 0.380 391 700 234 24 × 2 = 0 + 0.760 783 400 468 48;
  • 46) 0.760 783 400 468 48 × 2 = 1 + 0.521 566 800 936 96;
  • 47) 0.521 566 800 936 96 × 2 = 1 + 0.043 133 601 873 92;
  • 48) 0.043 133 601 873 92 × 2 = 0 + 0.086 267 203 747 84;
  • 49) 0.086 267 203 747 84 × 2 = 0 + 0.172 534 407 495 68;
  • 50) 0.172 534 407 495 68 × 2 = 0 + 0.345 068 814 991 36;
  • 51) 0.345 068 814 991 36 × 2 = 0 + 0.690 137 629 982 72;
  • 52) 0.690 137 629 982 72 × 2 = 1 + 0.380 275 259 965 44;
  • 53) 0.380 275 259 965 44 × 2 = 0 + 0.760 550 519 930 88;
  • 54) 0.760 550 519 930 88 × 2 = 1 + 0.521 101 039 861 76;
  • 55) 0.521 101 039 861 76 × 2 = 1 + 0.042 202 079 723 52;
  • 56) 0.042 202 079 723 52 × 2 = 0 + 0.084 404 159 447 04;
  • 57) 0.084 404 159 447 04 × 2 = 0 + 0.168 808 318 894 08;
  • 58) 0.168 808 318 894 08 × 2 = 0 + 0.337 616 637 788 16;
  • 59) 0.337 616 637 788 16 × 2 = 0 + 0.675 233 275 576 32;
  • 60) 0.675 233 275 576 32 × 2 = 1 + 0.350 466 551 152 64;
  • 61) 0.350 466 551 152 64 × 2 = 0 + 0.700 933 102 305 28;
  • 62) 0.700 933 102 305 28 × 2 = 1 + 0.401 866 204 610 56;
  • 63) 0.401 866 204 610 56 × 2 = 0 + 0.803 732 409 221 12;
  • 64) 0.803 732 409 221 12 × 2 = 1 + 0.607 464 818 442 24;

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:


0.000 282 005 913 64(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 0010 0110 0001 0110 0001 0101(2)

6. Positive number before normalization:

0.000 282 005 913 64(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 0010 0110 0001 0110 0001 0101(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:


0.000 282 005 913 64(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 0010 0110 0001 0110 0001 0101(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 0010 0110 0001 0110 0001 0101(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1101 0010 0110 0001 0110 0001 0101(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 1101 0010 0110 0001 0110 0001 0101


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 1101 0010 0110 0001 0110 0001 0101 =

0010 0111 1011 0100 0110 0010 1101 0010 0110 0001 0110 0001 0101


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 1101 0010 0110 0001 0110 0001 0101


Decimal number -0.000 282 005 913 64 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1101 0010 0110 0001 0110 0001 0101

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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