-0.000 282 005 912 24 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 912 24(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 24(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 24| = 0.000 282 005 912 24


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 24.

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 24 × 2 = 0 + 0.000 564 011 824 48;
  • 2) 0.000 564 011 824 48 × 2 = 0 + 0.001 128 023 648 96;
  • 3) 0.001 128 023 648 96 × 2 = 0 + 0.002 256 047 297 92;
  • 4) 0.002 256 047 297 92 × 2 = 0 + 0.004 512 094 595 84;
  • 5) 0.004 512 094 595 84 × 2 = 0 + 0.009 024 189 191 68;
  • 6) 0.009 024 189 191 68 × 2 = 0 + 0.018 048 378 383 36;
  • 7) 0.018 048 378 383 36 × 2 = 0 + 0.036 096 756 766 72;
  • 8) 0.036 096 756 766 72 × 2 = 0 + 0.072 193 513 533 44;
  • 9) 0.072 193 513 533 44 × 2 = 0 + 0.144 387 027 066 88;
  • 10) 0.144 387 027 066 88 × 2 = 0 + 0.288 774 054 133 76;
  • 11) 0.288 774 054 133 76 × 2 = 0 + 0.577 548 108 267 52;
  • 12) 0.577 548 108 267 52 × 2 = 1 + 0.155 096 216 535 04;
  • 13) 0.155 096 216 535 04 × 2 = 0 + 0.310 192 433 070 08;
  • 14) 0.310 192 433 070 08 × 2 = 0 + 0.620 384 866 140 16;
  • 15) 0.620 384 866 140 16 × 2 = 1 + 0.240 769 732 280 32;
  • 16) 0.240 769 732 280 32 × 2 = 0 + 0.481 539 464 560 64;
  • 17) 0.481 539 464 560 64 × 2 = 0 + 0.963 078 929 121 28;
  • 18) 0.963 078 929 121 28 × 2 = 1 + 0.926 157 858 242 56;
  • 19) 0.926 157 858 242 56 × 2 = 1 + 0.852 315 716 485 12;
  • 20) 0.852 315 716 485 12 × 2 = 1 + 0.704 631 432 970 24;
  • 21) 0.704 631 432 970 24 × 2 = 1 + 0.409 262 865 940 48;
  • 22) 0.409 262 865 940 48 × 2 = 0 + 0.818 525 731 880 96;
  • 23) 0.818 525 731 880 96 × 2 = 1 + 0.637 051 463 761 92;
  • 24) 0.637 051 463 761 92 × 2 = 1 + 0.274 102 927 523 84;
  • 25) 0.274 102 927 523 84 × 2 = 0 + 0.548 205 855 047 68;
  • 26) 0.548 205 855 047 68 × 2 = 1 + 0.096 411 710 095 36;
  • 27) 0.096 411 710 095 36 × 2 = 0 + 0.192 823 420 190 72;
  • 28) 0.192 823 420 190 72 × 2 = 0 + 0.385 646 840 381 44;
  • 29) 0.385 646 840 381 44 × 2 = 0 + 0.771 293 680 762 88;
  • 30) 0.771 293 680 762 88 × 2 = 1 + 0.542 587 361 525 76;
  • 31) 0.542 587 361 525 76 × 2 = 1 + 0.085 174 723 051 52;
  • 32) 0.085 174 723 051 52 × 2 = 0 + 0.170 349 446 103 04;
  • 33) 0.170 349 446 103 04 × 2 = 0 + 0.340 698 892 206 08;
  • 34) 0.340 698 892 206 08 × 2 = 0 + 0.681 397 784 412 16;
  • 35) 0.681 397 784 412 16 × 2 = 1 + 0.362 795 568 824 32;
  • 36) 0.362 795 568 824 32 × 2 = 0 + 0.725 591 137 648 64;
  • 37) 0.725 591 137 648 64 × 2 = 1 + 0.451 182 275 297 28;
  • 38) 0.451 182 275 297 28 × 2 = 0 + 0.902 364 550 594 56;
  • 39) 0.902 364 550 594 56 × 2 = 1 + 0.804 729 101 189 12;
  • 40) 0.804 729 101 189 12 × 2 = 1 + 0.609 458 202 378 24;
  • 41) 0.609 458 202 378 24 × 2 = 1 + 0.218 916 404 756 48;
  • 42) 0.218 916 404 756 48 × 2 = 0 + 0.437 832 809 512 96;
  • 43) 0.437 832 809 512 96 × 2 = 0 + 0.875 665 619 025 92;
  • 44) 0.875 665 619 025 92 × 2 = 1 + 0.751 331 238 051 84;
  • 45) 0.751 331 238 051 84 × 2 = 1 + 0.502 662 476 103 68;
  • 46) 0.502 662 476 103 68 × 2 = 1 + 0.005 324 952 207 36;
  • 47) 0.005 324 952 207 36 × 2 = 0 + 0.010 649 904 414 72;
  • 48) 0.010 649 904 414 72 × 2 = 0 + 0.021 299 808 829 44;
  • 49) 0.021 299 808 829 44 × 2 = 0 + 0.042 599 617 658 88;
  • 50) 0.042 599 617 658 88 × 2 = 0 + 0.085 199 235 317 76;
  • 51) 0.085 199 235 317 76 × 2 = 0 + 0.170 398 470 635 52;
  • 52) 0.170 398 470 635 52 × 2 = 0 + 0.340 796 941 271 04;
  • 53) 0.340 796 941 271 04 × 2 = 0 + 0.681 593 882 542 08;
  • 54) 0.681 593 882 542 08 × 2 = 1 + 0.363 187 765 084 16;
  • 55) 0.363 187 765 084 16 × 2 = 0 + 0.726 375 530 168 32;
  • 56) 0.726 375 530 168 32 × 2 = 1 + 0.452 751 060 336 64;
  • 57) 0.452 751 060 336 64 × 2 = 0 + 0.905 502 120 673 28;
  • 58) 0.905 502 120 673 28 × 2 = 1 + 0.811 004 241 346 56;
  • 59) 0.811 004 241 346 56 × 2 = 1 + 0.622 008 482 693 12;
  • 60) 0.622 008 482 693 12 × 2 = 1 + 0.244 016 965 386 24;
  • 61) 0.244 016 965 386 24 × 2 = 0 + 0.488 033 930 772 48;
  • 62) 0.488 033 930 772 48 × 2 = 0 + 0.976 067 861 544 96;
  • 63) 0.976 067 861 544 96 × 2 = 1 + 0.952 135 723 089 92;
  • 64) 0.952 135 723 089 92 × 2 = 1 + 0.904 271 446 179 84;

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 24(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1011 1001 1100 0000 0101 0111 0011(2)

6. Positive number before normalization:

0.000 282 005 912 24(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1011 1001 1100 0000 0101 0111 0011(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 24(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1011 1001 1100 0000 0101 0111 0011(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1011 1001 1100 0000 0101 0111 0011(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1011 1001 1100 0000 0101 0111 0011(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 1011 1001 1100 0000 0101 0111 0011


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 1011 1001 1100 0000 0101 0111 0011 =

0010 0111 1011 0100 0110 0010 1011 1001 1100 0000 0101 0111 0011


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 1011 1001 1100 0000 0101 0111 0011


Decimal number -0.000 282 005 912 24 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1011 1001 1100 0000 0101 0111 0011

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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