-0.000 282 005 913 44 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 913 44(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 44(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 44| = 0.000 282 005 913 44


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 44.

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 44 × 2 = 0 + 0.000 564 011 826 88;
  • 2) 0.000 564 011 826 88 × 2 = 0 + 0.001 128 023 653 76;
  • 3) 0.001 128 023 653 76 × 2 = 0 + 0.002 256 047 307 52;
  • 4) 0.002 256 047 307 52 × 2 = 0 + 0.004 512 094 615 04;
  • 5) 0.004 512 094 615 04 × 2 = 0 + 0.009 024 189 230 08;
  • 6) 0.009 024 189 230 08 × 2 = 0 + 0.018 048 378 460 16;
  • 7) 0.018 048 378 460 16 × 2 = 0 + 0.036 096 756 920 32;
  • 8) 0.036 096 756 920 32 × 2 = 0 + 0.072 193 513 840 64;
  • 9) 0.072 193 513 840 64 × 2 = 0 + 0.144 387 027 681 28;
  • 10) 0.144 387 027 681 28 × 2 = 0 + 0.288 774 055 362 56;
  • 11) 0.288 774 055 362 56 × 2 = 0 + 0.577 548 110 725 12;
  • 12) 0.577 548 110 725 12 × 2 = 1 + 0.155 096 221 450 24;
  • 13) 0.155 096 221 450 24 × 2 = 0 + 0.310 192 442 900 48;
  • 14) 0.310 192 442 900 48 × 2 = 0 + 0.620 384 885 800 96;
  • 15) 0.620 384 885 800 96 × 2 = 1 + 0.240 769 771 601 92;
  • 16) 0.240 769 771 601 92 × 2 = 0 + 0.481 539 543 203 84;
  • 17) 0.481 539 543 203 84 × 2 = 0 + 0.963 079 086 407 68;
  • 18) 0.963 079 086 407 68 × 2 = 1 + 0.926 158 172 815 36;
  • 19) 0.926 158 172 815 36 × 2 = 1 + 0.852 316 345 630 72;
  • 20) 0.852 316 345 630 72 × 2 = 1 + 0.704 632 691 261 44;
  • 21) 0.704 632 691 261 44 × 2 = 1 + 0.409 265 382 522 88;
  • 22) 0.409 265 382 522 88 × 2 = 0 + 0.818 530 765 045 76;
  • 23) 0.818 530 765 045 76 × 2 = 1 + 0.637 061 530 091 52;
  • 24) 0.637 061 530 091 52 × 2 = 1 + 0.274 123 060 183 04;
  • 25) 0.274 123 060 183 04 × 2 = 0 + 0.548 246 120 366 08;
  • 26) 0.548 246 120 366 08 × 2 = 1 + 0.096 492 240 732 16;
  • 27) 0.096 492 240 732 16 × 2 = 0 + 0.192 984 481 464 32;
  • 28) 0.192 984 481 464 32 × 2 = 0 + 0.385 968 962 928 64;
  • 29) 0.385 968 962 928 64 × 2 = 0 + 0.771 937 925 857 28;
  • 30) 0.771 937 925 857 28 × 2 = 1 + 0.543 875 851 714 56;
  • 31) 0.543 875 851 714 56 × 2 = 1 + 0.087 751 703 429 12;
  • 32) 0.087 751 703 429 12 × 2 = 0 + 0.175 503 406 858 24;
  • 33) 0.175 503 406 858 24 × 2 = 0 + 0.351 006 813 716 48;
  • 34) 0.351 006 813 716 48 × 2 = 0 + 0.702 013 627 432 96;
  • 35) 0.702 013 627 432 96 × 2 = 1 + 0.404 027 254 865 92;
  • 36) 0.404 027 254 865 92 × 2 = 0 + 0.808 054 509 731 84;
  • 37) 0.808 054 509 731 84 × 2 = 1 + 0.616 109 019 463 68;
  • 38) 0.616 109 019 463 68 × 2 = 1 + 0.232 218 038 927 36;
  • 39) 0.232 218 038 927 36 × 2 = 0 + 0.464 436 077 854 72;
  • 40) 0.464 436 077 854 72 × 2 = 0 + 0.928 872 155 709 44;
  • 41) 0.928 872 155 709 44 × 2 = 1 + 0.857 744 311 418 88;
  • 42) 0.857 744 311 418 88 × 2 = 1 + 0.715 488 622 837 76;
  • 43) 0.715 488 622 837 76 × 2 = 1 + 0.430 977 245 675 52;
  • 44) 0.430 977 245 675 52 × 2 = 0 + 0.861 954 491 351 04;
  • 45) 0.861 954 491 351 04 × 2 = 1 + 0.723 908 982 702 08;
  • 46) 0.723 908 982 702 08 × 2 = 1 + 0.447 817 965 404 16;
  • 47) 0.447 817 965 404 16 × 2 = 0 + 0.895 635 930 808 32;
  • 48) 0.895 635 930 808 32 × 2 = 1 + 0.791 271 861 616 64;
  • 49) 0.791 271 861 616 64 × 2 = 1 + 0.582 543 723 233 28;
  • 50) 0.582 543 723 233 28 × 2 = 1 + 0.165 087 446 466 56;
  • 51) 0.165 087 446 466 56 × 2 = 0 + 0.330 174 892 933 12;
  • 52) 0.330 174 892 933 12 × 2 = 0 + 0.660 349 785 866 24;
  • 53) 0.660 349 785 866 24 × 2 = 1 + 0.320 699 571 732 48;
  • 54) 0.320 699 571 732 48 × 2 = 0 + 0.641 399 143 464 96;
  • 55) 0.641 399 143 464 96 × 2 = 1 + 0.282 798 286 929 92;
  • 56) 0.282 798 286 929 92 × 2 = 0 + 0.565 596 573 859 84;
  • 57) 0.565 596 573 859 84 × 2 = 1 + 0.131 193 147 719 68;
  • 58) 0.131 193 147 719 68 × 2 = 0 + 0.262 386 295 439 36;
  • 59) 0.262 386 295 439 36 × 2 = 0 + 0.524 772 590 878 72;
  • 60) 0.524 772 590 878 72 × 2 = 1 + 0.049 545 181 757 44;
  • 61) 0.049 545 181 757 44 × 2 = 0 + 0.099 090 363 514 88;
  • 62) 0.099 090 363 514 88 × 2 = 0 + 0.198 180 727 029 76;
  • 63) 0.198 180 727 029 76 × 2 = 0 + 0.396 361 454 059 52;
  • 64) 0.396 361 454 059 52 × 2 = 0 + 0.792 722 908 119 04;

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 44(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 1110 1101 1100 1010 1001 0000(2)

6. Positive number before normalization:

0.000 282 005 913 44(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 1110 1101 1100 1010 1001 0000(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 44(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 1110 1101 1100 1010 1001 0000(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 1110 1101 1100 1010 1001 0000(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1100 1110 1101 1100 1010 1001 0000(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 1110 1101 1100 1010 1001 0000


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 1110 1101 1100 1010 1001 0000 =

0010 0111 1011 0100 0110 0010 1100 1110 1101 1100 1010 1001 0000


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 1110 1101 1100 1010 1001 0000


Decimal number -0.000 282 005 913 44 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1100 1110 1101 1100 1010 1001 0000

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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