-0.000 282 005 913 32 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 913 32(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 32(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 32| = 0.000 282 005 913 32


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 32.

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 32 × 2 = 0 + 0.000 564 011 826 64;
  • 2) 0.000 564 011 826 64 × 2 = 0 + 0.001 128 023 653 28;
  • 3) 0.001 128 023 653 28 × 2 = 0 + 0.002 256 047 306 56;
  • 4) 0.002 256 047 306 56 × 2 = 0 + 0.004 512 094 613 12;
  • 5) 0.004 512 094 613 12 × 2 = 0 + 0.009 024 189 226 24;
  • 6) 0.009 024 189 226 24 × 2 = 0 + 0.018 048 378 452 48;
  • 7) 0.018 048 378 452 48 × 2 = 0 + 0.036 096 756 904 96;
  • 8) 0.036 096 756 904 96 × 2 = 0 + 0.072 193 513 809 92;
  • 9) 0.072 193 513 809 92 × 2 = 0 + 0.144 387 027 619 84;
  • 10) 0.144 387 027 619 84 × 2 = 0 + 0.288 774 055 239 68;
  • 11) 0.288 774 055 239 68 × 2 = 0 + 0.577 548 110 479 36;
  • 12) 0.577 548 110 479 36 × 2 = 1 + 0.155 096 220 958 72;
  • 13) 0.155 096 220 958 72 × 2 = 0 + 0.310 192 441 917 44;
  • 14) 0.310 192 441 917 44 × 2 = 0 + 0.620 384 883 834 88;
  • 15) 0.620 384 883 834 88 × 2 = 1 + 0.240 769 767 669 76;
  • 16) 0.240 769 767 669 76 × 2 = 0 + 0.481 539 535 339 52;
  • 17) 0.481 539 535 339 52 × 2 = 0 + 0.963 079 070 679 04;
  • 18) 0.963 079 070 679 04 × 2 = 1 + 0.926 158 141 358 08;
  • 19) 0.926 158 141 358 08 × 2 = 1 + 0.852 316 282 716 16;
  • 20) 0.852 316 282 716 16 × 2 = 1 + 0.704 632 565 432 32;
  • 21) 0.704 632 565 432 32 × 2 = 1 + 0.409 265 130 864 64;
  • 22) 0.409 265 130 864 64 × 2 = 0 + 0.818 530 261 729 28;
  • 23) 0.818 530 261 729 28 × 2 = 1 + 0.637 060 523 458 56;
  • 24) 0.637 060 523 458 56 × 2 = 1 + 0.274 121 046 917 12;
  • 25) 0.274 121 046 917 12 × 2 = 0 + 0.548 242 093 834 24;
  • 26) 0.548 242 093 834 24 × 2 = 1 + 0.096 484 187 668 48;
  • 27) 0.096 484 187 668 48 × 2 = 0 + 0.192 968 375 336 96;
  • 28) 0.192 968 375 336 96 × 2 = 0 + 0.385 936 750 673 92;
  • 29) 0.385 936 750 673 92 × 2 = 0 + 0.771 873 501 347 84;
  • 30) 0.771 873 501 347 84 × 2 = 1 + 0.543 747 002 695 68;
  • 31) 0.543 747 002 695 68 × 2 = 1 + 0.087 494 005 391 36;
  • 32) 0.087 494 005 391 36 × 2 = 0 + 0.174 988 010 782 72;
  • 33) 0.174 988 010 782 72 × 2 = 0 + 0.349 976 021 565 44;
  • 34) 0.349 976 021 565 44 × 2 = 0 + 0.699 952 043 130 88;
  • 35) 0.699 952 043 130 88 × 2 = 1 + 0.399 904 086 261 76;
  • 36) 0.399 904 086 261 76 × 2 = 0 + 0.799 808 172 523 52;
  • 37) 0.799 808 172 523 52 × 2 = 1 + 0.599 616 345 047 04;
  • 38) 0.599 616 345 047 04 × 2 = 1 + 0.199 232 690 094 08;
  • 39) 0.199 232 690 094 08 × 2 = 0 + 0.398 465 380 188 16;
  • 40) 0.398 465 380 188 16 × 2 = 0 + 0.796 930 760 376 32;
  • 41) 0.796 930 760 376 32 × 2 = 1 + 0.593 861 520 752 64;
  • 42) 0.593 861 520 752 64 × 2 = 1 + 0.187 723 041 505 28;
  • 43) 0.187 723 041 505 28 × 2 = 0 + 0.375 446 083 010 56;
  • 44) 0.375 446 083 010 56 × 2 = 0 + 0.750 892 166 021 12;
  • 45) 0.750 892 166 021 12 × 2 = 1 + 0.501 784 332 042 24;
  • 46) 0.501 784 332 042 24 × 2 = 1 + 0.003 568 664 084 48;
  • 47) 0.003 568 664 084 48 × 2 = 0 + 0.007 137 328 168 96;
  • 48) 0.007 137 328 168 96 × 2 = 0 + 0.014 274 656 337 92;
  • 49) 0.014 274 656 337 92 × 2 = 0 + 0.028 549 312 675 84;
  • 50) 0.028 549 312 675 84 × 2 = 0 + 0.057 098 625 351 68;
  • 51) 0.057 098 625 351 68 × 2 = 0 + 0.114 197 250 703 36;
  • 52) 0.114 197 250 703 36 × 2 = 0 + 0.228 394 501 406 72;
  • 53) 0.228 394 501 406 72 × 2 = 0 + 0.456 789 002 813 44;
  • 54) 0.456 789 002 813 44 × 2 = 0 + 0.913 578 005 626 88;
  • 55) 0.913 578 005 626 88 × 2 = 1 + 0.827 156 011 253 76;
  • 56) 0.827 156 011 253 76 × 2 = 1 + 0.654 312 022 507 52;
  • 57) 0.654 312 022 507 52 × 2 = 1 + 0.308 624 045 015 04;
  • 58) 0.308 624 045 015 04 × 2 = 0 + 0.617 248 090 030 08;
  • 59) 0.617 248 090 030 08 × 2 = 1 + 0.234 496 180 060 16;
  • 60) 0.234 496 180 060 16 × 2 = 0 + 0.468 992 360 120 32;
  • 61) 0.468 992 360 120 32 × 2 = 0 + 0.937 984 720 240 64;
  • 62) 0.937 984 720 240 64 × 2 = 1 + 0.875 969 440 481 28;
  • 63) 0.875 969 440 481 28 × 2 = 1 + 0.751 938 880 962 56;
  • 64) 0.751 938 880 962 56 × 2 = 1 + 0.503 877 761 925 12;

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 32(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 1100 1100 0000 0011 1010 0111(2)

6. Positive number before normalization:

0.000 282 005 913 32(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 1100 1100 0000 0011 1010 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:


0.000 282 005 913 32(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 1100 1100 0000 0011 1010 0111(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 1100 1100 0000 0011 1010 0111(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1100 1100 1100 0000 0011 1010 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 0110 0010 1100 1100 1100 0000 0011 1010 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 0110 0010 1100 1100 1100 0000 0011 1010 0111 =

0010 0111 1011 0100 0110 0010 1100 1100 1100 0000 0011 1010 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 0110 0010 1100 1100 1100 0000 0011 1010 0111


Decimal number -0.000 282 005 913 32 converted to 64 bit double precision IEEE 754 binary floating point representation:

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

Share

» Convert decimal -0.000 282 005 913 3 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: The Hidden Requirement in Every Single Job Description. NotebookLM YouTube Video of 8.15 minutes

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