-0.000 282 005 912 77 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 912 77(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 77(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 77| = 0.000 282 005 912 77


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 77.

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 77 × 2 = 0 + 0.000 564 011 825 54;
  • 2) 0.000 564 011 825 54 × 2 = 0 + 0.001 128 023 651 08;
  • 3) 0.001 128 023 651 08 × 2 = 0 + 0.002 256 047 302 16;
  • 4) 0.002 256 047 302 16 × 2 = 0 + 0.004 512 094 604 32;
  • 5) 0.004 512 094 604 32 × 2 = 0 + 0.009 024 189 208 64;
  • 6) 0.009 024 189 208 64 × 2 = 0 + 0.018 048 378 417 28;
  • 7) 0.018 048 378 417 28 × 2 = 0 + 0.036 096 756 834 56;
  • 8) 0.036 096 756 834 56 × 2 = 0 + 0.072 193 513 669 12;
  • 9) 0.072 193 513 669 12 × 2 = 0 + 0.144 387 027 338 24;
  • 10) 0.144 387 027 338 24 × 2 = 0 + 0.288 774 054 676 48;
  • 11) 0.288 774 054 676 48 × 2 = 0 + 0.577 548 109 352 96;
  • 12) 0.577 548 109 352 96 × 2 = 1 + 0.155 096 218 705 92;
  • 13) 0.155 096 218 705 92 × 2 = 0 + 0.310 192 437 411 84;
  • 14) 0.310 192 437 411 84 × 2 = 0 + 0.620 384 874 823 68;
  • 15) 0.620 384 874 823 68 × 2 = 1 + 0.240 769 749 647 36;
  • 16) 0.240 769 749 647 36 × 2 = 0 + 0.481 539 499 294 72;
  • 17) 0.481 539 499 294 72 × 2 = 0 + 0.963 078 998 589 44;
  • 18) 0.963 078 998 589 44 × 2 = 1 + 0.926 157 997 178 88;
  • 19) 0.926 157 997 178 88 × 2 = 1 + 0.852 315 994 357 76;
  • 20) 0.852 315 994 357 76 × 2 = 1 + 0.704 631 988 715 52;
  • 21) 0.704 631 988 715 52 × 2 = 1 + 0.409 263 977 431 04;
  • 22) 0.409 263 977 431 04 × 2 = 0 + 0.818 527 954 862 08;
  • 23) 0.818 527 954 862 08 × 2 = 1 + 0.637 055 909 724 16;
  • 24) 0.637 055 909 724 16 × 2 = 1 + 0.274 111 819 448 32;
  • 25) 0.274 111 819 448 32 × 2 = 0 + 0.548 223 638 896 64;
  • 26) 0.548 223 638 896 64 × 2 = 1 + 0.096 447 277 793 28;
  • 27) 0.096 447 277 793 28 × 2 = 0 + 0.192 894 555 586 56;
  • 28) 0.192 894 555 586 56 × 2 = 0 + 0.385 789 111 173 12;
  • 29) 0.385 789 111 173 12 × 2 = 0 + 0.771 578 222 346 24;
  • 30) 0.771 578 222 346 24 × 2 = 1 + 0.543 156 444 692 48;
  • 31) 0.543 156 444 692 48 × 2 = 1 + 0.086 312 889 384 96;
  • 32) 0.086 312 889 384 96 × 2 = 0 + 0.172 625 778 769 92;
  • 33) 0.172 625 778 769 92 × 2 = 0 + 0.345 251 557 539 84;
  • 34) 0.345 251 557 539 84 × 2 = 0 + 0.690 503 115 079 68;
  • 35) 0.690 503 115 079 68 × 2 = 1 + 0.381 006 230 159 36;
  • 36) 0.381 006 230 159 36 × 2 = 0 + 0.762 012 460 318 72;
  • 37) 0.762 012 460 318 72 × 2 = 1 + 0.524 024 920 637 44;
  • 38) 0.524 024 920 637 44 × 2 = 1 + 0.048 049 841 274 88;
  • 39) 0.048 049 841 274 88 × 2 = 0 + 0.096 099 682 549 76;
  • 40) 0.096 099 682 549 76 × 2 = 0 + 0.192 199 365 099 52;
  • 41) 0.192 199 365 099 52 × 2 = 0 + 0.384 398 730 199 04;
  • 42) 0.384 398 730 199 04 × 2 = 0 + 0.768 797 460 398 08;
  • 43) 0.768 797 460 398 08 × 2 = 1 + 0.537 594 920 796 16;
  • 44) 0.537 594 920 796 16 × 2 = 1 + 0.075 189 841 592 32;
  • 45) 0.075 189 841 592 32 × 2 = 0 + 0.150 379 683 184 64;
  • 46) 0.150 379 683 184 64 × 2 = 0 + 0.300 759 366 369 28;
  • 47) 0.300 759 366 369 28 × 2 = 0 + 0.601 518 732 738 56;
  • 48) 0.601 518 732 738 56 × 2 = 1 + 0.203 037 465 477 12;
  • 49) 0.203 037 465 477 12 × 2 = 0 + 0.406 074 930 954 24;
  • 50) 0.406 074 930 954 24 × 2 = 0 + 0.812 149 861 908 48;
  • 51) 0.812 149 861 908 48 × 2 = 1 + 0.624 299 723 816 96;
  • 52) 0.624 299 723 816 96 × 2 = 1 + 0.248 599 447 633 92;
  • 53) 0.248 599 447 633 92 × 2 = 0 + 0.497 198 895 267 84;
  • 54) 0.497 198 895 267 84 × 2 = 0 + 0.994 397 790 535 68;
  • 55) 0.994 397 790 535 68 × 2 = 1 + 0.988 795 581 071 36;
  • 56) 0.988 795 581 071 36 × 2 = 1 + 0.977 591 162 142 72;
  • 57) 0.977 591 162 142 72 × 2 = 1 + 0.955 182 324 285 44;
  • 58) 0.955 182 324 285 44 × 2 = 1 + 0.910 364 648 570 88;
  • 59) 0.910 364 648 570 88 × 2 = 1 + 0.820 729 297 141 76;
  • 60) 0.820 729 297 141 76 × 2 = 1 + 0.641 458 594 283 52;
  • 61) 0.641 458 594 283 52 × 2 = 1 + 0.282 917 188 567 04;
  • 62) 0.282 917 188 567 04 × 2 = 0 + 0.565 834 377 134 08;
  • 63) 0.565 834 377 134 08 × 2 = 1 + 0.131 668 754 268 16;
  • 64) 0.131 668 754 268 16 × 2 = 0 + 0.263 337 508 536 32;

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 77(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 0011 0001 0011 0011 1111 1010(2)

6. Positive number before normalization:

0.000 282 005 912 77(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 0011 0001 0011 0011 1111 1010(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 77(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 0011 0001 0011 0011 1111 1010(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 0011 0001 0011 0011 1111 1010(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1100 0011 0001 0011 0011 1111 1010(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 0011 0001 0011 0011 1111 1010


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 0011 0001 0011 0011 1111 1010 =

0010 0111 1011 0100 0110 0010 1100 0011 0001 0011 0011 1111 1010


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 0011 0001 0011 0011 1111 1010


Decimal number -0.000 282 005 912 77 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1100 0011 0001 0011 0011 1111 1010

Share

» Convert decimal -0.000 282 005 913 06 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