-0.000 282 005 915 91 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 915 91(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
-0.000 282 005 915 91(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:

|-0.000 282 005 915 91| = 0.000 282 005 915 91


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: 0.000 282 005 915 91.

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) 0.000 282 005 915 91 × 2 = 0 + 0.000 564 011 831 82;
  • 2) 0.000 564 011 831 82 × 2 = 0 + 0.001 128 023 663 64;
  • 3) 0.001 128 023 663 64 × 2 = 0 + 0.002 256 047 327 28;
  • 4) 0.002 256 047 327 28 × 2 = 0 + 0.004 512 094 654 56;
  • 5) 0.004 512 094 654 56 × 2 = 0 + 0.009 024 189 309 12;
  • 6) 0.009 024 189 309 12 × 2 = 0 + 0.018 048 378 618 24;
  • 7) 0.018 048 378 618 24 × 2 = 0 + 0.036 096 757 236 48;
  • 8) 0.036 096 757 236 48 × 2 = 0 + 0.072 193 514 472 96;
  • 9) 0.072 193 514 472 96 × 2 = 0 + 0.144 387 028 945 92;
  • 10) 0.144 387 028 945 92 × 2 = 0 + 0.288 774 057 891 84;
  • 11) 0.288 774 057 891 84 × 2 = 0 + 0.577 548 115 783 68;
  • 12) 0.577 548 115 783 68 × 2 = 1 + 0.155 096 231 567 36;
  • 13) 0.155 096 231 567 36 × 2 = 0 + 0.310 192 463 134 72;
  • 14) 0.310 192 463 134 72 × 2 = 0 + 0.620 384 926 269 44;
  • 15) 0.620 384 926 269 44 × 2 = 1 + 0.240 769 852 538 88;
  • 16) 0.240 769 852 538 88 × 2 = 0 + 0.481 539 705 077 76;
  • 17) 0.481 539 705 077 76 × 2 = 0 + 0.963 079 410 155 52;
  • 18) 0.963 079 410 155 52 × 2 = 1 + 0.926 158 820 311 04;
  • 19) 0.926 158 820 311 04 × 2 = 1 + 0.852 317 640 622 08;
  • 20) 0.852 317 640 622 08 × 2 = 1 + 0.704 635 281 244 16;
  • 21) 0.704 635 281 244 16 × 2 = 1 + 0.409 270 562 488 32;
  • 22) 0.409 270 562 488 32 × 2 = 0 + 0.818 541 124 976 64;
  • 23) 0.818 541 124 976 64 × 2 = 1 + 0.637 082 249 953 28;
  • 24) 0.637 082 249 953 28 × 2 = 1 + 0.274 164 499 906 56;
  • 25) 0.274 164 499 906 56 × 2 = 0 + 0.548 328 999 813 12;
  • 26) 0.548 328 999 813 12 × 2 = 1 + 0.096 657 999 626 24;
  • 27) 0.096 657 999 626 24 × 2 = 0 + 0.193 315 999 252 48;
  • 28) 0.193 315 999 252 48 × 2 = 0 + 0.386 631 998 504 96;
  • 29) 0.386 631 998 504 96 × 2 = 0 + 0.773 263 997 009 92;
  • 30) 0.773 263 997 009 92 × 2 = 1 + 0.546 527 994 019 84;
  • 31) 0.546 527 994 019 84 × 2 = 1 + 0.093 055 988 039 68;
  • 32) 0.093 055 988 039 68 × 2 = 0 + 0.186 111 976 079 36;
  • 33) 0.186 111 976 079 36 × 2 = 0 + 0.372 223 952 158 72;
  • 34) 0.372 223 952 158 72 × 2 = 0 + 0.744 447 904 317 44;
  • 35) 0.744 447 904 317 44 × 2 = 1 + 0.488 895 808 634 88;
  • 36) 0.488 895 808 634 88 × 2 = 0 + 0.977 791 617 269 76;
  • 37) 0.977 791 617 269 76 × 2 = 1 + 0.955 583 234 539 52;
  • 38) 0.955 583 234 539 52 × 2 = 1 + 0.911 166 469 079 04;
  • 39) 0.911 166 469 079 04 × 2 = 1 + 0.822 332 938 158 08;
  • 40) 0.822 332 938 158 08 × 2 = 1 + 0.644 665 876 316 16;
  • 41) 0.644 665 876 316 16 × 2 = 1 + 0.289 331 752 632 32;
  • 42) 0.289 331 752 632 32 × 2 = 0 + 0.578 663 505 264 64;
  • 43) 0.578 663 505 264 64 × 2 = 1 + 0.157 327 010 529 28;
  • 44) 0.157 327 010 529 28 × 2 = 0 + 0.314 654 021 058 56;
  • 45) 0.314 654 021 058 56 × 2 = 0 + 0.629 308 042 117 12;
  • 46) 0.629 308 042 117 12 × 2 = 1 + 0.258 616 084 234 24;
  • 47) 0.258 616 084 234 24 × 2 = 0 + 0.517 232 168 468 48;
  • 48) 0.517 232 168 468 48 × 2 = 1 + 0.034 464 336 936 96;
  • 49) 0.034 464 336 936 96 × 2 = 0 + 0.068 928 673 873 92;
  • 50) 0.068 928 673 873 92 × 2 = 0 + 0.137 857 347 747 84;
  • 51) 0.137 857 347 747 84 × 2 = 0 + 0.275 714 695 495 68;
  • 52) 0.275 714 695 495 68 × 2 = 0 + 0.551 429 390 991 36;
  • 53) 0.551 429 390 991 36 × 2 = 1 + 0.102 858 781 982 72;
  • 54) 0.102 858 781 982 72 × 2 = 0 + 0.205 717 563 965 44;
  • 55) 0.205 717 563 965 44 × 2 = 0 + 0.411 435 127 930 88;
  • 56) 0.411 435 127 930 88 × 2 = 0 + 0.822 870 255 861 76;
  • 57) 0.822 870 255 861 76 × 2 = 1 + 0.645 740 511 723 52;
  • 58) 0.645 740 511 723 52 × 2 = 1 + 0.291 481 023 447 04;
  • 59) 0.291 481 023 447 04 × 2 = 0 + 0.582 962 046 894 08;
  • 60) 0.582 962 046 894 08 × 2 = 1 + 0.165 924 093 788 16;
  • 61) 0.165 924 093 788 16 × 2 = 0 + 0.331 848 187 576 32;
  • 62) 0.331 848 187 576 32 × 2 = 0 + 0.663 696 375 152 64;
  • 63) 0.663 696 375 152 64 × 2 = 1 + 0.327 392 750 305 28;
  • 64) 0.327 392 750 305 28 × 2 = 0 + 0.654 785 500 610 56;

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:


0.000 282 005 915 91(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1111 1010 0101 0000 1000 1101 0010(2)

6. Positive number before normalization:

0.000 282 005 915 91(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1111 1010 0101 0000 1000 1101 0010(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:


0.000 282 005 915 91(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1111 1010 0101 0000 1000 1101 0010(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1111 1010 0101 0000 1000 1101 0010(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1111 1010 0101 0000 1000 1101 0010(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 1111 1010 0101 0000 1000 1101 0010


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 1111 1010 0101 0000 1000 1101 0010 =

0010 0111 1011 0100 0110 0010 1111 1010 0101 0000 1000 1101 0010


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 1111 1010 0101 0000 1000 1101 0010


Decimal number -0.000 282 005 915 91 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1111 1010 0101 0000 1000 1101 0010

Share

» Convert decimal -0.000 282 005 916 56 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 Attention Economy - The Battlefield For Your Focus. NotebookLM YouTube Video of 7.50 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