-0.000 282 005 54 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 54(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 54(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 54| = 0.000 282 005 54


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 54.

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 54 × 2 = 0 + 0.000 564 011 08;
  • 2) 0.000 564 011 08 × 2 = 0 + 0.001 128 022 16;
  • 3) 0.001 128 022 16 × 2 = 0 + 0.002 256 044 32;
  • 4) 0.002 256 044 32 × 2 = 0 + 0.004 512 088 64;
  • 5) 0.004 512 088 64 × 2 = 0 + 0.009 024 177 28;
  • 6) 0.009 024 177 28 × 2 = 0 + 0.018 048 354 56;
  • 7) 0.018 048 354 56 × 2 = 0 + 0.036 096 709 12;
  • 8) 0.036 096 709 12 × 2 = 0 + 0.072 193 418 24;
  • 9) 0.072 193 418 24 × 2 = 0 + 0.144 386 836 48;
  • 10) 0.144 386 836 48 × 2 = 0 + 0.288 773 672 96;
  • 11) 0.288 773 672 96 × 2 = 0 + 0.577 547 345 92;
  • 12) 0.577 547 345 92 × 2 = 1 + 0.155 094 691 84;
  • 13) 0.155 094 691 84 × 2 = 0 + 0.310 189 383 68;
  • 14) 0.310 189 383 68 × 2 = 0 + 0.620 378 767 36;
  • 15) 0.620 378 767 36 × 2 = 1 + 0.240 757 534 72;
  • 16) 0.240 757 534 72 × 2 = 0 + 0.481 515 069 44;
  • 17) 0.481 515 069 44 × 2 = 0 + 0.963 030 138 88;
  • 18) 0.963 030 138 88 × 2 = 1 + 0.926 060 277 76;
  • 19) 0.926 060 277 76 × 2 = 1 + 0.852 120 555 52;
  • 20) 0.852 120 555 52 × 2 = 1 + 0.704 241 111 04;
  • 21) 0.704 241 111 04 × 2 = 1 + 0.408 482 222 08;
  • 22) 0.408 482 222 08 × 2 = 0 + 0.816 964 444 16;
  • 23) 0.816 964 444 16 × 2 = 1 + 0.633 928 888 32;
  • 24) 0.633 928 888 32 × 2 = 1 + 0.267 857 776 64;
  • 25) 0.267 857 776 64 × 2 = 0 + 0.535 715 553 28;
  • 26) 0.535 715 553 28 × 2 = 1 + 0.071 431 106 56;
  • 27) 0.071 431 106 56 × 2 = 0 + 0.142 862 213 12;
  • 28) 0.142 862 213 12 × 2 = 0 + 0.285 724 426 24;
  • 29) 0.285 724 426 24 × 2 = 0 + 0.571 448 852 48;
  • 30) 0.571 448 852 48 × 2 = 1 + 0.142 897 704 96;
  • 31) 0.142 897 704 96 × 2 = 0 + 0.285 795 409 92;
  • 32) 0.285 795 409 92 × 2 = 0 + 0.571 590 819 84;
  • 33) 0.571 590 819 84 × 2 = 1 + 0.143 181 639 68;
  • 34) 0.143 181 639 68 × 2 = 0 + 0.286 363 279 36;
  • 35) 0.286 363 279 36 × 2 = 0 + 0.572 726 558 72;
  • 36) 0.572 726 558 72 × 2 = 1 + 0.145 453 117 44;
  • 37) 0.145 453 117 44 × 2 = 0 + 0.290 906 234 88;
  • 38) 0.290 906 234 88 × 2 = 0 + 0.581 812 469 76;
  • 39) 0.581 812 469 76 × 2 = 1 + 0.163 624 939 52;
  • 40) 0.163 624 939 52 × 2 = 0 + 0.327 249 879 04;
  • 41) 0.327 249 879 04 × 2 = 0 + 0.654 499 758 08;
  • 42) 0.654 499 758 08 × 2 = 1 + 0.308 999 516 16;
  • 43) 0.308 999 516 16 × 2 = 0 + 0.617 999 032 32;
  • 44) 0.617 999 032 32 × 2 = 1 + 0.235 998 064 64;
  • 45) 0.235 998 064 64 × 2 = 0 + 0.471 996 129 28;
  • 46) 0.471 996 129 28 × 2 = 0 + 0.943 992 258 56;
  • 47) 0.943 992 258 56 × 2 = 1 + 0.887 984 517 12;
  • 48) 0.887 984 517 12 × 2 = 1 + 0.775 969 034 24;
  • 49) 0.775 969 034 24 × 2 = 1 + 0.551 938 068 48;
  • 50) 0.551 938 068 48 × 2 = 1 + 0.103 876 136 96;
  • 51) 0.103 876 136 96 × 2 = 0 + 0.207 752 273 92;
  • 52) 0.207 752 273 92 × 2 = 0 + 0.415 504 547 84;
  • 53) 0.415 504 547 84 × 2 = 0 + 0.831 009 095 68;
  • 54) 0.831 009 095 68 × 2 = 1 + 0.662 018 191 36;
  • 55) 0.662 018 191 36 × 2 = 1 + 0.324 036 382 72;
  • 56) 0.324 036 382 72 × 2 = 0 + 0.648 072 765 44;
  • 57) 0.648 072 765 44 × 2 = 1 + 0.296 145 530 88;
  • 58) 0.296 145 530 88 × 2 = 0 + 0.592 291 061 76;
  • 59) 0.592 291 061 76 × 2 = 1 + 0.184 582 123 52;
  • 60) 0.184 582 123 52 × 2 = 0 + 0.369 164 247 04;
  • 61) 0.369 164 247 04 × 2 = 0 + 0.738 328 494 08;
  • 62) 0.738 328 494 08 × 2 = 1 + 0.476 656 988 16;
  • 63) 0.476 656 988 16 × 2 = 0 + 0.953 313 976 32;
  • 64) 0.953 313 976 32 × 2 = 1 + 0.906 627 952 64;

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 54(10) =

0.0000 0000 0001 0010 0111 1011 0100 0100 1001 0010 0101 0011 1100 0110 1010 0101(2)

6. Positive number before normalization:

0.000 282 005 54(10) =

0.0000 0000 0001 0010 0111 1011 0100 0100 1001 0010 0101 0011 1100 0110 1010 0101(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 54(10) =

0.0000 0000 0001 0010 0111 1011 0100 0100 1001 0010 0101 0011 1100 0110 1010 0101(2) =

0.0000 0000 0001 0010 0111 1011 0100 0100 1001 0010 0101 0011 1100 0110 1010 0101(2) × 20 =

1.0010 0111 1011 0100 0100 1001 0010 0101 0011 1100 0110 1010 0101(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 0100 1001 0010 0101 0011 1100 0110 1010 0101


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 0100 1001 0010 0101 0011 1100 0110 1010 0101 =

0010 0111 1011 0100 0100 1001 0010 0101 0011 1100 0110 1010 0101


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 0100 1001 0010 0101 0011 1100 0110 1010 0101


Decimal number -0.000 282 005 54 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0100 1001 0010 0101 0011 1100 0110 1010 0101

Share

» Convert decimal -0.000 282 006 16 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