-0.000 283 42 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 283 42(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
-0.000 283 42(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:

|-0.000 283 42| = 0.000 283 42


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: 0.000 283 42.

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) 0.000 283 42 × 2 = 0 + 0.000 566 84;
  • 2) 0.000 566 84 × 2 = 0 + 0.001 133 68;
  • 3) 0.001 133 68 × 2 = 0 + 0.002 267 36;
  • 4) 0.002 267 36 × 2 = 0 + 0.004 534 72;
  • 5) 0.004 534 72 × 2 = 0 + 0.009 069 44;
  • 6) 0.009 069 44 × 2 = 0 + 0.018 138 88;
  • 7) 0.018 138 88 × 2 = 0 + 0.036 277 76;
  • 8) 0.036 277 76 × 2 = 0 + 0.072 555 52;
  • 9) 0.072 555 52 × 2 = 0 + 0.145 111 04;
  • 10) 0.145 111 04 × 2 = 0 + 0.290 222 08;
  • 11) 0.290 222 08 × 2 = 0 + 0.580 444 16;
  • 12) 0.580 444 16 × 2 = 1 + 0.160 888 32;
  • 13) 0.160 888 32 × 2 = 0 + 0.321 776 64;
  • 14) 0.321 776 64 × 2 = 0 + 0.643 553 28;
  • 15) 0.643 553 28 × 2 = 1 + 0.287 106 56;
  • 16) 0.287 106 56 × 2 = 0 + 0.574 213 12;
  • 17) 0.574 213 12 × 2 = 1 + 0.148 426 24;
  • 18) 0.148 426 24 × 2 = 0 + 0.296 852 48;
  • 19) 0.296 852 48 × 2 = 0 + 0.593 704 96;
  • 20) 0.593 704 96 × 2 = 1 + 0.187 409 92;
  • 21) 0.187 409 92 × 2 = 0 + 0.374 819 84;
  • 22) 0.374 819 84 × 2 = 0 + 0.749 639 68;
  • 23) 0.749 639 68 × 2 = 1 + 0.499 279 36;
  • 24) 0.499 279 36 × 2 = 0 + 0.998 558 72;
  • 25) 0.998 558 72 × 2 = 1 + 0.997 117 44;
  • 26) 0.997 117 44 × 2 = 1 + 0.994 234 88;
  • 27) 0.994 234 88 × 2 = 1 + 0.988 469 76;
  • 28) 0.988 469 76 × 2 = 1 + 0.976 939 52;
  • 29) 0.976 939 52 × 2 = 1 + 0.953 879 04;
  • 30) 0.953 879 04 × 2 = 1 + 0.907 758 08;
  • 31) 0.907 758 08 × 2 = 1 + 0.815 516 16;
  • 32) 0.815 516 16 × 2 = 1 + 0.631 032 32;
  • 33) 0.631 032 32 × 2 = 1 + 0.262 064 64;
  • 34) 0.262 064 64 × 2 = 0 + 0.524 129 28;
  • 35) 0.524 129 28 × 2 = 1 + 0.048 258 56;
  • 36) 0.048 258 56 × 2 = 0 + 0.096 517 12;
  • 37) 0.096 517 12 × 2 = 0 + 0.193 034 24;
  • 38) 0.193 034 24 × 2 = 0 + 0.386 068 48;
  • 39) 0.386 068 48 × 2 = 0 + 0.772 136 96;
  • 40) 0.772 136 96 × 2 = 1 + 0.544 273 92;
  • 41) 0.544 273 92 × 2 = 1 + 0.088 547 84;
  • 42) 0.088 547 84 × 2 = 0 + 0.177 095 68;
  • 43) 0.177 095 68 × 2 = 0 + 0.354 191 36;
  • 44) 0.354 191 36 × 2 = 0 + 0.708 382 72;
  • 45) 0.708 382 72 × 2 = 1 + 0.416 765 44;
  • 46) 0.416 765 44 × 2 = 0 + 0.833 530 88;
  • 47) 0.833 530 88 × 2 = 1 + 0.667 061 76;
  • 48) 0.667 061 76 × 2 = 1 + 0.334 123 52;
  • 49) 0.334 123 52 × 2 = 0 + 0.668 247 04;
  • 50) 0.668 247 04 × 2 = 1 + 0.336 494 08;
  • 51) 0.336 494 08 × 2 = 0 + 0.672 988 16;
  • 52) 0.672 988 16 × 2 = 1 + 0.345 976 32;
  • 53) 0.345 976 32 × 2 = 0 + 0.691 952 64;
  • 54) 0.691 952 64 × 2 = 1 + 0.383 905 28;
  • 55) 0.383 905 28 × 2 = 0 + 0.767 810 56;
  • 56) 0.767 810 56 × 2 = 1 + 0.535 621 12;
  • 57) 0.535 621 12 × 2 = 1 + 0.071 242 24;
  • 58) 0.071 242 24 × 2 = 0 + 0.142 484 48;
  • 59) 0.142 484 48 × 2 = 0 + 0.284 968 96;
  • 60) 0.284 968 96 × 2 = 0 + 0.569 937 92;
  • 61) 0.569 937 92 × 2 = 1 + 0.139 875 84;
  • 62) 0.139 875 84 × 2 = 0 + 0.279 751 68;
  • 63) 0.279 751 68 × 2 = 0 + 0.559 503 36;
  • 64) 0.559 503 36 × 2 = 1 + 0.119 006 72;

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:


0.000 283 42(10) =

0.0000 0000 0001 0010 1001 0010 1111 1111 1010 0001 1000 1011 0101 0101 1000 1001(2)

6. Positive number before normalization:

0.000 283 42(10) =

0.0000 0000 0001 0010 1001 0010 1111 1111 1010 0001 1000 1011 0101 0101 1000 1001(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:


0.000 283 42(10) =

0.0000 0000 0001 0010 1001 0010 1111 1111 1010 0001 1000 1011 0101 0101 1000 1001(2) =

0.0000 0000 0001 0010 1001 0010 1111 1111 1010 0001 1000 1011 0101 0101 1000 1001(2) × 20 =

1.0010 1001 0010 1111 1111 1010 0001 1000 1011 0101 0101 1000 1001(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 1001 0010 1111 1111 1010 0001 1000 1011 0101 0101 1000 1001


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 1001 0010 1111 1111 1010 0001 1000 1011 0101 0101 1000 1001 =

0010 1001 0010 1111 1111 1010 0001 1000 1011 0101 0101 1000 1001


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 1001 0010 1111 1111 1010 0001 1000 1011 0101 0101 1000 1001


Decimal number -0.000 283 42 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1001 0010 1111 1111 1010 0001 1000 1011 0101 0101 1000 1001

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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