-0.000 282 105 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 105(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
-0.000 282 105(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:

|-0.000 282 105| = 0.000 282 105


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: 0.000 282 105.

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) 0.000 282 105 × 2 = 0 + 0.000 564 21;
  • 2) 0.000 564 21 × 2 = 0 + 0.001 128 42;
  • 3) 0.001 128 42 × 2 = 0 + 0.002 256 84;
  • 4) 0.002 256 84 × 2 = 0 + 0.004 513 68;
  • 5) 0.004 513 68 × 2 = 0 + 0.009 027 36;
  • 6) 0.009 027 36 × 2 = 0 + 0.018 054 72;
  • 7) 0.018 054 72 × 2 = 0 + 0.036 109 44;
  • 8) 0.036 109 44 × 2 = 0 + 0.072 218 88;
  • 9) 0.072 218 88 × 2 = 0 + 0.144 437 76;
  • 10) 0.144 437 76 × 2 = 0 + 0.288 875 52;
  • 11) 0.288 875 52 × 2 = 0 + 0.577 751 04;
  • 12) 0.577 751 04 × 2 = 1 + 0.155 502 08;
  • 13) 0.155 502 08 × 2 = 0 + 0.311 004 16;
  • 14) 0.311 004 16 × 2 = 0 + 0.622 008 32;
  • 15) 0.622 008 32 × 2 = 1 + 0.244 016 64;
  • 16) 0.244 016 64 × 2 = 0 + 0.488 033 28;
  • 17) 0.488 033 28 × 2 = 0 + 0.976 066 56;
  • 18) 0.976 066 56 × 2 = 1 + 0.952 133 12;
  • 19) 0.952 133 12 × 2 = 1 + 0.904 266 24;
  • 20) 0.904 266 24 × 2 = 1 + 0.808 532 48;
  • 21) 0.808 532 48 × 2 = 1 + 0.617 064 96;
  • 22) 0.617 064 96 × 2 = 1 + 0.234 129 92;
  • 23) 0.234 129 92 × 2 = 0 + 0.468 259 84;
  • 24) 0.468 259 84 × 2 = 0 + 0.936 519 68;
  • 25) 0.936 519 68 × 2 = 1 + 0.873 039 36;
  • 26) 0.873 039 36 × 2 = 1 + 0.746 078 72;
  • 27) 0.746 078 72 × 2 = 1 + 0.492 157 44;
  • 28) 0.492 157 44 × 2 = 0 + 0.984 314 88;
  • 29) 0.984 314 88 × 2 = 1 + 0.968 629 76;
  • 30) 0.968 629 76 × 2 = 1 + 0.937 259 52;
  • 31) 0.937 259 52 × 2 = 1 + 0.874 519 04;
  • 32) 0.874 519 04 × 2 = 1 + 0.749 038 08;
  • 33) 0.749 038 08 × 2 = 1 + 0.498 076 16;
  • 34) 0.498 076 16 × 2 = 0 + 0.996 152 32;
  • 35) 0.996 152 32 × 2 = 1 + 0.992 304 64;
  • 36) 0.992 304 64 × 2 = 1 + 0.984 609 28;
  • 37) 0.984 609 28 × 2 = 1 + 0.969 218 56;
  • 38) 0.969 218 56 × 2 = 1 + 0.938 437 12;
  • 39) 0.938 437 12 × 2 = 1 + 0.876 874 24;
  • 40) 0.876 874 24 × 2 = 1 + 0.753 748 48;
  • 41) 0.753 748 48 × 2 = 1 + 0.507 496 96;
  • 42) 0.507 496 96 × 2 = 1 + 0.014 993 92;
  • 43) 0.014 993 92 × 2 = 0 + 0.029 987 84;
  • 44) 0.029 987 84 × 2 = 0 + 0.059 975 68;
  • 45) 0.059 975 68 × 2 = 0 + 0.119 951 36;
  • 46) 0.119 951 36 × 2 = 0 + 0.239 902 72;
  • 47) 0.239 902 72 × 2 = 0 + 0.479 805 44;
  • 48) 0.479 805 44 × 2 = 0 + 0.959 610 88;
  • 49) 0.959 610 88 × 2 = 1 + 0.919 221 76;
  • 50) 0.919 221 76 × 2 = 1 + 0.838 443 52;
  • 51) 0.838 443 52 × 2 = 1 + 0.676 887 04;
  • 52) 0.676 887 04 × 2 = 1 + 0.353 774 08;
  • 53) 0.353 774 08 × 2 = 0 + 0.707 548 16;
  • 54) 0.707 548 16 × 2 = 1 + 0.415 096 32;
  • 55) 0.415 096 32 × 2 = 0 + 0.830 192 64;
  • 56) 0.830 192 64 × 2 = 1 + 0.660 385 28;
  • 57) 0.660 385 28 × 2 = 1 + 0.320 770 56;
  • 58) 0.320 770 56 × 2 = 0 + 0.641 541 12;
  • 59) 0.641 541 12 × 2 = 1 + 0.283 082 24;
  • 60) 0.283 082 24 × 2 = 0 + 0.566 164 48;
  • 61) 0.566 164 48 × 2 = 1 + 0.132 328 96;
  • 62) 0.132 328 96 × 2 = 0 + 0.264 657 92;
  • 63) 0.264 657 92 × 2 = 0 + 0.529 315 84;
  • 64) 0.529 315 84 × 2 = 1 + 0.058 631 68;

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:


0.000 282 105(10) =

0.0000 0000 0001 0010 0111 1100 1110 1111 1011 1111 1100 0000 1111 0101 1010 1001(2)

6. Positive number before normalization:

0.000 282 105(10) =

0.0000 0000 0001 0010 0111 1100 1110 1111 1011 1111 1100 0000 1111 0101 1010 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:


0.000 282 105(10) =

0.0000 0000 0001 0010 0111 1100 1110 1111 1011 1111 1100 0000 1111 0101 1010 1001(2) =

0.0000 0000 0001 0010 0111 1100 1110 1111 1011 1111 1100 0000 1111 0101 1010 1001(2) × 20 =

1.0010 0111 1100 1110 1111 1011 1111 1100 0000 1111 0101 1010 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 0111 1100 1110 1111 1011 1111 1100 0000 1111 0101 1010 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 0111 1100 1110 1111 1011 1111 1100 0000 1111 0101 1010 1001 =

0010 0111 1100 1110 1111 1011 1111 1100 0000 1111 0101 1010 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 0111 1100 1110 1111 1011 1111 1100 0000 1111 0101 1010 1001


Decimal number -0.000 282 105 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1100 1110 1111 1011 1111 1100 0000 1111 0101 1010 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