-0.000 279 14 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 279 14(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
-0.000 279 14(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:

|-0.000 279 14| = 0.000 279 14


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: 0.000 279 14.

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) 0.000 279 14 × 2 = 0 + 0.000 558 28;
  • 2) 0.000 558 28 × 2 = 0 + 0.001 116 56;
  • 3) 0.001 116 56 × 2 = 0 + 0.002 233 12;
  • 4) 0.002 233 12 × 2 = 0 + 0.004 466 24;
  • 5) 0.004 466 24 × 2 = 0 + 0.008 932 48;
  • 6) 0.008 932 48 × 2 = 0 + 0.017 864 96;
  • 7) 0.017 864 96 × 2 = 0 + 0.035 729 92;
  • 8) 0.035 729 92 × 2 = 0 + 0.071 459 84;
  • 9) 0.071 459 84 × 2 = 0 + 0.142 919 68;
  • 10) 0.142 919 68 × 2 = 0 + 0.285 839 36;
  • 11) 0.285 839 36 × 2 = 0 + 0.571 678 72;
  • 12) 0.571 678 72 × 2 = 1 + 0.143 357 44;
  • 13) 0.143 357 44 × 2 = 0 + 0.286 714 88;
  • 14) 0.286 714 88 × 2 = 0 + 0.573 429 76;
  • 15) 0.573 429 76 × 2 = 1 + 0.146 859 52;
  • 16) 0.146 859 52 × 2 = 0 + 0.293 719 04;
  • 17) 0.293 719 04 × 2 = 0 + 0.587 438 08;
  • 18) 0.587 438 08 × 2 = 1 + 0.174 876 16;
  • 19) 0.174 876 16 × 2 = 0 + 0.349 752 32;
  • 20) 0.349 752 32 × 2 = 0 + 0.699 504 64;
  • 21) 0.699 504 64 × 2 = 1 + 0.399 009 28;
  • 22) 0.399 009 28 × 2 = 0 + 0.798 018 56;
  • 23) 0.798 018 56 × 2 = 1 + 0.596 037 12;
  • 24) 0.596 037 12 × 2 = 1 + 0.192 074 24;
  • 25) 0.192 074 24 × 2 = 0 + 0.384 148 48;
  • 26) 0.384 148 48 × 2 = 0 + 0.768 296 96;
  • 27) 0.768 296 96 × 2 = 1 + 0.536 593 92;
  • 28) 0.536 593 92 × 2 = 1 + 0.073 187 84;
  • 29) 0.073 187 84 × 2 = 0 + 0.146 375 68;
  • 30) 0.146 375 68 × 2 = 0 + 0.292 751 36;
  • 31) 0.292 751 36 × 2 = 0 + 0.585 502 72;
  • 32) 0.585 502 72 × 2 = 1 + 0.171 005 44;
  • 33) 0.171 005 44 × 2 = 0 + 0.342 010 88;
  • 34) 0.342 010 88 × 2 = 0 + 0.684 021 76;
  • 35) 0.684 021 76 × 2 = 1 + 0.368 043 52;
  • 36) 0.368 043 52 × 2 = 0 + 0.736 087 04;
  • 37) 0.736 087 04 × 2 = 1 + 0.472 174 08;
  • 38) 0.472 174 08 × 2 = 0 + 0.944 348 16;
  • 39) 0.944 348 16 × 2 = 1 + 0.888 696 32;
  • 40) 0.888 696 32 × 2 = 1 + 0.777 392 64;
  • 41) 0.777 392 64 × 2 = 1 + 0.554 785 28;
  • 42) 0.554 785 28 × 2 = 1 + 0.109 570 56;
  • 43) 0.109 570 56 × 2 = 0 + 0.219 141 12;
  • 44) 0.219 141 12 × 2 = 0 + 0.438 282 24;
  • 45) 0.438 282 24 × 2 = 0 + 0.876 564 48;
  • 46) 0.876 564 48 × 2 = 1 + 0.753 128 96;
  • 47) 0.753 128 96 × 2 = 1 + 0.506 257 92;
  • 48) 0.506 257 92 × 2 = 1 + 0.012 515 84;
  • 49) 0.012 515 84 × 2 = 0 + 0.025 031 68;
  • 50) 0.025 031 68 × 2 = 0 + 0.050 063 36;
  • 51) 0.050 063 36 × 2 = 0 + 0.100 126 72;
  • 52) 0.100 126 72 × 2 = 0 + 0.200 253 44;
  • 53) 0.200 253 44 × 2 = 0 + 0.400 506 88;
  • 54) 0.400 506 88 × 2 = 0 + 0.801 013 76;
  • 55) 0.801 013 76 × 2 = 1 + 0.602 027 52;
  • 56) 0.602 027 52 × 2 = 1 + 0.204 055 04;
  • 57) 0.204 055 04 × 2 = 0 + 0.408 110 08;
  • 58) 0.408 110 08 × 2 = 0 + 0.816 220 16;
  • 59) 0.816 220 16 × 2 = 1 + 0.632 440 32;
  • 60) 0.632 440 32 × 2 = 1 + 0.264 880 64;
  • 61) 0.264 880 64 × 2 = 0 + 0.529 761 28;
  • 62) 0.529 761 28 × 2 = 1 + 0.059 522 56;
  • 63) 0.059 522 56 × 2 = 0 + 0.119 045 12;
  • 64) 0.119 045 12 × 2 = 0 + 0.238 090 24;

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:


0.000 279 14(10) =

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

6. Positive number before normalization:

0.000 279 14(10) =

0.0000 0000 0001 0010 0100 1011 0011 0001 0010 1011 1100 0111 0000 0011 0011 0100(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:


0.000 279 14(10) =

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

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

1.0010 0100 1011 0011 0001 0010 1011 1100 0111 0000 0011 0011 0100(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 0100 1011 0011 0001 0010 1011 1100 0111 0000 0011 0011 0100


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 0100 1011 0011 0001 0010 1011 1100 0111 0000 0011 0011 0100 =

0010 0100 1011 0011 0001 0010 1011 1100 0111 0000 0011 0011 0100


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 0100 1011 0011 0001 0010 1011 1100 0111 0000 0011 0011 0100


Decimal number -0.000 279 14 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0100 1011 0011 0001 0010 1011 1100 0111 0000 0011 0011 0100

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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