-0.000 279 11 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 279 11(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 11(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 11| = 0.000 279 11


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 11.

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 11 × 2 = 0 + 0.000 558 22;
  • 2) 0.000 558 22 × 2 = 0 + 0.001 116 44;
  • 3) 0.001 116 44 × 2 = 0 + 0.002 232 88;
  • 4) 0.002 232 88 × 2 = 0 + 0.004 465 76;
  • 5) 0.004 465 76 × 2 = 0 + 0.008 931 52;
  • 6) 0.008 931 52 × 2 = 0 + 0.017 863 04;
  • 7) 0.017 863 04 × 2 = 0 + 0.035 726 08;
  • 8) 0.035 726 08 × 2 = 0 + 0.071 452 16;
  • 9) 0.071 452 16 × 2 = 0 + 0.142 904 32;
  • 10) 0.142 904 32 × 2 = 0 + 0.285 808 64;
  • 11) 0.285 808 64 × 2 = 0 + 0.571 617 28;
  • 12) 0.571 617 28 × 2 = 1 + 0.143 234 56;
  • 13) 0.143 234 56 × 2 = 0 + 0.286 469 12;
  • 14) 0.286 469 12 × 2 = 0 + 0.572 938 24;
  • 15) 0.572 938 24 × 2 = 1 + 0.145 876 48;
  • 16) 0.145 876 48 × 2 = 0 + 0.291 752 96;
  • 17) 0.291 752 96 × 2 = 0 + 0.583 505 92;
  • 18) 0.583 505 92 × 2 = 1 + 0.167 011 84;
  • 19) 0.167 011 84 × 2 = 0 + 0.334 023 68;
  • 20) 0.334 023 68 × 2 = 0 + 0.668 047 36;
  • 21) 0.668 047 36 × 2 = 1 + 0.336 094 72;
  • 22) 0.336 094 72 × 2 = 0 + 0.672 189 44;
  • 23) 0.672 189 44 × 2 = 1 + 0.344 378 88;
  • 24) 0.344 378 88 × 2 = 0 + 0.688 757 76;
  • 25) 0.688 757 76 × 2 = 1 + 0.377 515 52;
  • 26) 0.377 515 52 × 2 = 0 + 0.755 031 04;
  • 27) 0.755 031 04 × 2 = 1 + 0.510 062 08;
  • 28) 0.510 062 08 × 2 = 1 + 0.020 124 16;
  • 29) 0.020 124 16 × 2 = 0 + 0.040 248 32;
  • 30) 0.040 248 32 × 2 = 0 + 0.080 496 64;
  • 31) 0.080 496 64 × 2 = 0 + 0.160 993 28;
  • 32) 0.160 993 28 × 2 = 0 + 0.321 986 56;
  • 33) 0.321 986 56 × 2 = 0 + 0.643 973 12;
  • 34) 0.643 973 12 × 2 = 1 + 0.287 946 24;
  • 35) 0.287 946 24 × 2 = 0 + 0.575 892 48;
  • 36) 0.575 892 48 × 2 = 1 + 0.151 784 96;
  • 37) 0.151 784 96 × 2 = 0 + 0.303 569 92;
  • 38) 0.303 569 92 × 2 = 0 + 0.607 139 84;
  • 39) 0.607 139 84 × 2 = 1 + 0.214 279 68;
  • 40) 0.214 279 68 × 2 = 0 + 0.428 559 36;
  • 41) 0.428 559 36 × 2 = 0 + 0.857 118 72;
  • 42) 0.857 118 72 × 2 = 1 + 0.714 237 44;
  • 43) 0.714 237 44 × 2 = 1 + 0.428 474 88;
  • 44) 0.428 474 88 × 2 = 0 + 0.856 949 76;
  • 45) 0.856 949 76 × 2 = 1 + 0.713 899 52;
  • 46) 0.713 899 52 × 2 = 1 + 0.427 799 04;
  • 47) 0.427 799 04 × 2 = 0 + 0.855 598 08;
  • 48) 0.855 598 08 × 2 = 1 + 0.711 196 16;
  • 49) 0.711 196 16 × 2 = 1 + 0.422 392 32;
  • 50) 0.422 392 32 × 2 = 0 + 0.844 784 64;
  • 51) 0.844 784 64 × 2 = 1 + 0.689 569 28;
  • 52) 0.689 569 28 × 2 = 1 + 0.379 138 56;
  • 53) 0.379 138 56 × 2 = 0 + 0.758 277 12;
  • 54) 0.758 277 12 × 2 = 1 + 0.516 554 24;
  • 55) 0.516 554 24 × 2 = 1 + 0.033 108 48;
  • 56) 0.033 108 48 × 2 = 0 + 0.066 216 96;
  • 57) 0.066 216 96 × 2 = 0 + 0.132 433 92;
  • 58) 0.132 433 92 × 2 = 0 + 0.264 867 84;
  • 59) 0.264 867 84 × 2 = 0 + 0.529 735 68;
  • 60) 0.529 735 68 × 2 = 1 + 0.059 471 36;
  • 61) 0.059 471 36 × 2 = 0 + 0.118 942 72;
  • 62) 0.118 942 72 × 2 = 0 + 0.237 885 44;
  • 63) 0.237 885 44 × 2 = 0 + 0.475 770 88;
  • 64) 0.475 770 88 × 2 = 0 + 0.951 541 76;

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 11(10) =

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

6. Positive number before normalization:

0.000 279 11(10) =

0.0000 0000 0001 0010 0100 1010 1011 0000 0101 0010 0110 1101 1011 0110 0001 0000(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 11(10) =

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

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

1.0010 0100 1010 1011 0000 0101 0010 0110 1101 1011 0110 0001 0000(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 1010 1011 0000 0101 0010 0110 1101 1011 0110 0001 0000


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 1010 1011 0000 0101 0010 0110 1101 1011 0110 0001 0000 =

0010 0100 1010 1011 0000 0101 0010 0110 1101 1011 0110 0001 0000


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 1010 1011 0000 0101 0010 0110 1101 1011 0110 0001 0000


Decimal number -0.000 279 11 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0100 1010 1011 0000 0101 0010 0110 1101 1011 0110 0001 0000

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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