-0.057 050 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.057 050 6(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
-0.057 050 6(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:

|-0.057 050 6| = 0.057 050 6


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: 0.057 050 6.

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) 0.057 050 6 × 2 = 0 + 0.114 101 2;
  • 2) 0.114 101 2 × 2 = 0 + 0.228 202 4;
  • 3) 0.228 202 4 × 2 = 0 + 0.456 404 8;
  • 4) 0.456 404 8 × 2 = 0 + 0.912 809 6;
  • 5) 0.912 809 6 × 2 = 1 + 0.825 619 2;
  • 6) 0.825 619 2 × 2 = 1 + 0.651 238 4;
  • 7) 0.651 238 4 × 2 = 1 + 0.302 476 8;
  • 8) 0.302 476 8 × 2 = 0 + 0.604 953 6;
  • 9) 0.604 953 6 × 2 = 1 + 0.209 907 2;
  • 10) 0.209 907 2 × 2 = 0 + 0.419 814 4;
  • 11) 0.419 814 4 × 2 = 0 + 0.839 628 8;
  • 12) 0.839 628 8 × 2 = 1 + 0.679 257 6;
  • 13) 0.679 257 6 × 2 = 1 + 0.358 515 2;
  • 14) 0.358 515 2 × 2 = 0 + 0.717 030 4;
  • 15) 0.717 030 4 × 2 = 1 + 0.434 060 8;
  • 16) 0.434 060 8 × 2 = 0 + 0.868 121 6;
  • 17) 0.868 121 6 × 2 = 1 + 0.736 243 2;
  • 18) 0.736 243 2 × 2 = 1 + 0.472 486 4;
  • 19) 0.472 486 4 × 2 = 0 + 0.944 972 8;
  • 20) 0.944 972 8 × 2 = 1 + 0.889 945 6;
  • 21) 0.889 945 6 × 2 = 1 + 0.779 891 2;
  • 22) 0.779 891 2 × 2 = 1 + 0.559 782 4;
  • 23) 0.559 782 4 × 2 = 1 + 0.119 564 8;
  • 24) 0.119 564 8 × 2 = 0 + 0.239 129 6;
  • 25) 0.239 129 6 × 2 = 0 + 0.478 259 2;
  • 26) 0.478 259 2 × 2 = 0 + 0.956 518 4;
  • 27) 0.956 518 4 × 2 = 1 + 0.913 036 8;
  • 28) 0.913 036 8 × 2 = 1 + 0.826 073 6;
  • 29) 0.826 073 6 × 2 = 1 + 0.652 147 2;
  • 30) 0.652 147 2 × 2 = 1 + 0.304 294 4;
  • 31) 0.304 294 4 × 2 = 0 + 0.608 588 8;
  • 32) 0.608 588 8 × 2 = 1 + 0.217 177 6;
  • 33) 0.217 177 6 × 2 = 0 + 0.434 355 2;
  • 34) 0.434 355 2 × 2 = 0 + 0.868 710 4;
  • 35) 0.868 710 4 × 2 = 1 + 0.737 420 8;
  • 36) 0.737 420 8 × 2 = 1 + 0.474 841 6;
  • 37) 0.474 841 6 × 2 = 0 + 0.949 683 2;
  • 38) 0.949 683 2 × 2 = 1 + 0.899 366 4;
  • 39) 0.899 366 4 × 2 = 1 + 0.798 732 8;
  • 40) 0.798 732 8 × 2 = 1 + 0.597 465 6;
  • 41) 0.597 465 6 × 2 = 1 + 0.194 931 2;
  • 42) 0.194 931 2 × 2 = 0 + 0.389 862 4;
  • 43) 0.389 862 4 × 2 = 0 + 0.779 724 8;
  • 44) 0.779 724 8 × 2 = 1 + 0.559 449 6;
  • 45) 0.559 449 6 × 2 = 1 + 0.118 899 2;
  • 46) 0.118 899 2 × 2 = 0 + 0.237 798 4;
  • 47) 0.237 798 4 × 2 = 0 + 0.475 596 8;
  • 48) 0.475 596 8 × 2 = 0 + 0.951 193 6;
  • 49) 0.951 193 6 × 2 = 1 + 0.902 387 2;
  • 50) 0.902 387 2 × 2 = 1 + 0.804 774 4;
  • 51) 0.804 774 4 × 2 = 1 + 0.609 548 8;
  • 52) 0.609 548 8 × 2 = 1 + 0.219 097 6;
  • 53) 0.219 097 6 × 2 = 0 + 0.438 195 2;
  • 54) 0.438 195 2 × 2 = 0 + 0.876 390 4;
  • 55) 0.876 390 4 × 2 = 1 + 0.752 780 8;
  • 56) 0.752 780 8 × 2 = 1 + 0.505 561 6;
  • 57) 0.505 561 6 × 2 = 1 + 0.011 123 2;

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:


0.057 050 6(10) =

0.0000 1110 1001 1010 1101 1110 0011 1101 0011 0111 1001 1000 1111 0011 1(2)

6. Positive number before normalization:

0.057 050 6(10) =

0.0000 1110 1001 1010 1101 1110 0011 1101 0011 0111 1001 1000 1111 0011 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 5 positions to the right, so that only one non zero digit remains to the left of it:


0.057 050 6(10) =

0.0000 1110 1001 1010 1101 1110 0011 1101 0011 0111 1001 1000 1111 0011 1(2) =

0.0000 1110 1001 1010 1101 1110 0011 1101 0011 0111 1001 1000 1111 0011 1(2) × 20 =

1.1101 0011 0101 1011 1100 0111 1010 0110 1111 0011 0001 1110 0111(2) × 2-5


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): -5

Mantissa (not normalized):
1.1101 0011 0101 1011 1100 0111 1010 0110 1111 0011 0001 1110 0111


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-5 + 2(11-1) - 1 =

(-5 + 1 023)(10) =

1 018(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;
  • 1 018 ÷ 2 = 509 + 0;
  • 509 ÷ 2 = 254 + 1;
  • 254 ÷ 2 = 127 + 0;
  • 127 ÷ 2 = 63 + 1;
  • 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) =

1018(10) =

011 1111 1010(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. 1101 0011 0101 1011 1100 0111 1010 0110 1111 0011 0001 1110 0111 =

1101 0011 0101 1011 1100 0111 1010 0110 1111 0011 0001 1110 0111


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 1010

Mantissa (52 bits) =
1101 0011 0101 1011 1100 0111 1010 0110 1111 0011 0001 1110 0111


Decimal number -0.057 050 6 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1010 - 1101 0011 0101 1011 1100 0111 1010 0110 1111 0011 0001 1110 0111

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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