0.026 916 93 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.026 916 93(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
0.026 916 93(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. 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;

2. 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)


3. Convert to binary (base 2) the fractional part: 0.026 916 93.

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) 0.026 916 93 × 2 = 0 + 0.053 833 86;
  • 2) 0.053 833 86 × 2 = 0 + 0.107 667 72;
  • 3) 0.107 667 72 × 2 = 0 + 0.215 335 44;
  • 4) 0.215 335 44 × 2 = 0 + 0.430 670 88;
  • 5) 0.430 670 88 × 2 = 0 + 0.861 341 76;
  • 6) 0.861 341 76 × 2 = 1 + 0.722 683 52;
  • 7) 0.722 683 52 × 2 = 1 + 0.445 367 04;
  • 8) 0.445 367 04 × 2 = 0 + 0.890 734 08;
  • 9) 0.890 734 08 × 2 = 1 + 0.781 468 16;
  • 10) 0.781 468 16 × 2 = 1 + 0.562 936 32;
  • 11) 0.562 936 32 × 2 = 1 + 0.125 872 64;
  • 12) 0.125 872 64 × 2 = 0 + 0.251 745 28;
  • 13) 0.251 745 28 × 2 = 0 + 0.503 490 56;
  • 14) 0.503 490 56 × 2 = 1 + 0.006 981 12;
  • 15) 0.006 981 12 × 2 = 0 + 0.013 962 24;
  • 16) 0.013 962 24 × 2 = 0 + 0.027 924 48;
  • 17) 0.027 924 48 × 2 = 0 + 0.055 848 96;
  • 18) 0.055 848 96 × 2 = 0 + 0.111 697 92;
  • 19) 0.111 697 92 × 2 = 0 + 0.223 395 84;
  • 20) 0.223 395 84 × 2 = 0 + 0.446 791 68;
  • 21) 0.446 791 68 × 2 = 0 + 0.893 583 36;
  • 22) 0.893 583 36 × 2 = 1 + 0.787 166 72;
  • 23) 0.787 166 72 × 2 = 1 + 0.574 333 44;
  • 24) 0.574 333 44 × 2 = 1 + 0.148 666 88;
  • 25) 0.148 666 88 × 2 = 0 + 0.297 333 76;
  • 26) 0.297 333 76 × 2 = 0 + 0.594 667 52;
  • 27) 0.594 667 52 × 2 = 1 + 0.189 335 04;
  • 28) 0.189 335 04 × 2 = 0 + 0.378 670 08;
  • 29) 0.378 670 08 × 2 = 0 + 0.757 340 16;
  • 30) 0.757 340 16 × 2 = 1 + 0.514 680 32;
  • 31) 0.514 680 32 × 2 = 1 + 0.029 360 64;
  • 32) 0.029 360 64 × 2 = 0 + 0.058 721 28;
  • 33) 0.058 721 28 × 2 = 0 + 0.117 442 56;
  • 34) 0.117 442 56 × 2 = 0 + 0.234 885 12;
  • 35) 0.234 885 12 × 2 = 0 + 0.469 770 24;
  • 36) 0.469 770 24 × 2 = 0 + 0.939 540 48;
  • 37) 0.939 540 48 × 2 = 1 + 0.879 080 96;
  • 38) 0.879 080 96 × 2 = 1 + 0.758 161 92;
  • 39) 0.758 161 92 × 2 = 1 + 0.516 323 84;
  • 40) 0.516 323 84 × 2 = 1 + 0.032 647 68;
  • 41) 0.032 647 68 × 2 = 0 + 0.065 295 36;
  • 42) 0.065 295 36 × 2 = 0 + 0.130 590 72;
  • 43) 0.130 590 72 × 2 = 0 + 0.261 181 44;
  • 44) 0.261 181 44 × 2 = 0 + 0.522 362 88;
  • 45) 0.522 362 88 × 2 = 1 + 0.044 725 76;
  • 46) 0.044 725 76 × 2 = 0 + 0.089 451 52;
  • 47) 0.089 451 52 × 2 = 0 + 0.178 903 04;
  • 48) 0.178 903 04 × 2 = 0 + 0.357 806 08;
  • 49) 0.357 806 08 × 2 = 0 + 0.715 612 16;
  • 50) 0.715 612 16 × 2 = 1 + 0.431 224 32;
  • 51) 0.431 224 32 × 2 = 0 + 0.862 448 64;
  • 52) 0.862 448 64 × 2 = 1 + 0.724 897 28;
  • 53) 0.724 897 28 × 2 = 1 + 0.449 794 56;
  • 54) 0.449 794 56 × 2 = 0 + 0.899 589 12;
  • 55) 0.899 589 12 × 2 = 1 + 0.799 178 24;
  • 56) 0.799 178 24 × 2 = 1 + 0.598 356 48;
  • 57) 0.598 356 48 × 2 = 1 + 0.196 712 96;
  • 58) 0.196 712 96 × 2 = 0 + 0.393 425 92;

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).


4. 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:


0.026 916 93(10) =

0.0000 0110 1110 0100 0000 0111 0010 0110 0000 1111 0000 1000 0101 1011 10(2)

5. Positive number before normalization:

0.026 916 93(10) =

0.0000 0110 1110 0100 0000 0111 0010 0110 0000 1111 0000 1000 0101 1011 10(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 6 positions to the right, so that only one non zero digit remains to the left of it:


0.026 916 93(10) =

0.0000 0110 1110 0100 0000 0111 0010 0110 0000 1111 0000 1000 0101 1011 10(2) =

0.0000 0110 1110 0100 0000 0111 0010 0110 0000 1111 0000 1000 0101 1011 10(2) × 20 =

1.1011 1001 0000 0001 1100 1001 1000 0011 1100 0010 0001 0110 1110(2) × 2-6


7. 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 0 (a positive number)

Exponent (unadjusted): -6

Mantissa (not normalized):
1.1011 1001 0000 0001 1100 1001 1000 0011 1100 0010 0001 0110 1110


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-6 + 2(11-1) - 1 =

(-6 + 1 023)(10) =

1 017(10)


9. 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;
  • 1 017 ÷ 2 = 508 + 1;
  • 508 ÷ 2 = 254 + 0;
  • 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;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1017(10) =

011 1111 1001(2)


11. 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. 1011 1001 0000 0001 1100 1001 1000 0011 1100 0010 0001 0110 1110 =

1011 1001 0000 0001 1100 1001 1000 0011 1100 0010 0001 0110 1110


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
011 1111 1001

Mantissa (52 bits) =
1011 1001 0000 0001 1100 1001 1000 0011 1100 0010 0001 0110 1110


Decimal number 0.026 916 93 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1001 - 1011 1001 0000 0001 1100 1001 1000 0011 1100 0010 0001 0110 1110

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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