0.026 916 85 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.026 916 85(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 85(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 85.

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 85 × 2 = 0 + 0.053 833 7;
  • 2) 0.053 833 7 × 2 = 0 + 0.107 667 4;
  • 3) 0.107 667 4 × 2 = 0 + 0.215 334 8;
  • 4) 0.215 334 8 × 2 = 0 + 0.430 669 6;
  • 5) 0.430 669 6 × 2 = 0 + 0.861 339 2;
  • 6) 0.861 339 2 × 2 = 1 + 0.722 678 4;
  • 7) 0.722 678 4 × 2 = 1 + 0.445 356 8;
  • 8) 0.445 356 8 × 2 = 0 + 0.890 713 6;
  • 9) 0.890 713 6 × 2 = 1 + 0.781 427 2;
  • 10) 0.781 427 2 × 2 = 1 + 0.562 854 4;
  • 11) 0.562 854 4 × 2 = 1 + 0.125 708 8;
  • 12) 0.125 708 8 × 2 = 0 + 0.251 417 6;
  • 13) 0.251 417 6 × 2 = 0 + 0.502 835 2;
  • 14) 0.502 835 2 × 2 = 1 + 0.005 670 4;
  • 15) 0.005 670 4 × 2 = 0 + 0.011 340 8;
  • 16) 0.011 340 8 × 2 = 0 + 0.022 681 6;
  • 17) 0.022 681 6 × 2 = 0 + 0.045 363 2;
  • 18) 0.045 363 2 × 2 = 0 + 0.090 726 4;
  • 19) 0.090 726 4 × 2 = 0 + 0.181 452 8;
  • 20) 0.181 452 8 × 2 = 0 + 0.362 905 6;
  • 21) 0.362 905 6 × 2 = 0 + 0.725 811 2;
  • 22) 0.725 811 2 × 2 = 1 + 0.451 622 4;
  • 23) 0.451 622 4 × 2 = 0 + 0.903 244 8;
  • 24) 0.903 244 8 × 2 = 1 + 0.806 489 6;
  • 25) 0.806 489 6 × 2 = 1 + 0.612 979 2;
  • 26) 0.612 979 2 × 2 = 1 + 0.225 958 4;
  • 27) 0.225 958 4 × 2 = 0 + 0.451 916 8;
  • 28) 0.451 916 8 × 2 = 0 + 0.903 833 6;
  • 29) 0.903 833 6 × 2 = 1 + 0.807 667 2;
  • 30) 0.807 667 2 × 2 = 1 + 0.615 334 4;
  • 31) 0.615 334 4 × 2 = 1 + 0.230 668 8;
  • 32) 0.230 668 8 × 2 = 0 + 0.461 337 6;
  • 33) 0.461 337 6 × 2 = 0 + 0.922 675 2;
  • 34) 0.922 675 2 × 2 = 1 + 0.845 350 4;
  • 35) 0.845 350 4 × 2 = 1 + 0.690 700 8;
  • 36) 0.690 700 8 × 2 = 1 + 0.381 401 6;
  • 37) 0.381 401 6 × 2 = 0 + 0.762 803 2;
  • 38) 0.762 803 2 × 2 = 1 + 0.525 606 4;
  • 39) 0.525 606 4 × 2 = 1 + 0.051 212 8;
  • 40) 0.051 212 8 × 2 = 0 + 0.102 425 6;
  • 41) 0.102 425 6 × 2 = 0 + 0.204 851 2;
  • 42) 0.204 851 2 × 2 = 0 + 0.409 702 4;
  • 43) 0.409 702 4 × 2 = 0 + 0.819 404 8;
  • 44) 0.819 404 8 × 2 = 1 + 0.638 809 6;
  • 45) 0.638 809 6 × 2 = 1 + 0.277 619 2;
  • 46) 0.277 619 2 × 2 = 0 + 0.555 238 4;
  • 47) 0.555 238 4 × 2 = 1 + 0.110 476 8;
  • 48) 0.110 476 8 × 2 = 0 + 0.220 953 6;
  • 49) 0.220 953 6 × 2 = 0 + 0.441 907 2;
  • 50) 0.441 907 2 × 2 = 0 + 0.883 814 4;
  • 51) 0.883 814 4 × 2 = 1 + 0.767 628 8;
  • 52) 0.767 628 8 × 2 = 1 + 0.535 257 6;
  • 53) 0.535 257 6 × 2 = 1 + 0.070 515 2;
  • 54) 0.070 515 2 × 2 = 0 + 0.141 030 4;
  • 55) 0.141 030 4 × 2 = 0 + 0.282 060 8;
  • 56) 0.282 060 8 × 2 = 0 + 0.564 121 6;
  • 57) 0.564 121 6 × 2 = 1 + 0.128 243 2;
  • 58) 0.128 243 2 × 2 = 0 + 0.256 486 4;

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 85(10) =

0.0000 0110 1110 0100 0000 0101 1100 1110 0111 0110 0001 1010 0011 1000 10(2)

5. Positive number before normalization:

0.026 916 85(10) =

0.0000 0110 1110 0100 0000 0101 1100 1110 0111 0110 0001 1010 0011 1000 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 85(10) =

0.0000 0110 1110 0100 0000 0101 1100 1110 0111 0110 0001 1010 0011 1000 10(2) =

0.0000 0110 1110 0100 0000 0101 1100 1110 0111 0110 0001 1010 0011 1000 10(2) × 20 =

1.1011 1001 0000 0001 0111 0011 1001 1101 1000 0110 1000 1110 0010(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 0111 0011 1001 1101 1000 0110 1000 1110 0010


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 0111 0011 1001 1101 1000 0110 1000 1110 0010 =

1011 1001 0000 0001 0111 0011 1001 1101 1000 0110 1000 1110 0010


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 0111 0011 1001 1101 1000 0110 1000 1110 0010


Decimal number 0.026 916 85 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1001 - 1011 1001 0000 0001 0111 0011 1001 1101 1000 0110 1000 1110 0010

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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