0.026 917 37 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.026 917 37(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
0.026 917 37(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: 0.026 917 37.

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) 0.026 917 37 × 2 = 0 + 0.053 834 74;
  • 2) 0.053 834 74 × 2 = 0 + 0.107 669 48;
  • 3) 0.107 669 48 × 2 = 0 + 0.215 338 96;
  • 4) 0.215 338 96 × 2 = 0 + 0.430 677 92;
  • 5) 0.430 677 92 × 2 = 0 + 0.861 355 84;
  • 6) 0.861 355 84 × 2 = 1 + 0.722 711 68;
  • 7) 0.722 711 68 × 2 = 1 + 0.445 423 36;
  • 8) 0.445 423 36 × 2 = 0 + 0.890 846 72;
  • 9) 0.890 846 72 × 2 = 1 + 0.781 693 44;
  • 10) 0.781 693 44 × 2 = 1 + 0.563 386 88;
  • 11) 0.563 386 88 × 2 = 1 + 0.126 773 76;
  • 12) 0.126 773 76 × 2 = 0 + 0.253 547 52;
  • 13) 0.253 547 52 × 2 = 0 + 0.507 095 04;
  • 14) 0.507 095 04 × 2 = 1 + 0.014 190 08;
  • 15) 0.014 190 08 × 2 = 0 + 0.028 380 16;
  • 16) 0.028 380 16 × 2 = 0 + 0.056 760 32;
  • 17) 0.056 760 32 × 2 = 0 + 0.113 520 64;
  • 18) 0.113 520 64 × 2 = 0 + 0.227 041 28;
  • 19) 0.227 041 28 × 2 = 0 + 0.454 082 56;
  • 20) 0.454 082 56 × 2 = 0 + 0.908 165 12;
  • 21) 0.908 165 12 × 2 = 1 + 0.816 330 24;
  • 22) 0.816 330 24 × 2 = 1 + 0.632 660 48;
  • 23) 0.632 660 48 × 2 = 1 + 0.265 320 96;
  • 24) 0.265 320 96 × 2 = 0 + 0.530 641 92;
  • 25) 0.530 641 92 × 2 = 1 + 0.061 283 84;
  • 26) 0.061 283 84 × 2 = 0 + 0.122 567 68;
  • 27) 0.122 567 68 × 2 = 0 + 0.245 135 36;
  • 28) 0.245 135 36 × 2 = 0 + 0.490 270 72;
  • 29) 0.490 270 72 × 2 = 0 + 0.980 541 44;
  • 30) 0.980 541 44 × 2 = 1 + 0.961 082 88;
  • 31) 0.961 082 88 × 2 = 1 + 0.922 165 76;
  • 32) 0.922 165 76 × 2 = 1 + 0.844 331 52;
  • 33) 0.844 331 52 × 2 = 1 + 0.688 663 04;
  • 34) 0.688 663 04 × 2 = 1 + 0.377 326 08;
  • 35) 0.377 326 08 × 2 = 0 + 0.754 652 16;
  • 36) 0.754 652 16 × 2 = 1 + 0.509 304 32;
  • 37) 0.509 304 32 × 2 = 1 + 0.018 608 64;
  • 38) 0.018 608 64 × 2 = 0 + 0.037 217 28;
  • 39) 0.037 217 28 × 2 = 0 + 0.074 434 56;
  • 40) 0.074 434 56 × 2 = 0 + 0.148 869 12;
  • 41) 0.148 869 12 × 2 = 0 + 0.297 738 24;
  • 42) 0.297 738 24 × 2 = 0 + 0.595 476 48;
  • 43) 0.595 476 48 × 2 = 1 + 0.190 952 96;
  • 44) 0.190 952 96 × 2 = 0 + 0.381 905 92;
  • 45) 0.381 905 92 × 2 = 0 + 0.763 811 84;
  • 46) 0.763 811 84 × 2 = 1 + 0.527 623 68;
  • 47) 0.527 623 68 × 2 = 1 + 0.055 247 36;
  • 48) 0.055 247 36 × 2 = 0 + 0.110 494 72;
  • 49) 0.110 494 72 × 2 = 0 + 0.220 989 44;
  • 50) 0.220 989 44 × 2 = 0 + 0.441 978 88;
  • 51) 0.441 978 88 × 2 = 0 + 0.883 957 76;
  • 52) 0.883 957 76 × 2 = 1 + 0.767 915 52;
  • 53) 0.767 915 52 × 2 = 1 + 0.535 831 04;
  • 54) 0.535 831 04 × 2 = 1 + 0.071 662 08;
  • 55) 0.071 662 08 × 2 = 0 + 0.143 324 16;
  • 56) 0.143 324 16 × 2 = 0 + 0.286 648 32;
  • 57) 0.286 648 32 × 2 = 0 + 0.573 296 64;
  • 58) 0.573 296 64 × 2 = 1 + 0.146 593 28;

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:


0.026 917 37(10) =

0.0000 0110 1110 0100 0000 1110 1000 0111 1101 1000 0010 0110 0001 1100 01(2)

5. Positive number before normalization:

0.026 917 37(10) =

0.0000 0110 1110 0100 0000 1110 1000 0111 1101 1000 0010 0110 0001 1100 01(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:


0.026 917 37(10) =

0.0000 0110 1110 0100 0000 1110 1000 0111 1101 1000 0010 0110 0001 1100 01(2) =

0.0000 0110 1110 0100 0000 1110 1000 0111 1101 1000 0010 0110 0001 1100 01(2) × 20 =

1.1011 1001 0000 0011 1010 0001 1111 0110 0000 1001 1000 0111 0001(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 0011 1010 0001 1111 0110 0000 1001 1000 0111 0001


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 0011 1010 0001 1111 0110 0000 1001 1000 0111 0001 =

1011 1001 0000 0011 1010 0001 1111 0110 0000 1001 1000 0111 0001


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 0011 1010 0001 1111 0110 0000 1001 1000 0111 0001


Decimal number 0.026 917 37 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1001 - 1011 1001 0000 0011 1010 0001 1111 0110 0000 1001 1000 0111 0001

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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