0.111 111 111 111 111 111 104 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.111 111 111 111 111 111 104 7(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
0.111 111 111 111 111 111 104 7(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: 0.111 111 111 111 111 111 104 7.

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) 0.111 111 111 111 111 111 104 7 × 2 = 0 + 0.222 222 222 222 222 222 209 4;
  • 2) 0.222 222 222 222 222 222 209 4 × 2 = 0 + 0.444 444 444 444 444 444 418 8;
  • 3) 0.444 444 444 444 444 444 418 8 × 2 = 0 + 0.888 888 888 888 888 888 837 6;
  • 4) 0.888 888 888 888 888 888 837 6 × 2 = 1 + 0.777 777 777 777 777 777 675 2;
  • 5) 0.777 777 777 777 777 777 675 2 × 2 = 1 + 0.555 555 555 555 555 555 350 4;
  • 6) 0.555 555 555 555 555 555 350 4 × 2 = 1 + 0.111 111 111 111 111 110 700 8;
  • 7) 0.111 111 111 111 111 110 700 8 × 2 = 0 + 0.222 222 222 222 222 221 401 6;
  • 8) 0.222 222 222 222 222 221 401 6 × 2 = 0 + 0.444 444 444 444 444 442 803 2;
  • 9) 0.444 444 444 444 444 442 803 2 × 2 = 0 + 0.888 888 888 888 888 885 606 4;
  • 10) 0.888 888 888 888 888 885 606 4 × 2 = 1 + 0.777 777 777 777 777 771 212 8;
  • 11) 0.777 777 777 777 777 771 212 8 × 2 = 1 + 0.555 555 555 555 555 542 425 6;
  • 12) 0.555 555 555 555 555 542 425 6 × 2 = 1 + 0.111 111 111 111 111 084 851 2;
  • 13) 0.111 111 111 111 111 084 851 2 × 2 = 0 + 0.222 222 222 222 222 169 702 4;
  • 14) 0.222 222 222 222 222 169 702 4 × 2 = 0 + 0.444 444 444 444 444 339 404 8;
  • 15) 0.444 444 444 444 444 339 404 8 × 2 = 0 + 0.888 888 888 888 888 678 809 6;
  • 16) 0.888 888 888 888 888 678 809 6 × 2 = 1 + 0.777 777 777 777 777 357 619 2;
  • 17) 0.777 777 777 777 777 357 619 2 × 2 = 1 + 0.555 555 555 555 554 715 238 4;
  • 18) 0.555 555 555 555 554 715 238 4 × 2 = 1 + 0.111 111 111 111 109 430 476 8;
  • 19) 0.111 111 111 111 109 430 476 8 × 2 = 0 + 0.222 222 222 222 218 860 953 6;
  • 20) 0.222 222 222 222 218 860 953 6 × 2 = 0 + 0.444 444 444 444 437 721 907 2;
  • 21) 0.444 444 444 444 437 721 907 2 × 2 = 0 + 0.888 888 888 888 875 443 814 4;
  • 22) 0.888 888 888 888 875 443 814 4 × 2 = 1 + 0.777 777 777 777 750 887 628 8;
  • 23) 0.777 777 777 777 750 887 628 8 × 2 = 1 + 0.555 555 555 555 501 775 257 6;
  • 24) 0.555 555 555 555 501 775 257 6 × 2 = 1 + 0.111 111 111 111 003 550 515 2;
  • 25) 0.111 111 111 111 003 550 515 2 × 2 = 0 + 0.222 222 222 222 007 101 030 4;
  • 26) 0.222 222 222 222 007 101 030 4 × 2 = 0 + 0.444 444 444 444 014 202 060 8;
  • 27) 0.444 444 444 444 014 202 060 8 × 2 = 0 + 0.888 888 888 888 028 404 121 6;
  • 28) 0.888 888 888 888 028 404 121 6 × 2 = 1 + 0.777 777 777 776 056 808 243 2;
  • 29) 0.777 777 777 776 056 808 243 2 × 2 = 1 + 0.555 555 555 552 113 616 486 4;
  • 30) 0.555 555 555 552 113 616 486 4 × 2 = 1 + 0.111 111 111 104 227 232 972 8;
  • 31) 0.111 111 111 104 227 232 972 8 × 2 = 0 + 0.222 222 222 208 454 465 945 6;
  • 32) 0.222 222 222 208 454 465 945 6 × 2 = 0 + 0.444 444 444 416 908 931 891 2;
  • 33) 0.444 444 444 416 908 931 891 2 × 2 = 0 + 0.888 888 888 833 817 863 782 4;
  • 34) 0.888 888 888 833 817 863 782 4 × 2 = 1 + 0.777 777 777 667 635 727 564 8;
  • 35) 0.777 777 777 667 635 727 564 8 × 2 = 1 + 0.555 555 555 335 271 455 129 6;
  • 36) 0.555 555 555 335 271 455 129 6 × 2 = 1 + 0.111 111 110 670 542 910 259 2;
  • 37) 0.111 111 110 670 542 910 259 2 × 2 = 0 + 0.222 222 221 341 085 820 518 4;
  • 38) 0.222 222 221 341 085 820 518 4 × 2 = 0 + 0.444 444 442 682 171 641 036 8;
  • 39) 0.444 444 442 682 171 641 036 8 × 2 = 0 + 0.888 888 885 364 343 282 073 6;
  • 40) 0.888 888 885 364 343 282 073 6 × 2 = 1 + 0.777 777 770 728 686 564 147 2;
  • 41) 0.777 777 770 728 686 564 147 2 × 2 = 1 + 0.555 555 541 457 373 128 294 4;
  • 42) 0.555 555 541 457 373 128 294 4 × 2 = 1 + 0.111 111 082 914 746 256 588 8;
  • 43) 0.111 111 082 914 746 256 588 8 × 2 = 0 + 0.222 222 165 829 492 513 177 6;
  • 44) 0.222 222 165 829 492 513 177 6 × 2 = 0 + 0.444 444 331 658 985 026 355 2;
  • 45) 0.444 444 331 658 985 026 355 2 × 2 = 0 + 0.888 888 663 317 970 052 710 4;
  • 46) 0.888 888 663 317 970 052 710 4 × 2 = 1 + 0.777 777 326 635 940 105 420 8;
  • 47) 0.777 777 326 635 940 105 420 8 × 2 = 1 + 0.555 554 653 271 880 210 841 6;
  • 48) 0.555 554 653 271 880 210 841 6 × 2 = 1 + 0.111 109 306 543 760 421 683 2;
  • 49) 0.111 109 306 543 760 421 683 2 × 2 = 0 + 0.222 218 613 087 520 843 366 4;
  • 50) 0.222 218 613 087 520 843 366 4 × 2 = 0 + 0.444 437 226 175 041 686 732 8;
  • 51) 0.444 437 226 175 041 686 732 8 × 2 = 0 + 0.888 874 452 350 083 373 465 6;
  • 52) 0.888 874 452 350 083 373 465 6 × 2 = 1 + 0.777 748 904 700 166 746 931 2;
  • 53) 0.777 748 904 700 166 746 931 2 × 2 = 1 + 0.555 497 809 400 333 493 862 4;
  • 54) 0.555 497 809 400 333 493 862 4 × 2 = 1 + 0.110 995 618 800 666 987 724 8;
  • 55) 0.110 995 618 800 666 987 724 8 × 2 = 0 + 0.221 991 237 601 333 975 449 6;
  • 56) 0.221 991 237 601 333 975 449 6 × 2 = 0 + 0.443 982 475 202 667 950 899 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).


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:


0.111 111 111 111 111 111 104 7(10) =

0.0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100(2)

5. Positive number before normalization:

0.111 111 111 111 111 111 104 7(10) =

0.0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the right, so that only one non zero digit remains to the left of it:


0.111 111 111 111 111 111 104 7(10) =

0.0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100(2) =

0.0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100(2) × 20 =

1.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100(2) × 2-4


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): -4

Mantissa (not normalized):
1.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-4 + 2(11-1) - 1 =

(-4 + 1 023)(10) =

1 019(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;
  • 1 019 ÷ 2 = 509 + 1;
  • 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;

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) =

1019(10) =

011 1111 1011(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. 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 =

1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100


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 1011

Mantissa (52 bits) =
1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100


Decimal number 0.111 111 111 111 111 111 104 7 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1011 - 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100

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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