24.777 777 777 777 777 777 31 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 777 31(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
24.777 777 777 777 777 777 31(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: 24.
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;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

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.

24(10) =

1 1000(2)


3. Convert to binary (base 2) the fractional part: 0.777 777 777 777 777 777 31.

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) 0.777 777 777 777 777 777 31 × 2 = 1 + 0.555 555 555 555 555 554 62;
  • 2) 0.555 555 555 555 555 554 62 × 2 = 1 + 0.111 111 111 111 111 109 24;
  • 3) 0.111 111 111 111 111 109 24 × 2 = 0 + 0.222 222 222 222 222 218 48;
  • 4) 0.222 222 222 222 222 218 48 × 2 = 0 + 0.444 444 444 444 444 436 96;
  • 5) 0.444 444 444 444 444 436 96 × 2 = 0 + 0.888 888 888 888 888 873 92;
  • 6) 0.888 888 888 888 888 873 92 × 2 = 1 + 0.777 777 777 777 777 747 84;
  • 7) 0.777 777 777 777 777 747 84 × 2 = 1 + 0.555 555 555 555 555 495 68;
  • 8) 0.555 555 555 555 555 495 68 × 2 = 1 + 0.111 111 111 111 110 991 36;
  • 9) 0.111 111 111 111 110 991 36 × 2 = 0 + 0.222 222 222 222 221 982 72;
  • 10) 0.222 222 222 222 221 982 72 × 2 = 0 + 0.444 444 444 444 443 965 44;
  • 11) 0.444 444 444 444 443 965 44 × 2 = 0 + 0.888 888 888 888 887 930 88;
  • 12) 0.888 888 888 888 887 930 88 × 2 = 1 + 0.777 777 777 777 775 861 76;
  • 13) 0.777 777 777 777 775 861 76 × 2 = 1 + 0.555 555 555 555 551 723 52;
  • 14) 0.555 555 555 555 551 723 52 × 2 = 1 + 0.111 111 111 111 103 447 04;
  • 15) 0.111 111 111 111 103 447 04 × 2 = 0 + 0.222 222 222 222 206 894 08;
  • 16) 0.222 222 222 222 206 894 08 × 2 = 0 + 0.444 444 444 444 413 788 16;
  • 17) 0.444 444 444 444 413 788 16 × 2 = 0 + 0.888 888 888 888 827 576 32;
  • 18) 0.888 888 888 888 827 576 32 × 2 = 1 + 0.777 777 777 777 655 152 64;
  • 19) 0.777 777 777 777 655 152 64 × 2 = 1 + 0.555 555 555 555 310 305 28;
  • 20) 0.555 555 555 555 310 305 28 × 2 = 1 + 0.111 111 111 110 620 610 56;
  • 21) 0.111 111 111 110 620 610 56 × 2 = 0 + 0.222 222 222 221 241 221 12;
  • 22) 0.222 222 222 221 241 221 12 × 2 = 0 + 0.444 444 444 442 482 442 24;
  • 23) 0.444 444 444 442 482 442 24 × 2 = 0 + 0.888 888 888 884 964 884 48;
  • 24) 0.888 888 888 884 964 884 48 × 2 = 1 + 0.777 777 777 769 929 768 96;
  • 25) 0.777 777 777 769 929 768 96 × 2 = 1 + 0.555 555 555 539 859 537 92;
  • 26) 0.555 555 555 539 859 537 92 × 2 = 1 + 0.111 111 111 079 719 075 84;
  • 27) 0.111 111 111 079 719 075 84 × 2 = 0 + 0.222 222 222 159 438 151 68;
  • 28) 0.222 222 222 159 438 151 68 × 2 = 0 + 0.444 444 444 318 876 303 36;
  • 29) 0.444 444 444 318 876 303 36 × 2 = 0 + 0.888 888 888 637 752 606 72;
  • 30) 0.888 888 888 637 752 606 72 × 2 = 1 + 0.777 777 777 275 505 213 44;
  • 31) 0.777 777 777 275 505 213 44 × 2 = 1 + 0.555 555 554 551 010 426 88;
  • 32) 0.555 555 554 551 010 426 88 × 2 = 1 + 0.111 111 109 102 020 853 76;
  • 33) 0.111 111 109 102 020 853 76 × 2 = 0 + 0.222 222 218 204 041 707 52;
  • 34) 0.222 222 218 204 041 707 52 × 2 = 0 + 0.444 444 436 408 083 415 04;
  • 35) 0.444 444 436 408 083 415 04 × 2 = 0 + 0.888 888 872 816 166 830 08;
  • 36) 0.888 888 872 816 166 830 08 × 2 = 1 + 0.777 777 745 632 333 660 16;
  • 37) 0.777 777 745 632 333 660 16 × 2 = 1 + 0.555 555 491 264 667 320 32;
  • 38) 0.555 555 491 264 667 320 32 × 2 = 1 + 0.111 110 982 529 334 640 64;
  • 39) 0.111 110 982 529 334 640 64 × 2 = 0 + 0.222 221 965 058 669 281 28;
  • 40) 0.222 221 965 058 669 281 28 × 2 = 0 + 0.444 443 930 117 338 562 56;
  • 41) 0.444 443 930 117 338 562 56 × 2 = 0 + 0.888 887 860 234 677 125 12;
  • 42) 0.888 887 860 234 677 125 12 × 2 = 1 + 0.777 775 720 469 354 250 24;
  • 43) 0.777 775 720 469 354 250 24 × 2 = 1 + 0.555 551 440 938 708 500 48;
  • 44) 0.555 551 440 938 708 500 48 × 2 = 1 + 0.111 102 881 877 417 000 96;
  • 45) 0.111 102 881 877 417 000 96 × 2 = 0 + 0.222 205 763 754 834 001 92;
  • 46) 0.222 205 763 754 834 001 92 × 2 = 0 + 0.444 411 527 509 668 003 84;
  • 47) 0.444 411 527 509 668 003 84 × 2 = 0 + 0.888 823 055 019 336 007 68;
  • 48) 0.888 823 055 019 336 007 68 × 2 = 1 + 0.777 646 110 038 672 015 36;
  • 49) 0.777 646 110 038 672 015 36 × 2 = 1 + 0.555 292 220 077 344 030 72;
  • 50) 0.555 292 220 077 344 030 72 × 2 = 1 + 0.110 584 440 154 688 061 44;
  • 51) 0.110 584 440 154 688 061 44 × 2 = 0 + 0.221 168 880 309 376 122 88;
  • 52) 0.221 168 880 309 376 122 88 × 2 = 0 + 0.442 337 760 618 752 245 76;
  • 53) 0.442 337 760 618 752 245 76 × 2 = 0 + 0.884 675 521 237 504 491 52;

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:


0.777 777 777 777 777 777 31(10) =

0.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0(2)

5. Positive number before normalization:

24.777 777 777 777 777 777 31(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0(2)

6. Normalize the binary representation of the number.

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


24.777 777 777 777 777 777 31(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0(2) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0(2) × 20 =

1.1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0(2) × 24


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):
1.1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0


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) =

1 027(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;
  • 1 027 ÷ 2 = 513 + 1;
  • 513 ÷ 2 = 256 + 1;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

1027(10) =

100 0000 0011(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1 1000 =

1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 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) =
100 0000 0011

Mantissa (52 bits) =
1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001


Decimal number 24.777 777 777 777 777 777 31 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0011 - 1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 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