24.777 777 777 777 777 777 775 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 777 775 3(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
24.777 777 777 777 777 777 775 3(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: 0.777 777 777 777 777 777 775 3.

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) 0.777 777 777 777 777 777 775 3 × 2 = 1 + 0.555 555 555 555 555 555 550 6;
  • 2) 0.555 555 555 555 555 555 550 6 × 2 = 1 + 0.111 111 111 111 111 111 101 2;
  • 3) 0.111 111 111 111 111 111 101 2 × 2 = 0 + 0.222 222 222 222 222 222 202 4;
  • 4) 0.222 222 222 222 222 222 202 4 × 2 = 0 + 0.444 444 444 444 444 444 404 8;
  • 5) 0.444 444 444 444 444 444 404 8 × 2 = 0 + 0.888 888 888 888 888 888 809 6;
  • 6) 0.888 888 888 888 888 888 809 6 × 2 = 1 + 0.777 777 777 777 777 777 619 2;
  • 7) 0.777 777 777 777 777 777 619 2 × 2 = 1 + 0.555 555 555 555 555 555 238 4;
  • 8) 0.555 555 555 555 555 555 238 4 × 2 = 1 + 0.111 111 111 111 111 110 476 8;
  • 9) 0.111 111 111 111 111 110 476 8 × 2 = 0 + 0.222 222 222 222 222 220 953 6;
  • 10) 0.222 222 222 222 222 220 953 6 × 2 = 0 + 0.444 444 444 444 444 441 907 2;
  • 11) 0.444 444 444 444 444 441 907 2 × 2 = 0 + 0.888 888 888 888 888 883 814 4;
  • 12) 0.888 888 888 888 888 883 814 4 × 2 = 1 + 0.777 777 777 777 777 767 628 8;
  • 13) 0.777 777 777 777 777 767 628 8 × 2 = 1 + 0.555 555 555 555 555 535 257 6;
  • 14) 0.555 555 555 555 555 535 257 6 × 2 = 1 + 0.111 111 111 111 111 070 515 2;
  • 15) 0.111 111 111 111 111 070 515 2 × 2 = 0 + 0.222 222 222 222 222 141 030 4;
  • 16) 0.222 222 222 222 222 141 030 4 × 2 = 0 + 0.444 444 444 444 444 282 060 8;
  • 17) 0.444 444 444 444 444 282 060 8 × 2 = 0 + 0.888 888 888 888 888 564 121 6;
  • 18) 0.888 888 888 888 888 564 121 6 × 2 = 1 + 0.777 777 777 777 777 128 243 2;
  • 19) 0.777 777 777 777 777 128 243 2 × 2 = 1 + 0.555 555 555 555 554 256 486 4;
  • 20) 0.555 555 555 555 554 256 486 4 × 2 = 1 + 0.111 111 111 111 108 512 972 8;
  • 21) 0.111 111 111 111 108 512 972 8 × 2 = 0 + 0.222 222 222 222 217 025 945 6;
  • 22) 0.222 222 222 222 217 025 945 6 × 2 = 0 + 0.444 444 444 444 434 051 891 2;
  • 23) 0.444 444 444 444 434 051 891 2 × 2 = 0 + 0.888 888 888 888 868 103 782 4;
  • 24) 0.888 888 888 888 868 103 782 4 × 2 = 1 + 0.777 777 777 777 736 207 564 8;
  • 25) 0.777 777 777 777 736 207 564 8 × 2 = 1 + 0.555 555 555 555 472 415 129 6;
  • 26) 0.555 555 555 555 472 415 129 6 × 2 = 1 + 0.111 111 111 110 944 830 259 2;
  • 27) 0.111 111 111 110 944 830 259 2 × 2 = 0 + 0.222 222 222 221 889 660 518 4;
  • 28) 0.222 222 222 221 889 660 518 4 × 2 = 0 + 0.444 444 444 443 779 321 036 8;
  • 29) 0.444 444 444 443 779 321 036 8 × 2 = 0 + 0.888 888 888 887 558 642 073 6;
  • 30) 0.888 888 888 887 558 642 073 6 × 2 = 1 + 0.777 777 777 775 117 284 147 2;
  • 31) 0.777 777 777 775 117 284 147 2 × 2 = 1 + 0.555 555 555 550 234 568 294 4;
  • 32) 0.555 555 555 550 234 568 294 4 × 2 = 1 + 0.111 111 111 100 469 136 588 8;
  • 33) 0.111 111 111 100 469 136 588 8 × 2 = 0 + 0.222 222 222 200 938 273 177 6;
  • 34) 0.222 222 222 200 938 273 177 6 × 2 = 0 + 0.444 444 444 401 876 546 355 2;
  • 35) 0.444 444 444 401 876 546 355 2 × 2 = 0 + 0.888 888 888 803 753 092 710 4;
  • 36) 0.888 888 888 803 753 092 710 4 × 2 = 1 + 0.777 777 777 607 506 185 420 8;
  • 37) 0.777 777 777 607 506 185 420 8 × 2 = 1 + 0.555 555 555 215 012 370 841 6;
  • 38) 0.555 555 555 215 012 370 841 6 × 2 = 1 + 0.111 111 110 430 024 741 683 2;
  • 39) 0.111 111 110 430 024 741 683 2 × 2 = 0 + 0.222 222 220 860 049 483 366 4;
  • 40) 0.222 222 220 860 049 483 366 4 × 2 = 0 + 0.444 444 441 720 098 966 732 8;
  • 41) 0.444 444 441 720 098 966 732 8 × 2 = 0 + 0.888 888 883 440 197 933 465 6;
  • 42) 0.888 888 883 440 197 933 465 6 × 2 = 1 + 0.777 777 766 880 395 866 931 2;
  • 43) 0.777 777 766 880 395 866 931 2 × 2 = 1 + 0.555 555 533 760 791 733 862 4;
  • 44) 0.555 555 533 760 791 733 862 4 × 2 = 1 + 0.111 111 067 521 583 467 724 8;
  • 45) 0.111 111 067 521 583 467 724 8 × 2 = 0 + 0.222 222 135 043 166 935 449 6;
  • 46) 0.222 222 135 043 166 935 449 6 × 2 = 0 + 0.444 444 270 086 333 870 899 2;
  • 47) 0.444 444 270 086 333 870 899 2 × 2 = 0 + 0.888 888 540 172 667 741 798 4;
  • 48) 0.888 888 540 172 667 741 798 4 × 2 = 1 + 0.777 777 080 345 335 483 596 8;
  • 49) 0.777 777 080 345 335 483 596 8 × 2 = 1 + 0.555 554 160 690 670 967 193 6;
  • 50) 0.555 554 160 690 670 967 193 6 × 2 = 1 + 0.111 108 321 381 341 934 387 2;
  • 51) 0.111 108 321 381 341 934 387 2 × 2 = 0 + 0.222 216 642 762 683 868 774 4;
  • 52) 0.222 216 642 762 683 868 774 4 × 2 = 0 + 0.444 433 285 525 367 737 548 8;
  • 53) 0.444 433 285 525 367 737 548 8 × 2 = 0 + 0.888 866 571 050 735 475 097 6;

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:


0.777 777 777 777 777 777 775 3(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 777 775 3(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:


24.777 777 777 777 777 777 775 3(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 24.777 777 777 777 777 777 775 3 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