24.777 777 777 777 905 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 905(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
24.777 777 777 777 905(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: 0.777 777 777 777 905.

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) 0.777 777 777 777 905 × 2 = 1 + 0.555 555 555 555 81;
  • 2) 0.555 555 555 555 81 × 2 = 1 + 0.111 111 111 111 62;
  • 3) 0.111 111 111 111 62 × 2 = 0 + 0.222 222 222 223 24;
  • 4) 0.222 222 222 223 24 × 2 = 0 + 0.444 444 444 446 48;
  • 5) 0.444 444 444 446 48 × 2 = 0 + 0.888 888 888 892 96;
  • 6) 0.888 888 888 892 96 × 2 = 1 + 0.777 777 777 785 92;
  • 7) 0.777 777 777 785 92 × 2 = 1 + 0.555 555 555 571 84;
  • 8) 0.555 555 555 571 84 × 2 = 1 + 0.111 111 111 143 68;
  • 9) 0.111 111 111 143 68 × 2 = 0 + 0.222 222 222 287 36;
  • 10) 0.222 222 222 287 36 × 2 = 0 + 0.444 444 444 574 72;
  • 11) 0.444 444 444 574 72 × 2 = 0 + 0.888 888 889 149 44;
  • 12) 0.888 888 889 149 44 × 2 = 1 + 0.777 777 778 298 88;
  • 13) 0.777 777 778 298 88 × 2 = 1 + 0.555 555 556 597 76;
  • 14) 0.555 555 556 597 76 × 2 = 1 + 0.111 111 113 195 52;
  • 15) 0.111 111 113 195 52 × 2 = 0 + 0.222 222 226 391 04;
  • 16) 0.222 222 226 391 04 × 2 = 0 + 0.444 444 452 782 08;
  • 17) 0.444 444 452 782 08 × 2 = 0 + 0.888 888 905 564 16;
  • 18) 0.888 888 905 564 16 × 2 = 1 + 0.777 777 811 128 32;
  • 19) 0.777 777 811 128 32 × 2 = 1 + 0.555 555 622 256 64;
  • 20) 0.555 555 622 256 64 × 2 = 1 + 0.111 111 244 513 28;
  • 21) 0.111 111 244 513 28 × 2 = 0 + 0.222 222 489 026 56;
  • 22) 0.222 222 489 026 56 × 2 = 0 + 0.444 444 978 053 12;
  • 23) 0.444 444 978 053 12 × 2 = 0 + 0.888 889 956 106 24;
  • 24) 0.888 889 956 106 24 × 2 = 1 + 0.777 779 912 212 48;
  • 25) 0.777 779 912 212 48 × 2 = 1 + 0.555 559 824 424 96;
  • 26) 0.555 559 824 424 96 × 2 = 1 + 0.111 119 648 849 92;
  • 27) 0.111 119 648 849 92 × 2 = 0 + 0.222 239 297 699 84;
  • 28) 0.222 239 297 699 84 × 2 = 0 + 0.444 478 595 399 68;
  • 29) 0.444 478 595 399 68 × 2 = 0 + 0.888 957 190 799 36;
  • 30) 0.888 957 190 799 36 × 2 = 1 + 0.777 914 381 598 72;
  • 31) 0.777 914 381 598 72 × 2 = 1 + 0.555 828 763 197 44;
  • 32) 0.555 828 763 197 44 × 2 = 1 + 0.111 657 526 394 88;
  • 33) 0.111 657 526 394 88 × 2 = 0 + 0.223 315 052 789 76;
  • 34) 0.223 315 052 789 76 × 2 = 0 + 0.446 630 105 579 52;
  • 35) 0.446 630 105 579 52 × 2 = 0 + 0.893 260 211 159 04;
  • 36) 0.893 260 211 159 04 × 2 = 1 + 0.786 520 422 318 08;
  • 37) 0.786 520 422 318 08 × 2 = 1 + 0.573 040 844 636 16;
  • 38) 0.573 040 844 636 16 × 2 = 1 + 0.146 081 689 272 32;
  • 39) 0.146 081 689 272 32 × 2 = 0 + 0.292 163 378 544 64;
  • 40) 0.292 163 378 544 64 × 2 = 0 + 0.584 326 757 089 28;
  • 41) 0.584 326 757 089 28 × 2 = 1 + 0.168 653 514 178 56;
  • 42) 0.168 653 514 178 56 × 2 = 0 + 0.337 307 028 357 12;
  • 43) 0.337 307 028 357 12 × 2 = 0 + 0.674 614 056 714 24;
  • 44) 0.674 614 056 714 24 × 2 = 1 + 0.349 228 113 428 48;
  • 45) 0.349 228 113 428 48 × 2 = 0 + 0.698 456 226 856 96;
  • 46) 0.698 456 226 856 96 × 2 = 1 + 0.396 912 453 713 92;
  • 47) 0.396 912 453 713 92 × 2 = 0 + 0.793 824 907 427 84;
  • 48) 0.793 824 907 427 84 × 2 = 1 + 0.587 649 814 855 68;
  • 49) 0.587 649 814 855 68 × 2 = 1 + 0.175 299 629 711 36;
  • 50) 0.175 299 629 711 36 × 2 = 0 + 0.350 599 259 422 72;
  • 51) 0.350 599 259 422 72 × 2 = 0 + 0.701 198 518 845 44;
  • 52) 0.701 198 518 845 44 × 2 = 1 + 0.402 397 037 690 88;
  • 53) 0.402 397 037 690 88 × 2 = 0 + 0.804 794 075 381 76;

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:


0.777 777 777 777 905(10) =

0.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1001 0101 1001 0(2)

5. Positive number before normalization:

24.777 777 777 777 905(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1001 0101 1001 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:


24.777 777 777 777 905(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1001 0101 1001 0(2) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1001 0101 1001 0(2) × 20 =

1.1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1001 0101 1001 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 1001 0101 1001 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 1001 0101 1 0010 =

1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1001 0101


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 1001 0101


Decimal number 24.777 777 777 777 905 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 1001 0101

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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