24.777 777 777 777 777 770 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 770 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
24.777 777 777 777 777 770 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: 0.777 777 777 777 777 770 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) 0.777 777 777 777 777 770 3 × 2 = 1 + 0.555 555 555 555 555 540 6;
  • 2) 0.555 555 555 555 555 540 6 × 2 = 1 + 0.111 111 111 111 111 081 2;
  • 3) 0.111 111 111 111 111 081 2 × 2 = 0 + 0.222 222 222 222 222 162 4;
  • 4) 0.222 222 222 222 222 162 4 × 2 = 0 + 0.444 444 444 444 444 324 8;
  • 5) 0.444 444 444 444 444 324 8 × 2 = 0 + 0.888 888 888 888 888 649 6;
  • 6) 0.888 888 888 888 888 649 6 × 2 = 1 + 0.777 777 777 777 777 299 2;
  • 7) 0.777 777 777 777 777 299 2 × 2 = 1 + 0.555 555 555 555 554 598 4;
  • 8) 0.555 555 555 555 554 598 4 × 2 = 1 + 0.111 111 111 111 109 196 8;
  • 9) 0.111 111 111 111 109 196 8 × 2 = 0 + 0.222 222 222 222 218 393 6;
  • 10) 0.222 222 222 222 218 393 6 × 2 = 0 + 0.444 444 444 444 436 787 2;
  • 11) 0.444 444 444 444 436 787 2 × 2 = 0 + 0.888 888 888 888 873 574 4;
  • 12) 0.888 888 888 888 873 574 4 × 2 = 1 + 0.777 777 777 777 747 148 8;
  • 13) 0.777 777 777 777 747 148 8 × 2 = 1 + 0.555 555 555 555 494 297 6;
  • 14) 0.555 555 555 555 494 297 6 × 2 = 1 + 0.111 111 111 110 988 595 2;
  • 15) 0.111 111 111 110 988 595 2 × 2 = 0 + 0.222 222 222 221 977 190 4;
  • 16) 0.222 222 222 221 977 190 4 × 2 = 0 + 0.444 444 444 443 954 380 8;
  • 17) 0.444 444 444 443 954 380 8 × 2 = 0 + 0.888 888 888 887 908 761 6;
  • 18) 0.888 888 888 887 908 761 6 × 2 = 1 + 0.777 777 777 775 817 523 2;
  • 19) 0.777 777 777 775 817 523 2 × 2 = 1 + 0.555 555 555 551 635 046 4;
  • 20) 0.555 555 555 551 635 046 4 × 2 = 1 + 0.111 111 111 103 270 092 8;
  • 21) 0.111 111 111 103 270 092 8 × 2 = 0 + 0.222 222 222 206 540 185 6;
  • 22) 0.222 222 222 206 540 185 6 × 2 = 0 + 0.444 444 444 413 080 371 2;
  • 23) 0.444 444 444 413 080 371 2 × 2 = 0 + 0.888 888 888 826 160 742 4;
  • 24) 0.888 888 888 826 160 742 4 × 2 = 1 + 0.777 777 777 652 321 484 8;
  • 25) 0.777 777 777 652 321 484 8 × 2 = 1 + 0.555 555 555 304 642 969 6;
  • 26) 0.555 555 555 304 642 969 6 × 2 = 1 + 0.111 111 110 609 285 939 2;
  • 27) 0.111 111 110 609 285 939 2 × 2 = 0 + 0.222 222 221 218 571 878 4;
  • 28) 0.222 222 221 218 571 878 4 × 2 = 0 + 0.444 444 442 437 143 756 8;
  • 29) 0.444 444 442 437 143 756 8 × 2 = 0 + 0.888 888 884 874 287 513 6;
  • 30) 0.888 888 884 874 287 513 6 × 2 = 1 + 0.777 777 769 748 575 027 2;
  • 31) 0.777 777 769 748 575 027 2 × 2 = 1 + 0.555 555 539 497 150 054 4;
  • 32) 0.555 555 539 497 150 054 4 × 2 = 1 + 0.111 111 078 994 300 108 8;
  • 33) 0.111 111 078 994 300 108 8 × 2 = 0 + 0.222 222 157 988 600 217 6;
  • 34) 0.222 222 157 988 600 217 6 × 2 = 0 + 0.444 444 315 977 200 435 2;
  • 35) 0.444 444 315 977 200 435 2 × 2 = 0 + 0.888 888 631 954 400 870 4;
  • 36) 0.888 888 631 954 400 870 4 × 2 = 1 + 0.777 777 263 908 801 740 8;
  • 37) 0.777 777 263 908 801 740 8 × 2 = 1 + 0.555 554 527 817 603 481 6;
  • 38) 0.555 554 527 817 603 481 6 × 2 = 1 + 0.111 109 055 635 206 963 2;
  • 39) 0.111 109 055 635 206 963 2 × 2 = 0 + 0.222 218 111 270 413 926 4;
  • 40) 0.222 218 111 270 413 926 4 × 2 = 0 + 0.444 436 222 540 827 852 8;
  • 41) 0.444 436 222 540 827 852 8 × 2 = 0 + 0.888 872 445 081 655 705 6;
  • 42) 0.888 872 445 081 655 705 6 × 2 = 1 + 0.777 744 890 163 311 411 2;
  • 43) 0.777 744 890 163 311 411 2 × 2 = 1 + 0.555 489 780 326 622 822 4;
  • 44) 0.555 489 780 326 622 822 4 × 2 = 1 + 0.110 979 560 653 245 644 8;
  • 45) 0.110 979 560 653 245 644 8 × 2 = 0 + 0.221 959 121 306 491 289 6;
  • 46) 0.221 959 121 306 491 289 6 × 2 = 0 + 0.443 918 242 612 982 579 2;
  • 47) 0.443 918 242 612 982 579 2 × 2 = 0 + 0.887 836 485 225 965 158 4;
  • 48) 0.887 836 485 225 965 158 4 × 2 = 1 + 0.775 672 970 451 930 316 8;
  • 49) 0.775 672 970 451 930 316 8 × 2 = 1 + 0.551 345 940 903 860 633 6;
  • 50) 0.551 345 940 903 860 633 6 × 2 = 1 + 0.102 691 881 807 721 267 2;
  • 51) 0.102 691 881 807 721 267 2 × 2 = 0 + 0.205 383 763 615 442 534 4;
  • 52) 0.205 383 763 615 442 534 4 × 2 = 0 + 0.410 767 527 230 885 068 8;
  • 53) 0.410 767 527 230 885 068 8 × 2 = 0 + 0.821 535 054 461 770 137 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:


0.777 777 777 777 777 770 3(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 770 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:


24.777 777 777 777 777 770 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 24.777 777 777 777 777 770 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