24.777 777 777 777 777 765 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 765 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
24.777 777 777 777 777 765 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: 0.777 777 777 777 777 765 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) 0.777 777 777 777 777 765 3 × 2 = 1 + 0.555 555 555 555 555 530 6;
  • 2) 0.555 555 555 555 555 530 6 × 2 = 1 + 0.111 111 111 111 111 061 2;
  • 3) 0.111 111 111 111 111 061 2 × 2 = 0 + 0.222 222 222 222 222 122 4;
  • 4) 0.222 222 222 222 222 122 4 × 2 = 0 + 0.444 444 444 444 444 244 8;
  • 5) 0.444 444 444 444 444 244 8 × 2 = 0 + 0.888 888 888 888 888 489 6;
  • 6) 0.888 888 888 888 888 489 6 × 2 = 1 + 0.777 777 777 777 776 979 2;
  • 7) 0.777 777 777 777 776 979 2 × 2 = 1 + 0.555 555 555 555 553 958 4;
  • 8) 0.555 555 555 555 553 958 4 × 2 = 1 + 0.111 111 111 111 107 916 8;
  • 9) 0.111 111 111 111 107 916 8 × 2 = 0 + 0.222 222 222 222 215 833 6;
  • 10) 0.222 222 222 222 215 833 6 × 2 = 0 + 0.444 444 444 444 431 667 2;
  • 11) 0.444 444 444 444 431 667 2 × 2 = 0 + 0.888 888 888 888 863 334 4;
  • 12) 0.888 888 888 888 863 334 4 × 2 = 1 + 0.777 777 777 777 726 668 8;
  • 13) 0.777 777 777 777 726 668 8 × 2 = 1 + 0.555 555 555 555 453 337 6;
  • 14) 0.555 555 555 555 453 337 6 × 2 = 1 + 0.111 111 111 110 906 675 2;
  • 15) 0.111 111 111 110 906 675 2 × 2 = 0 + 0.222 222 222 221 813 350 4;
  • 16) 0.222 222 222 221 813 350 4 × 2 = 0 + 0.444 444 444 443 626 700 8;
  • 17) 0.444 444 444 443 626 700 8 × 2 = 0 + 0.888 888 888 887 253 401 6;
  • 18) 0.888 888 888 887 253 401 6 × 2 = 1 + 0.777 777 777 774 506 803 2;
  • 19) 0.777 777 777 774 506 803 2 × 2 = 1 + 0.555 555 555 549 013 606 4;
  • 20) 0.555 555 555 549 013 606 4 × 2 = 1 + 0.111 111 111 098 027 212 8;
  • 21) 0.111 111 111 098 027 212 8 × 2 = 0 + 0.222 222 222 196 054 425 6;
  • 22) 0.222 222 222 196 054 425 6 × 2 = 0 + 0.444 444 444 392 108 851 2;
  • 23) 0.444 444 444 392 108 851 2 × 2 = 0 + 0.888 888 888 784 217 702 4;
  • 24) 0.888 888 888 784 217 702 4 × 2 = 1 + 0.777 777 777 568 435 404 8;
  • 25) 0.777 777 777 568 435 404 8 × 2 = 1 + 0.555 555 555 136 870 809 6;
  • 26) 0.555 555 555 136 870 809 6 × 2 = 1 + 0.111 111 110 273 741 619 2;
  • 27) 0.111 111 110 273 741 619 2 × 2 = 0 + 0.222 222 220 547 483 238 4;
  • 28) 0.222 222 220 547 483 238 4 × 2 = 0 + 0.444 444 441 094 966 476 8;
  • 29) 0.444 444 441 094 966 476 8 × 2 = 0 + 0.888 888 882 189 932 953 6;
  • 30) 0.888 888 882 189 932 953 6 × 2 = 1 + 0.777 777 764 379 865 907 2;
  • 31) 0.777 777 764 379 865 907 2 × 2 = 1 + 0.555 555 528 759 731 814 4;
  • 32) 0.555 555 528 759 731 814 4 × 2 = 1 + 0.111 111 057 519 463 628 8;
  • 33) 0.111 111 057 519 463 628 8 × 2 = 0 + 0.222 222 115 038 927 257 6;
  • 34) 0.222 222 115 038 927 257 6 × 2 = 0 + 0.444 444 230 077 854 515 2;
  • 35) 0.444 444 230 077 854 515 2 × 2 = 0 + 0.888 888 460 155 709 030 4;
  • 36) 0.888 888 460 155 709 030 4 × 2 = 1 + 0.777 776 920 311 418 060 8;
  • 37) 0.777 776 920 311 418 060 8 × 2 = 1 + 0.555 553 840 622 836 121 6;
  • 38) 0.555 553 840 622 836 121 6 × 2 = 1 + 0.111 107 681 245 672 243 2;
  • 39) 0.111 107 681 245 672 243 2 × 2 = 0 + 0.222 215 362 491 344 486 4;
  • 40) 0.222 215 362 491 344 486 4 × 2 = 0 + 0.444 430 724 982 688 972 8;
  • 41) 0.444 430 724 982 688 972 8 × 2 = 0 + 0.888 861 449 965 377 945 6;
  • 42) 0.888 861 449 965 377 945 6 × 2 = 1 + 0.777 722 899 930 755 891 2;
  • 43) 0.777 722 899 930 755 891 2 × 2 = 1 + 0.555 445 799 861 511 782 4;
  • 44) 0.555 445 799 861 511 782 4 × 2 = 1 + 0.110 891 599 723 023 564 8;
  • 45) 0.110 891 599 723 023 564 8 × 2 = 0 + 0.221 783 199 446 047 129 6;
  • 46) 0.221 783 199 446 047 129 6 × 2 = 0 + 0.443 566 398 892 094 259 2;
  • 47) 0.443 566 398 892 094 259 2 × 2 = 0 + 0.887 132 797 784 188 518 4;
  • 48) 0.887 132 797 784 188 518 4 × 2 = 1 + 0.774 265 595 568 377 036 8;
  • 49) 0.774 265 595 568 377 036 8 × 2 = 1 + 0.548 531 191 136 754 073 6;
  • 50) 0.548 531 191 136 754 073 6 × 2 = 1 + 0.097 062 382 273 508 147 2;
  • 51) 0.097 062 382 273 508 147 2 × 2 = 0 + 0.194 124 764 547 016 294 4;
  • 52) 0.194 124 764 547 016 294 4 × 2 = 0 + 0.388 249 529 094 032 588 8;
  • 53) 0.388 249 529 094 032 588 8 × 2 = 0 + 0.776 499 058 188 065 177 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:


0.777 777 777 777 777 765 3(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 765 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:


24.777 777 777 777 777 765 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 24.777 777 777 777 777 765 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