24.777 777 777 777 777 777 777 861 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 777 777 861(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
24.777 777 777 777 777 777 777 861(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: 0.777 777 777 777 777 777 777 861.

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) 0.777 777 777 777 777 777 777 861 × 2 = 1 + 0.555 555 555 555 555 555 555 722;
  • 2) 0.555 555 555 555 555 555 555 722 × 2 = 1 + 0.111 111 111 111 111 111 111 444;
  • 3) 0.111 111 111 111 111 111 111 444 × 2 = 0 + 0.222 222 222 222 222 222 222 888;
  • 4) 0.222 222 222 222 222 222 222 888 × 2 = 0 + 0.444 444 444 444 444 444 445 776;
  • 5) 0.444 444 444 444 444 444 445 776 × 2 = 0 + 0.888 888 888 888 888 888 891 552;
  • 6) 0.888 888 888 888 888 888 891 552 × 2 = 1 + 0.777 777 777 777 777 777 783 104;
  • 7) 0.777 777 777 777 777 777 783 104 × 2 = 1 + 0.555 555 555 555 555 555 566 208;
  • 8) 0.555 555 555 555 555 555 566 208 × 2 = 1 + 0.111 111 111 111 111 111 132 416;
  • 9) 0.111 111 111 111 111 111 132 416 × 2 = 0 + 0.222 222 222 222 222 222 264 832;
  • 10) 0.222 222 222 222 222 222 264 832 × 2 = 0 + 0.444 444 444 444 444 444 529 664;
  • 11) 0.444 444 444 444 444 444 529 664 × 2 = 0 + 0.888 888 888 888 888 889 059 328;
  • 12) 0.888 888 888 888 888 889 059 328 × 2 = 1 + 0.777 777 777 777 777 778 118 656;
  • 13) 0.777 777 777 777 777 778 118 656 × 2 = 1 + 0.555 555 555 555 555 556 237 312;
  • 14) 0.555 555 555 555 555 556 237 312 × 2 = 1 + 0.111 111 111 111 111 112 474 624;
  • 15) 0.111 111 111 111 111 112 474 624 × 2 = 0 + 0.222 222 222 222 222 224 949 248;
  • 16) 0.222 222 222 222 222 224 949 248 × 2 = 0 + 0.444 444 444 444 444 449 898 496;
  • 17) 0.444 444 444 444 444 449 898 496 × 2 = 0 + 0.888 888 888 888 888 899 796 992;
  • 18) 0.888 888 888 888 888 899 796 992 × 2 = 1 + 0.777 777 777 777 777 799 593 984;
  • 19) 0.777 777 777 777 777 799 593 984 × 2 = 1 + 0.555 555 555 555 555 599 187 968;
  • 20) 0.555 555 555 555 555 599 187 968 × 2 = 1 + 0.111 111 111 111 111 198 375 936;
  • 21) 0.111 111 111 111 111 198 375 936 × 2 = 0 + 0.222 222 222 222 222 396 751 872;
  • 22) 0.222 222 222 222 222 396 751 872 × 2 = 0 + 0.444 444 444 444 444 793 503 744;
  • 23) 0.444 444 444 444 444 793 503 744 × 2 = 0 + 0.888 888 888 888 889 587 007 488;
  • 24) 0.888 888 888 888 889 587 007 488 × 2 = 1 + 0.777 777 777 777 779 174 014 976;
  • 25) 0.777 777 777 777 779 174 014 976 × 2 = 1 + 0.555 555 555 555 558 348 029 952;
  • 26) 0.555 555 555 555 558 348 029 952 × 2 = 1 + 0.111 111 111 111 116 696 059 904;
  • 27) 0.111 111 111 111 116 696 059 904 × 2 = 0 + 0.222 222 222 222 233 392 119 808;
  • 28) 0.222 222 222 222 233 392 119 808 × 2 = 0 + 0.444 444 444 444 466 784 239 616;
  • 29) 0.444 444 444 444 466 784 239 616 × 2 = 0 + 0.888 888 888 888 933 568 479 232;
  • 30) 0.888 888 888 888 933 568 479 232 × 2 = 1 + 0.777 777 777 777 867 136 958 464;
  • 31) 0.777 777 777 777 867 136 958 464 × 2 = 1 + 0.555 555 555 555 734 273 916 928;
  • 32) 0.555 555 555 555 734 273 916 928 × 2 = 1 + 0.111 111 111 111 468 547 833 856;
  • 33) 0.111 111 111 111 468 547 833 856 × 2 = 0 + 0.222 222 222 222 937 095 667 712;
  • 34) 0.222 222 222 222 937 095 667 712 × 2 = 0 + 0.444 444 444 445 874 191 335 424;
  • 35) 0.444 444 444 445 874 191 335 424 × 2 = 0 + 0.888 888 888 891 748 382 670 848;
  • 36) 0.888 888 888 891 748 382 670 848 × 2 = 1 + 0.777 777 777 783 496 765 341 696;
  • 37) 0.777 777 777 783 496 765 341 696 × 2 = 1 + 0.555 555 555 566 993 530 683 392;
  • 38) 0.555 555 555 566 993 530 683 392 × 2 = 1 + 0.111 111 111 133 987 061 366 784;
  • 39) 0.111 111 111 133 987 061 366 784 × 2 = 0 + 0.222 222 222 267 974 122 733 568;
  • 40) 0.222 222 222 267 974 122 733 568 × 2 = 0 + 0.444 444 444 535 948 245 467 136;
  • 41) 0.444 444 444 535 948 245 467 136 × 2 = 0 + 0.888 888 889 071 896 490 934 272;
  • 42) 0.888 888 889 071 896 490 934 272 × 2 = 1 + 0.777 777 778 143 792 981 868 544;
  • 43) 0.777 777 778 143 792 981 868 544 × 2 = 1 + 0.555 555 556 287 585 963 737 088;
  • 44) 0.555 555 556 287 585 963 737 088 × 2 = 1 + 0.111 111 112 575 171 927 474 176;
  • 45) 0.111 111 112 575 171 927 474 176 × 2 = 0 + 0.222 222 225 150 343 854 948 352;
  • 46) 0.222 222 225 150 343 854 948 352 × 2 = 0 + 0.444 444 450 300 687 709 896 704;
  • 47) 0.444 444 450 300 687 709 896 704 × 2 = 0 + 0.888 888 900 601 375 419 793 408;
  • 48) 0.888 888 900 601 375 419 793 408 × 2 = 1 + 0.777 777 801 202 750 839 586 816;
  • 49) 0.777 777 801 202 750 839 586 816 × 2 = 1 + 0.555 555 602 405 501 679 173 632;
  • 50) 0.555 555 602 405 501 679 173 632 × 2 = 1 + 0.111 111 204 811 003 358 347 264;
  • 51) 0.111 111 204 811 003 358 347 264 × 2 = 0 + 0.222 222 409 622 006 716 694 528;
  • 52) 0.222 222 409 622 006 716 694 528 × 2 = 0 + 0.444 444 819 244 013 433 389 056;
  • 53) 0.444 444 819 244 013 433 389 056 × 2 = 0 + 0.888 889 638 488 026 866 778 112;

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:


0.777 777 777 777 777 777 777 861(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 777 777 861(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:


24.777 777 777 777 777 777 777 861(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 24.777 777 777 777 777 777 777 861 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