24.777 777 777 777 777 777 779 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 777 779 6(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
24.777 777 777 777 777 777 779 6(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: 0.777 777 777 777 777 777 779 6.

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) 0.777 777 777 777 777 777 779 6 × 2 = 1 + 0.555 555 555 555 555 555 559 2;
  • 2) 0.555 555 555 555 555 555 559 2 × 2 = 1 + 0.111 111 111 111 111 111 118 4;
  • 3) 0.111 111 111 111 111 111 118 4 × 2 = 0 + 0.222 222 222 222 222 222 236 8;
  • 4) 0.222 222 222 222 222 222 236 8 × 2 = 0 + 0.444 444 444 444 444 444 473 6;
  • 5) 0.444 444 444 444 444 444 473 6 × 2 = 0 + 0.888 888 888 888 888 888 947 2;
  • 6) 0.888 888 888 888 888 888 947 2 × 2 = 1 + 0.777 777 777 777 777 777 894 4;
  • 7) 0.777 777 777 777 777 777 894 4 × 2 = 1 + 0.555 555 555 555 555 555 788 8;
  • 8) 0.555 555 555 555 555 555 788 8 × 2 = 1 + 0.111 111 111 111 111 111 577 6;
  • 9) 0.111 111 111 111 111 111 577 6 × 2 = 0 + 0.222 222 222 222 222 223 155 2;
  • 10) 0.222 222 222 222 222 223 155 2 × 2 = 0 + 0.444 444 444 444 444 446 310 4;
  • 11) 0.444 444 444 444 444 446 310 4 × 2 = 0 + 0.888 888 888 888 888 892 620 8;
  • 12) 0.888 888 888 888 888 892 620 8 × 2 = 1 + 0.777 777 777 777 777 785 241 6;
  • 13) 0.777 777 777 777 777 785 241 6 × 2 = 1 + 0.555 555 555 555 555 570 483 2;
  • 14) 0.555 555 555 555 555 570 483 2 × 2 = 1 + 0.111 111 111 111 111 140 966 4;
  • 15) 0.111 111 111 111 111 140 966 4 × 2 = 0 + 0.222 222 222 222 222 281 932 8;
  • 16) 0.222 222 222 222 222 281 932 8 × 2 = 0 + 0.444 444 444 444 444 563 865 6;
  • 17) 0.444 444 444 444 444 563 865 6 × 2 = 0 + 0.888 888 888 888 889 127 731 2;
  • 18) 0.888 888 888 888 889 127 731 2 × 2 = 1 + 0.777 777 777 777 778 255 462 4;
  • 19) 0.777 777 777 777 778 255 462 4 × 2 = 1 + 0.555 555 555 555 556 510 924 8;
  • 20) 0.555 555 555 555 556 510 924 8 × 2 = 1 + 0.111 111 111 111 113 021 849 6;
  • 21) 0.111 111 111 111 113 021 849 6 × 2 = 0 + 0.222 222 222 222 226 043 699 2;
  • 22) 0.222 222 222 222 226 043 699 2 × 2 = 0 + 0.444 444 444 444 452 087 398 4;
  • 23) 0.444 444 444 444 452 087 398 4 × 2 = 0 + 0.888 888 888 888 904 174 796 8;
  • 24) 0.888 888 888 888 904 174 796 8 × 2 = 1 + 0.777 777 777 777 808 349 593 6;
  • 25) 0.777 777 777 777 808 349 593 6 × 2 = 1 + 0.555 555 555 555 616 699 187 2;
  • 26) 0.555 555 555 555 616 699 187 2 × 2 = 1 + 0.111 111 111 111 233 398 374 4;
  • 27) 0.111 111 111 111 233 398 374 4 × 2 = 0 + 0.222 222 222 222 466 796 748 8;
  • 28) 0.222 222 222 222 466 796 748 8 × 2 = 0 + 0.444 444 444 444 933 593 497 6;
  • 29) 0.444 444 444 444 933 593 497 6 × 2 = 0 + 0.888 888 888 889 867 186 995 2;
  • 30) 0.888 888 888 889 867 186 995 2 × 2 = 1 + 0.777 777 777 779 734 373 990 4;
  • 31) 0.777 777 777 779 734 373 990 4 × 2 = 1 + 0.555 555 555 559 468 747 980 8;
  • 32) 0.555 555 555 559 468 747 980 8 × 2 = 1 + 0.111 111 111 118 937 495 961 6;
  • 33) 0.111 111 111 118 937 495 961 6 × 2 = 0 + 0.222 222 222 237 874 991 923 2;
  • 34) 0.222 222 222 237 874 991 923 2 × 2 = 0 + 0.444 444 444 475 749 983 846 4;
  • 35) 0.444 444 444 475 749 983 846 4 × 2 = 0 + 0.888 888 888 951 499 967 692 8;
  • 36) 0.888 888 888 951 499 967 692 8 × 2 = 1 + 0.777 777 777 902 999 935 385 6;
  • 37) 0.777 777 777 902 999 935 385 6 × 2 = 1 + 0.555 555 555 805 999 870 771 2;
  • 38) 0.555 555 555 805 999 870 771 2 × 2 = 1 + 0.111 111 111 611 999 741 542 4;
  • 39) 0.111 111 111 611 999 741 542 4 × 2 = 0 + 0.222 222 223 223 999 483 084 8;
  • 40) 0.222 222 223 223 999 483 084 8 × 2 = 0 + 0.444 444 446 447 998 966 169 6;
  • 41) 0.444 444 446 447 998 966 169 6 × 2 = 0 + 0.888 888 892 895 997 932 339 2;
  • 42) 0.888 888 892 895 997 932 339 2 × 2 = 1 + 0.777 777 785 791 995 864 678 4;
  • 43) 0.777 777 785 791 995 864 678 4 × 2 = 1 + 0.555 555 571 583 991 729 356 8;
  • 44) 0.555 555 571 583 991 729 356 8 × 2 = 1 + 0.111 111 143 167 983 458 713 6;
  • 45) 0.111 111 143 167 983 458 713 6 × 2 = 0 + 0.222 222 286 335 966 917 427 2;
  • 46) 0.222 222 286 335 966 917 427 2 × 2 = 0 + 0.444 444 572 671 933 834 854 4;
  • 47) 0.444 444 572 671 933 834 854 4 × 2 = 0 + 0.888 889 145 343 867 669 708 8;
  • 48) 0.888 889 145 343 867 669 708 8 × 2 = 1 + 0.777 778 290 687 735 339 417 6;
  • 49) 0.777 778 290 687 735 339 417 6 × 2 = 1 + 0.555 556 581 375 470 678 835 2;
  • 50) 0.555 556 581 375 470 678 835 2 × 2 = 1 + 0.111 113 162 750 941 357 670 4;
  • 51) 0.111 113 162 750 941 357 670 4 × 2 = 0 + 0.222 226 325 501 882 715 340 8;
  • 52) 0.222 226 325 501 882 715 340 8 × 2 = 0 + 0.444 452 651 003 765 430 681 6;
  • 53) 0.444 452 651 003 765 430 681 6 × 2 = 0 + 0.888 905 302 007 530 861 363 2;

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:


0.777 777 777 777 777 777 779 6(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 777 779 6(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:


24.777 777 777 777 777 777 779 6(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 24.777 777 777 777 777 777 779 6 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

Share

» Convert decimal 24.777 777 777 777 777 777 784 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 06, 2026 [June]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: June - by our visitors

» Improve your social skills: Are you using communication by design, or by default? NotebookLM YouTube Video of 6.33 minutes

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