24.777 777 777 777 777 777 778 096 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 777 778 096(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
24.777 777 777 777 777 777 778 096(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: 0.777 777 777 777 777 777 778 096.

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) 0.777 777 777 777 777 777 778 096 × 2 = 1 + 0.555 555 555 555 555 555 556 192;
  • 2) 0.555 555 555 555 555 555 556 192 × 2 = 1 + 0.111 111 111 111 111 111 112 384;
  • 3) 0.111 111 111 111 111 111 112 384 × 2 = 0 + 0.222 222 222 222 222 222 224 768;
  • 4) 0.222 222 222 222 222 222 224 768 × 2 = 0 + 0.444 444 444 444 444 444 449 536;
  • 5) 0.444 444 444 444 444 444 449 536 × 2 = 0 + 0.888 888 888 888 888 888 899 072;
  • 6) 0.888 888 888 888 888 888 899 072 × 2 = 1 + 0.777 777 777 777 777 777 798 144;
  • 7) 0.777 777 777 777 777 777 798 144 × 2 = 1 + 0.555 555 555 555 555 555 596 288;
  • 8) 0.555 555 555 555 555 555 596 288 × 2 = 1 + 0.111 111 111 111 111 111 192 576;
  • 9) 0.111 111 111 111 111 111 192 576 × 2 = 0 + 0.222 222 222 222 222 222 385 152;
  • 10) 0.222 222 222 222 222 222 385 152 × 2 = 0 + 0.444 444 444 444 444 444 770 304;
  • 11) 0.444 444 444 444 444 444 770 304 × 2 = 0 + 0.888 888 888 888 888 889 540 608;
  • 12) 0.888 888 888 888 888 889 540 608 × 2 = 1 + 0.777 777 777 777 777 779 081 216;
  • 13) 0.777 777 777 777 777 779 081 216 × 2 = 1 + 0.555 555 555 555 555 558 162 432;
  • 14) 0.555 555 555 555 555 558 162 432 × 2 = 1 + 0.111 111 111 111 111 116 324 864;
  • 15) 0.111 111 111 111 111 116 324 864 × 2 = 0 + 0.222 222 222 222 222 232 649 728;
  • 16) 0.222 222 222 222 222 232 649 728 × 2 = 0 + 0.444 444 444 444 444 465 299 456;
  • 17) 0.444 444 444 444 444 465 299 456 × 2 = 0 + 0.888 888 888 888 888 930 598 912;
  • 18) 0.888 888 888 888 888 930 598 912 × 2 = 1 + 0.777 777 777 777 777 861 197 824;
  • 19) 0.777 777 777 777 777 861 197 824 × 2 = 1 + 0.555 555 555 555 555 722 395 648;
  • 20) 0.555 555 555 555 555 722 395 648 × 2 = 1 + 0.111 111 111 111 111 444 791 296;
  • 21) 0.111 111 111 111 111 444 791 296 × 2 = 0 + 0.222 222 222 222 222 889 582 592;
  • 22) 0.222 222 222 222 222 889 582 592 × 2 = 0 + 0.444 444 444 444 445 779 165 184;
  • 23) 0.444 444 444 444 445 779 165 184 × 2 = 0 + 0.888 888 888 888 891 558 330 368;
  • 24) 0.888 888 888 888 891 558 330 368 × 2 = 1 + 0.777 777 777 777 783 116 660 736;
  • 25) 0.777 777 777 777 783 116 660 736 × 2 = 1 + 0.555 555 555 555 566 233 321 472;
  • 26) 0.555 555 555 555 566 233 321 472 × 2 = 1 + 0.111 111 111 111 132 466 642 944;
  • 27) 0.111 111 111 111 132 466 642 944 × 2 = 0 + 0.222 222 222 222 264 933 285 888;
  • 28) 0.222 222 222 222 264 933 285 888 × 2 = 0 + 0.444 444 444 444 529 866 571 776;
  • 29) 0.444 444 444 444 529 866 571 776 × 2 = 0 + 0.888 888 888 889 059 733 143 552;
  • 30) 0.888 888 888 889 059 733 143 552 × 2 = 1 + 0.777 777 777 778 119 466 287 104;
  • 31) 0.777 777 777 778 119 466 287 104 × 2 = 1 + 0.555 555 555 556 238 932 574 208;
  • 32) 0.555 555 555 556 238 932 574 208 × 2 = 1 + 0.111 111 111 112 477 865 148 416;
  • 33) 0.111 111 111 112 477 865 148 416 × 2 = 0 + 0.222 222 222 224 955 730 296 832;
  • 34) 0.222 222 222 224 955 730 296 832 × 2 = 0 + 0.444 444 444 449 911 460 593 664;
  • 35) 0.444 444 444 449 911 460 593 664 × 2 = 0 + 0.888 888 888 899 822 921 187 328;
  • 36) 0.888 888 888 899 822 921 187 328 × 2 = 1 + 0.777 777 777 799 645 842 374 656;
  • 37) 0.777 777 777 799 645 842 374 656 × 2 = 1 + 0.555 555 555 599 291 684 749 312;
  • 38) 0.555 555 555 599 291 684 749 312 × 2 = 1 + 0.111 111 111 198 583 369 498 624;
  • 39) 0.111 111 111 198 583 369 498 624 × 2 = 0 + 0.222 222 222 397 166 738 997 248;
  • 40) 0.222 222 222 397 166 738 997 248 × 2 = 0 + 0.444 444 444 794 333 477 994 496;
  • 41) 0.444 444 444 794 333 477 994 496 × 2 = 0 + 0.888 888 889 588 666 955 988 992;
  • 42) 0.888 888 889 588 666 955 988 992 × 2 = 1 + 0.777 777 779 177 333 911 977 984;
  • 43) 0.777 777 779 177 333 911 977 984 × 2 = 1 + 0.555 555 558 354 667 823 955 968;
  • 44) 0.555 555 558 354 667 823 955 968 × 2 = 1 + 0.111 111 116 709 335 647 911 936;
  • 45) 0.111 111 116 709 335 647 911 936 × 2 = 0 + 0.222 222 233 418 671 295 823 872;
  • 46) 0.222 222 233 418 671 295 823 872 × 2 = 0 + 0.444 444 466 837 342 591 647 744;
  • 47) 0.444 444 466 837 342 591 647 744 × 2 = 0 + 0.888 888 933 674 685 183 295 488;
  • 48) 0.888 888 933 674 685 183 295 488 × 2 = 1 + 0.777 777 867 349 370 366 590 976;
  • 49) 0.777 777 867 349 370 366 590 976 × 2 = 1 + 0.555 555 734 698 740 733 181 952;
  • 50) 0.555 555 734 698 740 733 181 952 × 2 = 1 + 0.111 111 469 397 481 466 363 904;
  • 51) 0.111 111 469 397 481 466 363 904 × 2 = 0 + 0.222 222 938 794 962 932 727 808;
  • 52) 0.222 222 938 794 962 932 727 808 × 2 = 0 + 0.444 445 877 589 925 865 455 616;
  • 53) 0.444 445 877 589 925 865 455 616 × 2 = 0 + 0.888 891 755 179 851 730 911 232;

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:


0.777 777 777 777 777 777 778 096(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 777 778 096(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:


24.777 777 777 777 777 777 778 096(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 24.777 777 777 777 777 777 778 096 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 24.777 777 777 777 777 777 778 04 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: The Hidden Requirement in Every Single Job Description. NotebookLM YouTube Video of 8.15 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