24.777 777 777 777 777 777 802 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 777 802 9(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
24.777 777 777 777 777 777 802 9(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: 0.777 777 777 777 777 777 802 9.

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) 0.777 777 777 777 777 777 802 9 × 2 = 1 + 0.555 555 555 555 555 555 605 8;
  • 2) 0.555 555 555 555 555 555 605 8 × 2 = 1 + 0.111 111 111 111 111 111 211 6;
  • 3) 0.111 111 111 111 111 111 211 6 × 2 = 0 + 0.222 222 222 222 222 222 423 2;
  • 4) 0.222 222 222 222 222 222 423 2 × 2 = 0 + 0.444 444 444 444 444 444 846 4;
  • 5) 0.444 444 444 444 444 444 846 4 × 2 = 0 + 0.888 888 888 888 888 889 692 8;
  • 6) 0.888 888 888 888 888 889 692 8 × 2 = 1 + 0.777 777 777 777 777 779 385 6;
  • 7) 0.777 777 777 777 777 779 385 6 × 2 = 1 + 0.555 555 555 555 555 558 771 2;
  • 8) 0.555 555 555 555 555 558 771 2 × 2 = 1 + 0.111 111 111 111 111 117 542 4;
  • 9) 0.111 111 111 111 111 117 542 4 × 2 = 0 + 0.222 222 222 222 222 235 084 8;
  • 10) 0.222 222 222 222 222 235 084 8 × 2 = 0 + 0.444 444 444 444 444 470 169 6;
  • 11) 0.444 444 444 444 444 470 169 6 × 2 = 0 + 0.888 888 888 888 888 940 339 2;
  • 12) 0.888 888 888 888 888 940 339 2 × 2 = 1 + 0.777 777 777 777 777 880 678 4;
  • 13) 0.777 777 777 777 777 880 678 4 × 2 = 1 + 0.555 555 555 555 555 761 356 8;
  • 14) 0.555 555 555 555 555 761 356 8 × 2 = 1 + 0.111 111 111 111 111 522 713 6;
  • 15) 0.111 111 111 111 111 522 713 6 × 2 = 0 + 0.222 222 222 222 223 045 427 2;
  • 16) 0.222 222 222 222 223 045 427 2 × 2 = 0 + 0.444 444 444 444 446 090 854 4;
  • 17) 0.444 444 444 444 446 090 854 4 × 2 = 0 + 0.888 888 888 888 892 181 708 8;
  • 18) 0.888 888 888 888 892 181 708 8 × 2 = 1 + 0.777 777 777 777 784 363 417 6;
  • 19) 0.777 777 777 777 784 363 417 6 × 2 = 1 + 0.555 555 555 555 568 726 835 2;
  • 20) 0.555 555 555 555 568 726 835 2 × 2 = 1 + 0.111 111 111 111 137 453 670 4;
  • 21) 0.111 111 111 111 137 453 670 4 × 2 = 0 + 0.222 222 222 222 274 907 340 8;
  • 22) 0.222 222 222 222 274 907 340 8 × 2 = 0 + 0.444 444 444 444 549 814 681 6;
  • 23) 0.444 444 444 444 549 814 681 6 × 2 = 0 + 0.888 888 888 889 099 629 363 2;
  • 24) 0.888 888 888 889 099 629 363 2 × 2 = 1 + 0.777 777 777 778 199 258 726 4;
  • 25) 0.777 777 777 778 199 258 726 4 × 2 = 1 + 0.555 555 555 556 398 517 452 8;
  • 26) 0.555 555 555 556 398 517 452 8 × 2 = 1 + 0.111 111 111 112 797 034 905 6;
  • 27) 0.111 111 111 112 797 034 905 6 × 2 = 0 + 0.222 222 222 225 594 069 811 2;
  • 28) 0.222 222 222 225 594 069 811 2 × 2 = 0 + 0.444 444 444 451 188 139 622 4;
  • 29) 0.444 444 444 451 188 139 622 4 × 2 = 0 + 0.888 888 888 902 376 279 244 8;
  • 30) 0.888 888 888 902 376 279 244 8 × 2 = 1 + 0.777 777 777 804 752 558 489 6;
  • 31) 0.777 777 777 804 752 558 489 6 × 2 = 1 + 0.555 555 555 609 505 116 979 2;
  • 32) 0.555 555 555 609 505 116 979 2 × 2 = 1 + 0.111 111 111 219 010 233 958 4;
  • 33) 0.111 111 111 219 010 233 958 4 × 2 = 0 + 0.222 222 222 438 020 467 916 8;
  • 34) 0.222 222 222 438 020 467 916 8 × 2 = 0 + 0.444 444 444 876 040 935 833 6;
  • 35) 0.444 444 444 876 040 935 833 6 × 2 = 0 + 0.888 888 889 752 081 871 667 2;
  • 36) 0.888 888 889 752 081 871 667 2 × 2 = 1 + 0.777 777 779 504 163 743 334 4;
  • 37) 0.777 777 779 504 163 743 334 4 × 2 = 1 + 0.555 555 559 008 327 486 668 8;
  • 38) 0.555 555 559 008 327 486 668 8 × 2 = 1 + 0.111 111 118 016 654 973 337 6;
  • 39) 0.111 111 118 016 654 973 337 6 × 2 = 0 + 0.222 222 236 033 309 946 675 2;
  • 40) 0.222 222 236 033 309 946 675 2 × 2 = 0 + 0.444 444 472 066 619 893 350 4;
  • 41) 0.444 444 472 066 619 893 350 4 × 2 = 0 + 0.888 888 944 133 239 786 700 8;
  • 42) 0.888 888 944 133 239 786 700 8 × 2 = 1 + 0.777 777 888 266 479 573 401 6;
  • 43) 0.777 777 888 266 479 573 401 6 × 2 = 1 + 0.555 555 776 532 959 146 803 2;
  • 44) 0.555 555 776 532 959 146 803 2 × 2 = 1 + 0.111 111 553 065 918 293 606 4;
  • 45) 0.111 111 553 065 918 293 606 4 × 2 = 0 + 0.222 223 106 131 836 587 212 8;
  • 46) 0.222 223 106 131 836 587 212 8 × 2 = 0 + 0.444 446 212 263 673 174 425 6;
  • 47) 0.444 446 212 263 673 174 425 6 × 2 = 0 + 0.888 892 424 527 346 348 851 2;
  • 48) 0.888 892 424 527 346 348 851 2 × 2 = 1 + 0.777 784 849 054 692 697 702 4;
  • 49) 0.777 784 849 054 692 697 702 4 × 2 = 1 + 0.555 569 698 109 385 395 404 8;
  • 50) 0.555 569 698 109 385 395 404 8 × 2 = 1 + 0.111 139 396 218 770 790 809 6;
  • 51) 0.111 139 396 218 770 790 809 6 × 2 = 0 + 0.222 278 792 437 541 581 619 2;
  • 52) 0.222 278 792 437 541 581 619 2 × 2 = 0 + 0.444 557 584 875 083 163 238 4;
  • 53) 0.444 557 584 875 083 163 238 4 × 2 = 0 + 0.889 115 169 750 166 326 476 8;

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:


0.777 777 777 777 777 777 802 9(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 777 802 9(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:


24.777 777 777 777 777 777 802 9(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 24.777 777 777 777 777 777 802 9 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 24.777 777 777 777 777 777 804 6 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: Your greatest rival, your most powerful ally. NotebookLM YouTube Video of 7.54 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