24.777 777 777 777 777 777 787 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 777 787 8(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
24.777 777 777 777 777 777 787 8(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: 0.777 777 777 777 777 777 787 8.

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) 0.777 777 777 777 777 777 787 8 × 2 = 1 + 0.555 555 555 555 555 555 575 6;
  • 2) 0.555 555 555 555 555 555 575 6 × 2 = 1 + 0.111 111 111 111 111 111 151 2;
  • 3) 0.111 111 111 111 111 111 151 2 × 2 = 0 + 0.222 222 222 222 222 222 302 4;
  • 4) 0.222 222 222 222 222 222 302 4 × 2 = 0 + 0.444 444 444 444 444 444 604 8;
  • 5) 0.444 444 444 444 444 444 604 8 × 2 = 0 + 0.888 888 888 888 888 889 209 6;
  • 6) 0.888 888 888 888 888 889 209 6 × 2 = 1 + 0.777 777 777 777 777 778 419 2;
  • 7) 0.777 777 777 777 777 778 419 2 × 2 = 1 + 0.555 555 555 555 555 556 838 4;
  • 8) 0.555 555 555 555 555 556 838 4 × 2 = 1 + 0.111 111 111 111 111 113 676 8;
  • 9) 0.111 111 111 111 111 113 676 8 × 2 = 0 + 0.222 222 222 222 222 227 353 6;
  • 10) 0.222 222 222 222 222 227 353 6 × 2 = 0 + 0.444 444 444 444 444 454 707 2;
  • 11) 0.444 444 444 444 444 454 707 2 × 2 = 0 + 0.888 888 888 888 888 909 414 4;
  • 12) 0.888 888 888 888 888 909 414 4 × 2 = 1 + 0.777 777 777 777 777 818 828 8;
  • 13) 0.777 777 777 777 777 818 828 8 × 2 = 1 + 0.555 555 555 555 555 637 657 6;
  • 14) 0.555 555 555 555 555 637 657 6 × 2 = 1 + 0.111 111 111 111 111 275 315 2;
  • 15) 0.111 111 111 111 111 275 315 2 × 2 = 0 + 0.222 222 222 222 222 550 630 4;
  • 16) 0.222 222 222 222 222 550 630 4 × 2 = 0 + 0.444 444 444 444 445 101 260 8;
  • 17) 0.444 444 444 444 445 101 260 8 × 2 = 0 + 0.888 888 888 888 890 202 521 6;
  • 18) 0.888 888 888 888 890 202 521 6 × 2 = 1 + 0.777 777 777 777 780 405 043 2;
  • 19) 0.777 777 777 777 780 405 043 2 × 2 = 1 + 0.555 555 555 555 560 810 086 4;
  • 20) 0.555 555 555 555 560 810 086 4 × 2 = 1 + 0.111 111 111 111 121 620 172 8;
  • 21) 0.111 111 111 111 121 620 172 8 × 2 = 0 + 0.222 222 222 222 243 240 345 6;
  • 22) 0.222 222 222 222 243 240 345 6 × 2 = 0 + 0.444 444 444 444 486 480 691 2;
  • 23) 0.444 444 444 444 486 480 691 2 × 2 = 0 + 0.888 888 888 888 972 961 382 4;
  • 24) 0.888 888 888 888 972 961 382 4 × 2 = 1 + 0.777 777 777 777 945 922 764 8;
  • 25) 0.777 777 777 777 945 922 764 8 × 2 = 1 + 0.555 555 555 555 891 845 529 6;
  • 26) 0.555 555 555 555 891 845 529 6 × 2 = 1 + 0.111 111 111 111 783 691 059 2;
  • 27) 0.111 111 111 111 783 691 059 2 × 2 = 0 + 0.222 222 222 223 567 382 118 4;
  • 28) 0.222 222 222 223 567 382 118 4 × 2 = 0 + 0.444 444 444 447 134 764 236 8;
  • 29) 0.444 444 444 447 134 764 236 8 × 2 = 0 + 0.888 888 888 894 269 528 473 6;
  • 30) 0.888 888 888 894 269 528 473 6 × 2 = 1 + 0.777 777 777 788 539 056 947 2;
  • 31) 0.777 777 777 788 539 056 947 2 × 2 = 1 + 0.555 555 555 577 078 113 894 4;
  • 32) 0.555 555 555 577 078 113 894 4 × 2 = 1 + 0.111 111 111 154 156 227 788 8;
  • 33) 0.111 111 111 154 156 227 788 8 × 2 = 0 + 0.222 222 222 308 312 455 577 6;
  • 34) 0.222 222 222 308 312 455 577 6 × 2 = 0 + 0.444 444 444 616 624 911 155 2;
  • 35) 0.444 444 444 616 624 911 155 2 × 2 = 0 + 0.888 888 889 233 249 822 310 4;
  • 36) 0.888 888 889 233 249 822 310 4 × 2 = 1 + 0.777 777 778 466 499 644 620 8;
  • 37) 0.777 777 778 466 499 644 620 8 × 2 = 1 + 0.555 555 556 932 999 289 241 6;
  • 38) 0.555 555 556 932 999 289 241 6 × 2 = 1 + 0.111 111 113 865 998 578 483 2;
  • 39) 0.111 111 113 865 998 578 483 2 × 2 = 0 + 0.222 222 227 731 997 156 966 4;
  • 40) 0.222 222 227 731 997 156 966 4 × 2 = 0 + 0.444 444 455 463 994 313 932 8;
  • 41) 0.444 444 455 463 994 313 932 8 × 2 = 0 + 0.888 888 910 927 988 627 865 6;
  • 42) 0.888 888 910 927 988 627 865 6 × 2 = 1 + 0.777 777 821 855 977 255 731 2;
  • 43) 0.777 777 821 855 977 255 731 2 × 2 = 1 + 0.555 555 643 711 954 511 462 4;
  • 44) 0.555 555 643 711 954 511 462 4 × 2 = 1 + 0.111 111 287 423 909 022 924 8;
  • 45) 0.111 111 287 423 909 022 924 8 × 2 = 0 + 0.222 222 574 847 818 045 849 6;
  • 46) 0.222 222 574 847 818 045 849 6 × 2 = 0 + 0.444 445 149 695 636 091 699 2;
  • 47) 0.444 445 149 695 636 091 699 2 × 2 = 0 + 0.888 890 299 391 272 183 398 4;
  • 48) 0.888 890 299 391 272 183 398 4 × 2 = 1 + 0.777 780 598 782 544 366 796 8;
  • 49) 0.777 780 598 782 544 366 796 8 × 2 = 1 + 0.555 561 197 565 088 733 593 6;
  • 50) 0.555 561 197 565 088 733 593 6 × 2 = 1 + 0.111 122 395 130 177 467 187 2;
  • 51) 0.111 122 395 130 177 467 187 2 × 2 = 0 + 0.222 244 790 260 354 934 374 4;
  • 52) 0.222 244 790 260 354 934 374 4 × 2 = 0 + 0.444 489 580 520 709 868 748 8;
  • 53) 0.444 489 580 520 709 868 748 8 × 2 = 0 + 0.888 979 161 041 419 737 497 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:


0.777 777 777 777 777 777 787 8(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 777 787 8(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:


24.777 777 777 777 777 777 787 8(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 24.777 777 777 777 777 777 787 8 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 24.777 777 777 777 777 777 780 4 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