24.777 777 777 777 777 777 777 926 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 777 777 926(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
24.777 777 777 777 777 777 777 926(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: 0.777 777 777 777 777 777 777 926.

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) 0.777 777 777 777 777 777 777 926 × 2 = 1 + 0.555 555 555 555 555 555 555 852;
  • 2) 0.555 555 555 555 555 555 555 852 × 2 = 1 + 0.111 111 111 111 111 111 111 704;
  • 3) 0.111 111 111 111 111 111 111 704 × 2 = 0 + 0.222 222 222 222 222 222 223 408;
  • 4) 0.222 222 222 222 222 222 223 408 × 2 = 0 + 0.444 444 444 444 444 444 446 816;
  • 5) 0.444 444 444 444 444 444 446 816 × 2 = 0 + 0.888 888 888 888 888 888 893 632;
  • 6) 0.888 888 888 888 888 888 893 632 × 2 = 1 + 0.777 777 777 777 777 777 787 264;
  • 7) 0.777 777 777 777 777 777 787 264 × 2 = 1 + 0.555 555 555 555 555 555 574 528;
  • 8) 0.555 555 555 555 555 555 574 528 × 2 = 1 + 0.111 111 111 111 111 111 149 056;
  • 9) 0.111 111 111 111 111 111 149 056 × 2 = 0 + 0.222 222 222 222 222 222 298 112;
  • 10) 0.222 222 222 222 222 222 298 112 × 2 = 0 + 0.444 444 444 444 444 444 596 224;
  • 11) 0.444 444 444 444 444 444 596 224 × 2 = 0 + 0.888 888 888 888 888 889 192 448;
  • 12) 0.888 888 888 888 888 889 192 448 × 2 = 1 + 0.777 777 777 777 777 778 384 896;
  • 13) 0.777 777 777 777 777 778 384 896 × 2 = 1 + 0.555 555 555 555 555 556 769 792;
  • 14) 0.555 555 555 555 555 556 769 792 × 2 = 1 + 0.111 111 111 111 111 113 539 584;
  • 15) 0.111 111 111 111 111 113 539 584 × 2 = 0 + 0.222 222 222 222 222 227 079 168;
  • 16) 0.222 222 222 222 222 227 079 168 × 2 = 0 + 0.444 444 444 444 444 454 158 336;
  • 17) 0.444 444 444 444 444 454 158 336 × 2 = 0 + 0.888 888 888 888 888 908 316 672;
  • 18) 0.888 888 888 888 888 908 316 672 × 2 = 1 + 0.777 777 777 777 777 816 633 344;
  • 19) 0.777 777 777 777 777 816 633 344 × 2 = 1 + 0.555 555 555 555 555 633 266 688;
  • 20) 0.555 555 555 555 555 633 266 688 × 2 = 1 + 0.111 111 111 111 111 266 533 376;
  • 21) 0.111 111 111 111 111 266 533 376 × 2 = 0 + 0.222 222 222 222 222 533 066 752;
  • 22) 0.222 222 222 222 222 533 066 752 × 2 = 0 + 0.444 444 444 444 445 066 133 504;
  • 23) 0.444 444 444 444 445 066 133 504 × 2 = 0 + 0.888 888 888 888 890 132 267 008;
  • 24) 0.888 888 888 888 890 132 267 008 × 2 = 1 + 0.777 777 777 777 780 264 534 016;
  • 25) 0.777 777 777 777 780 264 534 016 × 2 = 1 + 0.555 555 555 555 560 529 068 032;
  • 26) 0.555 555 555 555 560 529 068 032 × 2 = 1 + 0.111 111 111 111 121 058 136 064;
  • 27) 0.111 111 111 111 121 058 136 064 × 2 = 0 + 0.222 222 222 222 242 116 272 128;
  • 28) 0.222 222 222 222 242 116 272 128 × 2 = 0 + 0.444 444 444 444 484 232 544 256;
  • 29) 0.444 444 444 444 484 232 544 256 × 2 = 0 + 0.888 888 888 888 968 465 088 512;
  • 30) 0.888 888 888 888 968 465 088 512 × 2 = 1 + 0.777 777 777 777 936 930 177 024;
  • 31) 0.777 777 777 777 936 930 177 024 × 2 = 1 + 0.555 555 555 555 873 860 354 048;
  • 32) 0.555 555 555 555 873 860 354 048 × 2 = 1 + 0.111 111 111 111 747 720 708 096;
  • 33) 0.111 111 111 111 747 720 708 096 × 2 = 0 + 0.222 222 222 223 495 441 416 192;
  • 34) 0.222 222 222 223 495 441 416 192 × 2 = 0 + 0.444 444 444 446 990 882 832 384;
  • 35) 0.444 444 444 446 990 882 832 384 × 2 = 0 + 0.888 888 888 893 981 765 664 768;
  • 36) 0.888 888 888 893 981 765 664 768 × 2 = 1 + 0.777 777 777 787 963 531 329 536;
  • 37) 0.777 777 777 787 963 531 329 536 × 2 = 1 + 0.555 555 555 575 927 062 659 072;
  • 38) 0.555 555 555 575 927 062 659 072 × 2 = 1 + 0.111 111 111 151 854 125 318 144;
  • 39) 0.111 111 111 151 854 125 318 144 × 2 = 0 + 0.222 222 222 303 708 250 636 288;
  • 40) 0.222 222 222 303 708 250 636 288 × 2 = 0 + 0.444 444 444 607 416 501 272 576;
  • 41) 0.444 444 444 607 416 501 272 576 × 2 = 0 + 0.888 888 889 214 833 002 545 152;
  • 42) 0.888 888 889 214 833 002 545 152 × 2 = 1 + 0.777 777 778 429 666 005 090 304;
  • 43) 0.777 777 778 429 666 005 090 304 × 2 = 1 + 0.555 555 556 859 332 010 180 608;
  • 44) 0.555 555 556 859 332 010 180 608 × 2 = 1 + 0.111 111 113 718 664 020 361 216;
  • 45) 0.111 111 113 718 664 020 361 216 × 2 = 0 + 0.222 222 227 437 328 040 722 432;
  • 46) 0.222 222 227 437 328 040 722 432 × 2 = 0 + 0.444 444 454 874 656 081 444 864;
  • 47) 0.444 444 454 874 656 081 444 864 × 2 = 0 + 0.888 888 909 749 312 162 889 728;
  • 48) 0.888 888 909 749 312 162 889 728 × 2 = 1 + 0.777 777 819 498 624 325 779 456;
  • 49) 0.777 777 819 498 624 325 779 456 × 2 = 1 + 0.555 555 638 997 248 651 558 912;
  • 50) 0.555 555 638 997 248 651 558 912 × 2 = 1 + 0.111 111 277 994 497 303 117 824;
  • 51) 0.111 111 277 994 497 303 117 824 × 2 = 0 + 0.222 222 555 988 994 606 235 648;
  • 52) 0.222 222 555 988 994 606 235 648 × 2 = 0 + 0.444 445 111 977 989 212 471 296;
  • 53) 0.444 445 111 977 989 212 471 296 × 2 = 0 + 0.888 890 223 955 978 424 942 592;

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:


0.777 777 777 777 777 777 777 926(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 777 777 926(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:


24.777 777 777 777 777 777 777 926(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 24.777 777 777 777 777 777 777 926 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 24.777 777 777 777 777 777 778 016 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 Attention Economy - The Battlefield For Your Focus. NotebookLM YouTube Video of 7.50 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