24.777 777 777 777 892 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 892(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
24.777 777 777 777 892(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: 0.777 777 777 777 892.

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) 0.777 777 777 777 892 × 2 = 1 + 0.555 555 555 555 784;
  • 2) 0.555 555 555 555 784 × 2 = 1 + 0.111 111 111 111 568;
  • 3) 0.111 111 111 111 568 × 2 = 0 + 0.222 222 222 223 136;
  • 4) 0.222 222 222 223 136 × 2 = 0 + 0.444 444 444 446 272;
  • 5) 0.444 444 444 446 272 × 2 = 0 + 0.888 888 888 892 544;
  • 6) 0.888 888 888 892 544 × 2 = 1 + 0.777 777 777 785 088;
  • 7) 0.777 777 777 785 088 × 2 = 1 + 0.555 555 555 570 176;
  • 8) 0.555 555 555 570 176 × 2 = 1 + 0.111 111 111 140 352;
  • 9) 0.111 111 111 140 352 × 2 = 0 + 0.222 222 222 280 704;
  • 10) 0.222 222 222 280 704 × 2 = 0 + 0.444 444 444 561 408;
  • 11) 0.444 444 444 561 408 × 2 = 0 + 0.888 888 889 122 816;
  • 12) 0.888 888 889 122 816 × 2 = 1 + 0.777 777 778 245 632;
  • 13) 0.777 777 778 245 632 × 2 = 1 + 0.555 555 556 491 264;
  • 14) 0.555 555 556 491 264 × 2 = 1 + 0.111 111 112 982 528;
  • 15) 0.111 111 112 982 528 × 2 = 0 + 0.222 222 225 965 056;
  • 16) 0.222 222 225 965 056 × 2 = 0 + 0.444 444 451 930 112;
  • 17) 0.444 444 451 930 112 × 2 = 0 + 0.888 888 903 860 224;
  • 18) 0.888 888 903 860 224 × 2 = 1 + 0.777 777 807 720 448;
  • 19) 0.777 777 807 720 448 × 2 = 1 + 0.555 555 615 440 896;
  • 20) 0.555 555 615 440 896 × 2 = 1 + 0.111 111 230 881 792;
  • 21) 0.111 111 230 881 792 × 2 = 0 + 0.222 222 461 763 584;
  • 22) 0.222 222 461 763 584 × 2 = 0 + 0.444 444 923 527 168;
  • 23) 0.444 444 923 527 168 × 2 = 0 + 0.888 889 847 054 336;
  • 24) 0.888 889 847 054 336 × 2 = 1 + 0.777 779 694 108 672;
  • 25) 0.777 779 694 108 672 × 2 = 1 + 0.555 559 388 217 344;
  • 26) 0.555 559 388 217 344 × 2 = 1 + 0.111 118 776 434 688;
  • 27) 0.111 118 776 434 688 × 2 = 0 + 0.222 237 552 869 376;
  • 28) 0.222 237 552 869 376 × 2 = 0 + 0.444 475 105 738 752;
  • 29) 0.444 475 105 738 752 × 2 = 0 + 0.888 950 211 477 504;
  • 30) 0.888 950 211 477 504 × 2 = 1 + 0.777 900 422 955 008;
  • 31) 0.777 900 422 955 008 × 2 = 1 + 0.555 800 845 910 016;
  • 32) 0.555 800 845 910 016 × 2 = 1 + 0.111 601 691 820 032;
  • 33) 0.111 601 691 820 032 × 2 = 0 + 0.223 203 383 640 064;
  • 34) 0.223 203 383 640 064 × 2 = 0 + 0.446 406 767 280 128;
  • 35) 0.446 406 767 280 128 × 2 = 0 + 0.892 813 534 560 256;
  • 36) 0.892 813 534 560 256 × 2 = 1 + 0.785 627 069 120 512;
  • 37) 0.785 627 069 120 512 × 2 = 1 + 0.571 254 138 241 024;
  • 38) 0.571 254 138 241 024 × 2 = 1 + 0.142 508 276 482 048;
  • 39) 0.142 508 276 482 048 × 2 = 0 + 0.285 016 552 964 096;
  • 40) 0.285 016 552 964 096 × 2 = 0 + 0.570 033 105 928 192;
  • 41) 0.570 033 105 928 192 × 2 = 1 + 0.140 066 211 856 384;
  • 42) 0.140 066 211 856 384 × 2 = 0 + 0.280 132 423 712 768;
  • 43) 0.280 132 423 712 768 × 2 = 0 + 0.560 264 847 425 536;
  • 44) 0.560 264 847 425 536 × 2 = 1 + 0.120 529 694 851 072;
  • 45) 0.120 529 694 851 072 × 2 = 0 + 0.241 059 389 702 144;
  • 46) 0.241 059 389 702 144 × 2 = 0 + 0.482 118 779 404 288;
  • 47) 0.482 118 779 404 288 × 2 = 0 + 0.964 237 558 808 576;
  • 48) 0.964 237 558 808 576 × 2 = 1 + 0.928 475 117 617 152;
  • 49) 0.928 475 117 617 152 × 2 = 1 + 0.856 950 235 234 304;
  • 50) 0.856 950 235 234 304 × 2 = 1 + 0.713 900 470 468 608;
  • 51) 0.713 900 470 468 608 × 2 = 1 + 0.427 800 940 937 216;
  • 52) 0.427 800 940 937 216 × 2 = 0 + 0.855 601 881 874 432;
  • 53) 0.855 601 881 874 432 × 2 = 1 + 0.711 203 763 748 864;

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:


0.777 777 777 777 892(10) =

0.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1001 0001 1110 1(2)

5. Positive number before normalization:

24.777 777 777 777 892(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1001 0001 1110 1(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:


24.777 777 777 777 892(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1001 0001 1110 1(2) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1001 0001 1110 1(2) × 20 =

1.1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1001 0001 1110 1(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 1001 0001 1110 1


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 1001 0001 1 1101 =

1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1001 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 1001 0001


Decimal number 24.777 777 777 777 892 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 1001 0001

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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