24.777 777 777 777 777 763 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 763 2(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
24.777 777 777 777 777 763 2(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: 0.777 777 777 777 777 763 2.

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) 0.777 777 777 777 777 763 2 × 2 = 1 + 0.555 555 555 555 555 526 4;
  • 2) 0.555 555 555 555 555 526 4 × 2 = 1 + 0.111 111 111 111 111 052 8;
  • 3) 0.111 111 111 111 111 052 8 × 2 = 0 + 0.222 222 222 222 222 105 6;
  • 4) 0.222 222 222 222 222 105 6 × 2 = 0 + 0.444 444 444 444 444 211 2;
  • 5) 0.444 444 444 444 444 211 2 × 2 = 0 + 0.888 888 888 888 888 422 4;
  • 6) 0.888 888 888 888 888 422 4 × 2 = 1 + 0.777 777 777 777 776 844 8;
  • 7) 0.777 777 777 777 776 844 8 × 2 = 1 + 0.555 555 555 555 553 689 6;
  • 8) 0.555 555 555 555 553 689 6 × 2 = 1 + 0.111 111 111 111 107 379 2;
  • 9) 0.111 111 111 111 107 379 2 × 2 = 0 + 0.222 222 222 222 214 758 4;
  • 10) 0.222 222 222 222 214 758 4 × 2 = 0 + 0.444 444 444 444 429 516 8;
  • 11) 0.444 444 444 444 429 516 8 × 2 = 0 + 0.888 888 888 888 859 033 6;
  • 12) 0.888 888 888 888 859 033 6 × 2 = 1 + 0.777 777 777 777 718 067 2;
  • 13) 0.777 777 777 777 718 067 2 × 2 = 1 + 0.555 555 555 555 436 134 4;
  • 14) 0.555 555 555 555 436 134 4 × 2 = 1 + 0.111 111 111 110 872 268 8;
  • 15) 0.111 111 111 110 872 268 8 × 2 = 0 + 0.222 222 222 221 744 537 6;
  • 16) 0.222 222 222 221 744 537 6 × 2 = 0 + 0.444 444 444 443 489 075 2;
  • 17) 0.444 444 444 443 489 075 2 × 2 = 0 + 0.888 888 888 886 978 150 4;
  • 18) 0.888 888 888 886 978 150 4 × 2 = 1 + 0.777 777 777 773 956 300 8;
  • 19) 0.777 777 777 773 956 300 8 × 2 = 1 + 0.555 555 555 547 912 601 6;
  • 20) 0.555 555 555 547 912 601 6 × 2 = 1 + 0.111 111 111 095 825 203 2;
  • 21) 0.111 111 111 095 825 203 2 × 2 = 0 + 0.222 222 222 191 650 406 4;
  • 22) 0.222 222 222 191 650 406 4 × 2 = 0 + 0.444 444 444 383 300 812 8;
  • 23) 0.444 444 444 383 300 812 8 × 2 = 0 + 0.888 888 888 766 601 625 6;
  • 24) 0.888 888 888 766 601 625 6 × 2 = 1 + 0.777 777 777 533 203 251 2;
  • 25) 0.777 777 777 533 203 251 2 × 2 = 1 + 0.555 555 555 066 406 502 4;
  • 26) 0.555 555 555 066 406 502 4 × 2 = 1 + 0.111 111 110 132 813 004 8;
  • 27) 0.111 111 110 132 813 004 8 × 2 = 0 + 0.222 222 220 265 626 009 6;
  • 28) 0.222 222 220 265 626 009 6 × 2 = 0 + 0.444 444 440 531 252 019 2;
  • 29) 0.444 444 440 531 252 019 2 × 2 = 0 + 0.888 888 881 062 504 038 4;
  • 30) 0.888 888 881 062 504 038 4 × 2 = 1 + 0.777 777 762 125 008 076 8;
  • 31) 0.777 777 762 125 008 076 8 × 2 = 1 + 0.555 555 524 250 016 153 6;
  • 32) 0.555 555 524 250 016 153 6 × 2 = 1 + 0.111 111 048 500 032 307 2;
  • 33) 0.111 111 048 500 032 307 2 × 2 = 0 + 0.222 222 097 000 064 614 4;
  • 34) 0.222 222 097 000 064 614 4 × 2 = 0 + 0.444 444 194 000 129 228 8;
  • 35) 0.444 444 194 000 129 228 8 × 2 = 0 + 0.888 888 388 000 258 457 6;
  • 36) 0.888 888 388 000 258 457 6 × 2 = 1 + 0.777 776 776 000 516 915 2;
  • 37) 0.777 776 776 000 516 915 2 × 2 = 1 + 0.555 553 552 001 033 830 4;
  • 38) 0.555 553 552 001 033 830 4 × 2 = 1 + 0.111 107 104 002 067 660 8;
  • 39) 0.111 107 104 002 067 660 8 × 2 = 0 + 0.222 214 208 004 135 321 6;
  • 40) 0.222 214 208 004 135 321 6 × 2 = 0 + 0.444 428 416 008 270 643 2;
  • 41) 0.444 428 416 008 270 643 2 × 2 = 0 + 0.888 856 832 016 541 286 4;
  • 42) 0.888 856 832 016 541 286 4 × 2 = 1 + 0.777 713 664 033 082 572 8;
  • 43) 0.777 713 664 033 082 572 8 × 2 = 1 + 0.555 427 328 066 165 145 6;
  • 44) 0.555 427 328 066 165 145 6 × 2 = 1 + 0.110 854 656 132 330 291 2;
  • 45) 0.110 854 656 132 330 291 2 × 2 = 0 + 0.221 709 312 264 660 582 4;
  • 46) 0.221 709 312 264 660 582 4 × 2 = 0 + 0.443 418 624 529 321 164 8;
  • 47) 0.443 418 624 529 321 164 8 × 2 = 0 + 0.886 837 249 058 642 329 6;
  • 48) 0.886 837 249 058 642 329 6 × 2 = 1 + 0.773 674 498 117 284 659 2;
  • 49) 0.773 674 498 117 284 659 2 × 2 = 1 + 0.547 348 996 234 569 318 4;
  • 50) 0.547 348 996 234 569 318 4 × 2 = 1 + 0.094 697 992 469 138 636 8;
  • 51) 0.094 697 992 469 138 636 8 × 2 = 0 + 0.189 395 984 938 277 273 6;
  • 52) 0.189 395 984 938 277 273 6 × 2 = 0 + 0.378 791 969 876 554 547 2;
  • 53) 0.378 791 969 876 554 547 2 × 2 = 0 + 0.757 583 939 753 109 094 4;

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:


0.777 777 777 777 777 763 2(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 763 2(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:


24.777 777 777 777 777 763 2(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 24.777 777 777 777 777 763 2 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

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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