24.777 777 777 777 780 17 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 780 17(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
24.777 777 777 777 780 17(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: 0.777 777 777 777 780 17.

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) 0.777 777 777 777 780 17 × 2 = 1 + 0.555 555 555 555 560 34;
  • 2) 0.555 555 555 555 560 34 × 2 = 1 + 0.111 111 111 111 120 68;
  • 3) 0.111 111 111 111 120 68 × 2 = 0 + 0.222 222 222 222 241 36;
  • 4) 0.222 222 222 222 241 36 × 2 = 0 + 0.444 444 444 444 482 72;
  • 5) 0.444 444 444 444 482 72 × 2 = 0 + 0.888 888 888 888 965 44;
  • 6) 0.888 888 888 888 965 44 × 2 = 1 + 0.777 777 777 777 930 88;
  • 7) 0.777 777 777 777 930 88 × 2 = 1 + 0.555 555 555 555 861 76;
  • 8) 0.555 555 555 555 861 76 × 2 = 1 + 0.111 111 111 111 723 52;
  • 9) 0.111 111 111 111 723 52 × 2 = 0 + 0.222 222 222 223 447 04;
  • 10) 0.222 222 222 223 447 04 × 2 = 0 + 0.444 444 444 446 894 08;
  • 11) 0.444 444 444 446 894 08 × 2 = 0 + 0.888 888 888 893 788 16;
  • 12) 0.888 888 888 893 788 16 × 2 = 1 + 0.777 777 777 787 576 32;
  • 13) 0.777 777 777 787 576 32 × 2 = 1 + 0.555 555 555 575 152 64;
  • 14) 0.555 555 555 575 152 64 × 2 = 1 + 0.111 111 111 150 305 28;
  • 15) 0.111 111 111 150 305 28 × 2 = 0 + 0.222 222 222 300 610 56;
  • 16) 0.222 222 222 300 610 56 × 2 = 0 + 0.444 444 444 601 221 12;
  • 17) 0.444 444 444 601 221 12 × 2 = 0 + 0.888 888 889 202 442 24;
  • 18) 0.888 888 889 202 442 24 × 2 = 1 + 0.777 777 778 404 884 48;
  • 19) 0.777 777 778 404 884 48 × 2 = 1 + 0.555 555 556 809 768 96;
  • 20) 0.555 555 556 809 768 96 × 2 = 1 + 0.111 111 113 619 537 92;
  • 21) 0.111 111 113 619 537 92 × 2 = 0 + 0.222 222 227 239 075 84;
  • 22) 0.222 222 227 239 075 84 × 2 = 0 + 0.444 444 454 478 151 68;
  • 23) 0.444 444 454 478 151 68 × 2 = 0 + 0.888 888 908 956 303 36;
  • 24) 0.888 888 908 956 303 36 × 2 = 1 + 0.777 777 817 912 606 72;
  • 25) 0.777 777 817 912 606 72 × 2 = 1 + 0.555 555 635 825 213 44;
  • 26) 0.555 555 635 825 213 44 × 2 = 1 + 0.111 111 271 650 426 88;
  • 27) 0.111 111 271 650 426 88 × 2 = 0 + 0.222 222 543 300 853 76;
  • 28) 0.222 222 543 300 853 76 × 2 = 0 + 0.444 445 086 601 707 52;
  • 29) 0.444 445 086 601 707 52 × 2 = 0 + 0.888 890 173 203 415 04;
  • 30) 0.888 890 173 203 415 04 × 2 = 1 + 0.777 780 346 406 830 08;
  • 31) 0.777 780 346 406 830 08 × 2 = 1 + 0.555 560 692 813 660 16;
  • 32) 0.555 560 692 813 660 16 × 2 = 1 + 0.111 121 385 627 320 32;
  • 33) 0.111 121 385 627 320 32 × 2 = 0 + 0.222 242 771 254 640 64;
  • 34) 0.222 242 771 254 640 64 × 2 = 0 + 0.444 485 542 509 281 28;
  • 35) 0.444 485 542 509 281 28 × 2 = 0 + 0.888 971 085 018 562 56;
  • 36) 0.888 971 085 018 562 56 × 2 = 1 + 0.777 942 170 037 125 12;
  • 37) 0.777 942 170 037 125 12 × 2 = 1 + 0.555 884 340 074 250 24;
  • 38) 0.555 884 340 074 250 24 × 2 = 1 + 0.111 768 680 148 500 48;
  • 39) 0.111 768 680 148 500 48 × 2 = 0 + 0.223 537 360 297 000 96;
  • 40) 0.223 537 360 297 000 96 × 2 = 0 + 0.447 074 720 594 001 92;
  • 41) 0.447 074 720 594 001 92 × 2 = 0 + 0.894 149 441 188 003 84;
  • 42) 0.894 149 441 188 003 84 × 2 = 1 + 0.788 298 882 376 007 68;
  • 43) 0.788 298 882 376 007 68 × 2 = 1 + 0.576 597 764 752 015 36;
  • 44) 0.576 597 764 752 015 36 × 2 = 1 + 0.153 195 529 504 030 72;
  • 45) 0.153 195 529 504 030 72 × 2 = 0 + 0.306 391 059 008 061 44;
  • 46) 0.306 391 059 008 061 44 × 2 = 0 + 0.612 782 118 016 122 88;
  • 47) 0.612 782 118 016 122 88 × 2 = 1 + 0.225 564 236 032 245 76;
  • 48) 0.225 564 236 032 245 76 × 2 = 0 + 0.451 128 472 064 491 52;
  • 49) 0.451 128 472 064 491 52 × 2 = 0 + 0.902 256 944 128 983 04;
  • 50) 0.902 256 944 128 983 04 × 2 = 1 + 0.804 513 888 257 966 08;
  • 51) 0.804 513 888 257 966 08 × 2 = 1 + 0.609 027 776 515 932 16;
  • 52) 0.609 027 776 515 932 16 × 2 = 1 + 0.218 055 553 031 864 32;
  • 53) 0.218 055 553 031 864 32 × 2 = 0 + 0.436 111 106 063 728 64;

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:


0.777 777 777 777 780 17(10) =

0.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0010 0111 0(2)

5. Positive number before normalization:

24.777 777 777 777 780 17(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0010 0111 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:


24.777 777 777 777 780 17(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0010 0111 0(2) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0010 0111 0(2) × 20 =

1.1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0010 0111 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 0010 0111 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 0010 0 1110 =

1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0010


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 0010


Decimal number 24.777 777 777 777 780 17 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 0010

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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