24.777 777 777 777 777 781 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 781 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
24.777 777 777 777 777 781 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: 0.777 777 777 777 777 781 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) 0.777 777 777 777 777 781 8 × 2 = 1 + 0.555 555 555 555 555 563 6;
  • 2) 0.555 555 555 555 555 563 6 × 2 = 1 + 0.111 111 111 111 111 127 2;
  • 3) 0.111 111 111 111 111 127 2 × 2 = 0 + 0.222 222 222 222 222 254 4;
  • 4) 0.222 222 222 222 222 254 4 × 2 = 0 + 0.444 444 444 444 444 508 8;
  • 5) 0.444 444 444 444 444 508 8 × 2 = 0 + 0.888 888 888 888 889 017 6;
  • 6) 0.888 888 888 888 889 017 6 × 2 = 1 + 0.777 777 777 777 778 035 2;
  • 7) 0.777 777 777 777 778 035 2 × 2 = 1 + 0.555 555 555 555 556 070 4;
  • 8) 0.555 555 555 555 556 070 4 × 2 = 1 + 0.111 111 111 111 112 140 8;
  • 9) 0.111 111 111 111 112 140 8 × 2 = 0 + 0.222 222 222 222 224 281 6;
  • 10) 0.222 222 222 222 224 281 6 × 2 = 0 + 0.444 444 444 444 448 563 2;
  • 11) 0.444 444 444 444 448 563 2 × 2 = 0 + 0.888 888 888 888 897 126 4;
  • 12) 0.888 888 888 888 897 126 4 × 2 = 1 + 0.777 777 777 777 794 252 8;
  • 13) 0.777 777 777 777 794 252 8 × 2 = 1 + 0.555 555 555 555 588 505 6;
  • 14) 0.555 555 555 555 588 505 6 × 2 = 1 + 0.111 111 111 111 177 011 2;
  • 15) 0.111 111 111 111 177 011 2 × 2 = 0 + 0.222 222 222 222 354 022 4;
  • 16) 0.222 222 222 222 354 022 4 × 2 = 0 + 0.444 444 444 444 708 044 8;
  • 17) 0.444 444 444 444 708 044 8 × 2 = 0 + 0.888 888 888 889 416 089 6;
  • 18) 0.888 888 888 889 416 089 6 × 2 = 1 + 0.777 777 777 778 832 179 2;
  • 19) 0.777 777 777 778 832 179 2 × 2 = 1 + 0.555 555 555 557 664 358 4;
  • 20) 0.555 555 555 557 664 358 4 × 2 = 1 + 0.111 111 111 115 328 716 8;
  • 21) 0.111 111 111 115 328 716 8 × 2 = 0 + 0.222 222 222 230 657 433 6;
  • 22) 0.222 222 222 230 657 433 6 × 2 = 0 + 0.444 444 444 461 314 867 2;
  • 23) 0.444 444 444 461 314 867 2 × 2 = 0 + 0.888 888 888 922 629 734 4;
  • 24) 0.888 888 888 922 629 734 4 × 2 = 1 + 0.777 777 777 845 259 468 8;
  • 25) 0.777 777 777 845 259 468 8 × 2 = 1 + 0.555 555 555 690 518 937 6;
  • 26) 0.555 555 555 690 518 937 6 × 2 = 1 + 0.111 111 111 381 037 875 2;
  • 27) 0.111 111 111 381 037 875 2 × 2 = 0 + 0.222 222 222 762 075 750 4;
  • 28) 0.222 222 222 762 075 750 4 × 2 = 0 + 0.444 444 445 524 151 500 8;
  • 29) 0.444 444 445 524 151 500 8 × 2 = 0 + 0.888 888 891 048 303 001 6;
  • 30) 0.888 888 891 048 303 001 6 × 2 = 1 + 0.777 777 782 096 606 003 2;
  • 31) 0.777 777 782 096 606 003 2 × 2 = 1 + 0.555 555 564 193 212 006 4;
  • 32) 0.555 555 564 193 212 006 4 × 2 = 1 + 0.111 111 128 386 424 012 8;
  • 33) 0.111 111 128 386 424 012 8 × 2 = 0 + 0.222 222 256 772 848 025 6;
  • 34) 0.222 222 256 772 848 025 6 × 2 = 0 + 0.444 444 513 545 696 051 2;
  • 35) 0.444 444 513 545 696 051 2 × 2 = 0 + 0.888 889 027 091 392 102 4;
  • 36) 0.888 889 027 091 392 102 4 × 2 = 1 + 0.777 778 054 182 784 204 8;
  • 37) 0.777 778 054 182 784 204 8 × 2 = 1 + 0.555 556 108 365 568 409 6;
  • 38) 0.555 556 108 365 568 409 6 × 2 = 1 + 0.111 112 216 731 136 819 2;
  • 39) 0.111 112 216 731 136 819 2 × 2 = 0 + 0.222 224 433 462 273 638 4;
  • 40) 0.222 224 433 462 273 638 4 × 2 = 0 + 0.444 448 866 924 547 276 8;
  • 41) 0.444 448 866 924 547 276 8 × 2 = 0 + 0.888 897 733 849 094 553 6;
  • 42) 0.888 897 733 849 094 553 6 × 2 = 1 + 0.777 795 467 698 189 107 2;
  • 43) 0.777 795 467 698 189 107 2 × 2 = 1 + 0.555 590 935 396 378 214 4;
  • 44) 0.555 590 935 396 378 214 4 × 2 = 1 + 0.111 181 870 792 756 428 8;
  • 45) 0.111 181 870 792 756 428 8 × 2 = 0 + 0.222 363 741 585 512 857 6;
  • 46) 0.222 363 741 585 512 857 6 × 2 = 0 + 0.444 727 483 171 025 715 2;
  • 47) 0.444 727 483 171 025 715 2 × 2 = 0 + 0.889 454 966 342 051 430 4;
  • 48) 0.889 454 966 342 051 430 4 × 2 = 1 + 0.778 909 932 684 102 860 8;
  • 49) 0.778 909 932 684 102 860 8 × 2 = 1 + 0.557 819 865 368 205 721 6;
  • 50) 0.557 819 865 368 205 721 6 × 2 = 1 + 0.115 639 730 736 411 443 2;
  • 51) 0.115 639 730 736 411 443 2 × 2 = 0 + 0.231 279 461 472 822 886 4;
  • 52) 0.231 279 461 472 822 886 4 × 2 = 0 + 0.462 558 922 945 645 772 8;
  • 53) 0.462 558 922 945 645 772 8 × 2 = 0 + 0.925 117 845 891 291 545 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:


0.777 777 777 777 777 781 8(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 781 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:


24.777 777 777 777 777 781 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 24.777 777 777 777 777 781 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

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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