24.777 777 777 777 777 744 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 744 1(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
24.777 777 777 777 777 744 1(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: 0.777 777 777 777 777 744 1.

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) 0.777 777 777 777 777 744 1 × 2 = 1 + 0.555 555 555 555 555 488 2;
  • 2) 0.555 555 555 555 555 488 2 × 2 = 1 + 0.111 111 111 111 110 976 4;
  • 3) 0.111 111 111 111 110 976 4 × 2 = 0 + 0.222 222 222 222 221 952 8;
  • 4) 0.222 222 222 222 221 952 8 × 2 = 0 + 0.444 444 444 444 443 905 6;
  • 5) 0.444 444 444 444 443 905 6 × 2 = 0 + 0.888 888 888 888 887 811 2;
  • 6) 0.888 888 888 888 887 811 2 × 2 = 1 + 0.777 777 777 777 775 622 4;
  • 7) 0.777 777 777 777 775 622 4 × 2 = 1 + 0.555 555 555 555 551 244 8;
  • 8) 0.555 555 555 555 551 244 8 × 2 = 1 + 0.111 111 111 111 102 489 6;
  • 9) 0.111 111 111 111 102 489 6 × 2 = 0 + 0.222 222 222 222 204 979 2;
  • 10) 0.222 222 222 222 204 979 2 × 2 = 0 + 0.444 444 444 444 409 958 4;
  • 11) 0.444 444 444 444 409 958 4 × 2 = 0 + 0.888 888 888 888 819 916 8;
  • 12) 0.888 888 888 888 819 916 8 × 2 = 1 + 0.777 777 777 777 639 833 6;
  • 13) 0.777 777 777 777 639 833 6 × 2 = 1 + 0.555 555 555 555 279 667 2;
  • 14) 0.555 555 555 555 279 667 2 × 2 = 1 + 0.111 111 111 110 559 334 4;
  • 15) 0.111 111 111 110 559 334 4 × 2 = 0 + 0.222 222 222 221 118 668 8;
  • 16) 0.222 222 222 221 118 668 8 × 2 = 0 + 0.444 444 444 442 237 337 6;
  • 17) 0.444 444 444 442 237 337 6 × 2 = 0 + 0.888 888 888 884 474 675 2;
  • 18) 0.888 888 888 884 474 675 2 × 2 = 1 + 0.777 777 777 768 949 350 4;
  • 19) 0.777 777 777 768 949 350 4 × 2 = 1 + 0.555 555 555 537 898 700 8;
  • 20) 0.555 555 555 537 898 700 8 × 2 = 1 + 0.111 111 111 075 797 401 6;
  • 21) 0.111 111 111 075 797 401 6 × 2 = 0 + 0.222 222 222 151 594 803 2;
  • 22) 0.222 222 222 151 594 803 2 × 2 = 0 + 0.444 444 444 303 189 606 4;
  • 23) 0.444 444 444 303 189 606 4 × 2 = 0 + 0.888 888 888 606 379 212 8;
  • 24) 0.888 888 888 606 379 212 8 × 2 = 1 + 0.777 777 777 212 758 425 6;
  • 25) 0.777 777 777 212 758 425 6 × 2 = 1 + 0.555 555 554 425 516 851 2;
  • 26) 0.555 555 554 425 516 851 2 × 2 = 1 + 0.111 111 108 851 033 702 4;
  • 27) 0.111 111 108 851 033 702 4 × 2 = 0 + 0.222 222 217 702 067 404 8;
  • 28) 0.222 222 217 702 067 404 8 × 2 = 0 + 0.444 444 435 404 134 809 6;
  • 29) 0.444 444 435 404 134 809 6 × 2 = 0 + 0.888 888 870 808 269 619 2;
  • 30) 0.888 888 870 808 269 619 2 × 2 = 1 + 0.777 777 741 616 539 238 4;
  • 31) 0.777 777 741 616 539 238 4 × 2 = 1 + 0.555 555 483 233 078 476 8;
  • 32) 0.555 555 483 233 078 476 8 × 2 = 1 + 0.111 110 966 466 156 953 6;
  • 33) 0.111 110 966 466 156 953 6 × 2 = 0 + 0.222 221 932 932 313 907 2;
  • 34) 0.222 221 932 932 313 907 2 × 2 = 0 + 0.444 443 865 864 627 814 4;
  • 35) 0.444 443 865 864 627 814 4 × 2 = 0 + 0.888 887 731 729 255 628 8;
  • 36) 0.888 887 731 729 255 628 8 × 2 = 1 + 0.777 775 463 458 511 257 6;
  • 37) 0.777 775 463 458 511 257 6 × 2 = 1 + 0.555 550 926 917 022 515 2;
  • 38) 0.555 550 926 917 022 515 2 × 2 = 1 + 0.111 101 853 834 045 030 4;
  • 39) 0.111 101 853 834 045 030 4 × 2 = 0 + 0.222 203 707 668 090 060 8;
  • 40) 0.222 203 707 668 090 060 8 × 2 = 0 + 0.444 407 415 336 180 121 6;
  • 41) 0.444 407 415 336 180 121 6 × 2 = 0 + 0.888 814 830 672 360 243 2;
  • 42) 0.888 814 830 672 360 243 2 × 2 = 1 + 0.777 629 661 344 720 486 4;
  • 43) 0.777 629 661 344 720 486 4 × 2 = 1 + 0.555 259 322 689 440 972 8;
  • 44) 0.555 259 322 689 440 972 8 × 2 = 1 + 0.110 518 645 378 881 945 6;
  • 45) 0.110 518 645 378 881 945 6 × 2 = 0 + 0.221 037 290 757 763 891 2;
  • 46) 0.221 037 290 757 763 891 2 × 2 = 0 + 0.442 074 581 515 527 782 4;
  • 47) 0.442 074 581 515 527 782 4 × 2 = 0 + 0.884 149 163 031 055 564 8;
  • 48) 0.884 149 163 031 055 564 8 × 2 = 1 + 0.768 298 326 062 111 129 6;
  • 49) 0.768 298 326 062 111 129 6 × 2 = 1 + 0.536 596 652 124 222 259 2;
  • 50) 0.536 596 652 124 222 259 2 × 2 = 1 + 0.073 193 304 248 444 518 4;
  • 51) 0.073 193 304 248 444 518 4 × 2 = 0 + 0.146 386 608 496 889 036 8;
  • 52) 0.146 386 608 496 889 036 8 × 2 = 0 + 0.292 773 216 993 778 073 6;
  • 53) 0.292 773 216 993 778 073 6 × 2 = 0 + 0.585 546 433 987 556 147 2;

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:


0.777 777 777 777 777 744 1(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 744 1(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:


24.777 777 777 777 777 744 1(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 24.777 777 777 777 777 744 1 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