24.777 777 777 777 777 776 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 776 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
24.777 777 777 777 777 776 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: 0.777 777 777 777 777 776 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) 0.777 777 777 777 777 776 1 × 2 = 1 + 0.555 555 555 555 555 552 2;
  • 2) 0.555 555 555 555 555 552 2 × 2 = 1 + 0.111 111 111 111 111 104 4;
  • 3) 0.111 111 111 111 111 104 4 × 2 = 0 + 0.222 222 222 222 222 208 8;
  • 4) 0.222 222 222 222 222 208 8 × 2 = 0 + 0.444 444 444 444 444 417 6;
  • 5) 0.444 444 444 444 444 417 6 × 2 = 0 + 0.888 888 888 888 888 835 2;
  • 6) 0.888 888 888 888 888 835 2 × 2 = 1 + 0.777 777 777 777 777 670 4;
  • 7) 0.777 777 777 777 777 670 4 × 2 = 1 + 0.555 555 555 555 555 340 8;
  • 8) 0.555 555 555 555 555 340 8 × 2 = 1 + 0.111 111 111 111 110 681 6;
  • 9) 0.111 111 111 111 110 681 6 × 2 = 0 + 0.222 222 222 222 221 363 2;
  • 10) 0.222 222 222 222 221 363 2 × 2 = 0 + 0.444 444 444 444 442 726 4;
  • 11) 0.444 444 444 444 442 726 4 × 2 = 0 + 0.888 888 888 888 885 452 8;
  • 12) 0.888 888 888 888 885 452 8 × 2 = 1 + 0.777 777 777 777 770 905 6;
  • 13) 0.777 777 777 777 770 905 6 × 2 = 1 + 0.555 555 555 555 541 811 2;
  • 14) 0.555 555 555 555 541 811 2 × 2 = 1 + 0.111 111 111 111 083 622 4;
  • 15) 0.111 111 111 111 083 622 4 × 2 = 0 + 0.222 222 222 222 167 244 8;
  • 16) 0.222 222 222 222 167 244 8 × 2 = 0 + 0.444 444 444 444 334 489 6;
  • 17) 0.444 444 444 444 334 489 6 × 2 = 0 + 0.888 888 888 888 668 979 2;
  • 18) 0.888 888 888 888 668 979 2 × 2 = 1 + 0.777 777 777 777 337 958 4;
  • 19) 0.777 777 777 777 337 958 4 × 2 = 1 + 0.555 555 555 554 675 916 8;
  • 20) 0.555 555 555 554 675 916 8 × 2 = 1 + 0.111 111 111 109 351 833 6;
  • 21) 0.111 111 111 109 351 833 6 × 2 = 0 + 0.222 222 222 218 703 667 2;
  • 22) 0.222 222 222 218 703 667 2 × 2 = 0 + 0.444 444 444 437 407 334 4;
  • 23) 0.444 444 444 437 407 334 4 × 2 = 0 + 0.888 888 888 874 814 668 8;
  • 24) 0.888 888 888 874 814 668 8 × 2 = 1 + 0.777 777 777 749 629 337 6;
  • 25) 0.777 777 777 749 629 337 6 × 2 = 1 + 0.555 555 555 499 258 675 2;
  • 26) 0.555 555 555 499 258 675 2 × 2 = 1 + 0.111 111 110 998 517 350 4;
  • 27) 0.111 111 110 998 517 350 4 × 2 = 0 + 0.222 222 221 997 034 700 8;
  • 28) 0.222 222 221 997 034 700 8 × 2 = 0 + 0.444 444 443 994 069 401 6;
  • 29) 0.444 444 443 994 069 401 6 × 2 = 0 + 0.888 888 887 988 138 803 2;
  • 30) 0.888 888 887 988 138 803 2 × 2 = 1 + 0.777 777 775 976 277 606 4;
  • 31) 0.777 777 775 976 277 606 4 × 2 = 1 + 0.555 555 551 952 555 212 8;
  • 32) 0.555 555 551 952 555 212 8 × 2 = 1 + 0.111 111 103 905 110 425 6;
  • 33) 0.111 111 103 905 110 425 6 × 2 = 0 + 0.222 222 207 810 220 851 2;
  • 34) 0.222 222 207 810 220 851 2 × 2 = 0 + 0.444 444 415 620 441 702 4;
  • 35) 0.444 444 415 620 441 702 4 × 2 = 0 + 0.888 888 831 240 883 404 8;
  • 36) 0.888 888 831 240 883 404 8 × 2 = 1 + 0.777 777 662 481 766 809 6;
  • 37) 0.777 777 662 481 766 809 6 × 2 = 1 + 0.555 555 324 963 533 619 2;
  • 38) 0.555 555 324 963 533 619 2 × 2 = 1 + 0.111 110 649 927 067 238 4;
  • 39) 0.111 110 649 927 067 238 4 × 2 = 0 + 0.222 221 299 854 134 476 8;
  • 40) 0.222 221 299 854 134 476 8 × 2 = 0 + 0.444 442 599 708 268 953 6;
  • 41) 0.444 442 599 708 268 953 6 × 2 = 0 + 0.888 885 199 416 537 907 2;
  • 42) 0.888 885 199 416 537 907 2 × 2 = 1 + 0.777 770 398 833 075 814 4;
  • 43) 0.777 770 398 833 075 814 4 × 2 = 1 + 0.555 540 797 666 151 628 8;
  • 44) 0.555 540 797 666 151 628 8 × 2 = 1 + 0.111 081 595 332 303 257 6;
  • 45) 0.111 081 595 332 303 257 6 × 2 = 0 + 0.222 163 190 664 606 515 2;
  • 46) 0.222 163 190 664 606 515 2 × 2 = 0 + 0.444 326 381 329 213 030 4;
  • 47) 0.444 326 381 329 213 030 4 × 2 = 0 + 0.888 652 762 658 426 060 8;
  • 48) 0.888 652 762 658 426 060 8 × 2 = 1 + 0.777 305 525 316 852 121 6;
  • 49) 0.777 305 525 316 852 121 6 × 2 = 1 + 0.554 611 050 633 704 243 2;
  • 50) 0.554 611 050 633 704 243 2 × 2 = 1 + 0.109 222 101 267 408 486 4;
  • 51) 0.109 222 101 267 408 486 4 × 2 = 0 + 0.218 444 202 534 816 972 8;
  • 52) 0.218 444 202 534 816 972 8 × 2 = 0 + 0.436 888 405 069 633 945 6;
  • 53) 0.436 888 405 069 633 945 6 × 2 = 0 + 0.873 776 810 139 267 891 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:


0.777 777 777 777 777 776 1(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 776 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:


24.777 777 777 777 777 776 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 24.777 777 777 777 777 776 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