24.777 777 777 777 779 15 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 779 15(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
24.777 777 777 777 779 15(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: 0.777 777 777 777 779 15.

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) 0.777 777 777 777 779 15 × 2 = 1 + 0.555 555 555 555 558 3;
  • 2) 0.555 555 555 555 558 3 × 2 = 1 + 0.111 111 111 111 116 6;
  • 3) 0.111 111 111 111 116 6 × 2 = 0 + 0.222 222 222 222 233 2;
  • 4) 0.222 222 222 222 233 2 × 2 = 0 + 0.444 444 444 444 466 4;
  • 5) 0.444 444 444 444 466 4 × 2 = 0 + 0.888 888 888 888 932 8;
  • 6) 0.888 888 888 888 932 8 × 2 = 1 + 0.777 777 777 777 865 6;
  • 7) 0.777 777 777 777 865 6 × 2 = 1 + 0.555 555 555 555 731 2;
  • 8) 0.555 555 555 555 731 2 × 2 = 1 + 0.111 111 111 111 462 4;
  • 9) 0.111 111 111 111 462 4 × 2 = 0 + 0.222 222 222 222 924 8;
  • 10) 0.222 222 222 222 924 8 × 2 = 0 + 0.444 444 444 445 849 6;
  • 11) 0.444 444 444 445 849 6 × 2 = 0 + 0.888 888 888 891 699 2;
  • 12) 0.888 888 888 891 699 2 × 2 = 1 + 0.777 777 777 783 398 4;
  • 13) 0.777 777 777 783 398 4 × 2 = 1 + 0.555 555 555 566 796 8;
  • 14) 0.555 555 555 566 796 8 × 2 = 1 + 0.111 111 111 133 593 6;
  • 15) 0.111 111 111 133 593 6 × 2 = 0 + 0.222 222 222 267 187 2;
  • 16) 0.222 222 222 267 187 2 × 2 = 0 + 0.444 444 444 534 374 4;
  • 17) 0.444 444 444 534 374 4 × 2 = 0 + 0.888 888 889 068 748 8;
  • 18) 0.888 888 889 068 748 8 × 2 = 1 + 0.777 777 778 137 497 6;
  • 19) 0.777 777 778 137 497 6 × 2 = 1 + 0.555 555 556 274 995 2;
  • 20) 0.555 555 556 274 995 2 × 2 = 1 + 0.111 111 112 549 990 4;
  • 21) 0.111 111 112 549 990 4 × 2 = 0 + 0.222 222 225 099 980 8;
  • 22) 0.222 222 225 099 980 8 × 2 = 0 + 0.444 444 450 199 961 6;
  • 23) 0.444 444 450 199 961 6 × 2 = 0 + 0.888 888 900 399 923 2;
  • 24) 0.888 888 900 399 923 2 × 2 = 1 + 0.777 777 800 799 846 4;
  • 25) 0.777 777 800 799 846 4 × 2 = 1 + 0.555 555 601 599 692 8;
  • 26) 0.555 555 601 599 692 8 × 2 = 1 + 0.111 111 203 199 385 6;
  • 27) 0.111 111 203 199 385 6 × 2 = 0 + 0.222 222 406 398 771 2;
  • 28) 0.222 222 406 398 771 2 × 2 = 0 + 0.444 444 812 797 542 4;
  • 29) 0.444 444 812 797 542 4 × 2 = 0 + 0.888 889 625 595 084 8;
  • 30) 0.888 889 625 595 084 8 × 2 = 1 + 0.777 779 251 190 169 6;
  • 31) 0.777 779 251 190 169 6 × 2 = 1 + 0.555 558 502 380 339 2;
  • 32) 0.555 558 502 380 339 2 × 2 = 1 + 0.111 117 004 760 678 4;
  • 33) 0.111 117 004 760 678 4 × 2 = 0 + 0.222 234 009 521 356 8;
  • 34) 0.222 234 009 521 356 8 × 2 = 0 + 0.444 468 019 042 713 6;
  • 35) 0.444 468 019 042 713 6 × 2 = 0 + 0.888 936 038 085 427 2;
  • 36) 0.888 936 038 085 427 2 × 2 = 1 + 0.777 872 076 170 854 4;
  • 37) 0.777 872 076 170 854 4 × 2 = 1 + 0.555 744 152 341 708 8;
  • 38) 0.555 744 152 341 708 8 × 2 = 1 + 0.111 488 304 683 417 6;
  • 39) 0.111 488 304 683 417 6 × 2 = 0 + 0.222 976 609 366 835 2;
  • 40) 0.222 976 609 366 835 2 × 2 = 0 + 0.445 953 218 733 670 4;
  • 41) 0.445 953 218 733 670 4 × 2 = 0 + 0.891 906 437 467 340 8;
  • 42) 0.891 906 437 467 340 8 × 2 = 1 + 0.783 812 874 934 681 6;
  • 43) 0.783 812 874 934 681 6 × 2 = 1 + 0.567 625 749 869 363 2;
  • 44) 0.567 625 749 869 363 2 × 2 = 1 + 0.135 251 499 738 726 4;
  • 45) 0.135 251 499 738 726 4 × 2 = 0 + 0.270 502 999 477 452 8;
  • 46) 0.270 502 999 477 452 8 × 2 = 0 + 0.541 005 998 954 905 6;
  • 47) 0.541 005 998 954 905 6 × 2 = 1 + 0.082 011 997 909 811 2;
  • 48) 0.082 011 997 909 811 2 × 2 = 0 + 0.164 023 995 819 622 4;
  • 49) 0.164 023 995 819 622 4 × 2 = 0 + 0.328 047 991 639 244 8;
  • 50) 0.328 047 991 639 244 8 × 2 = 0 + 0.656 095 983 278 489 6;
  • 51) 0.656 095 983 278 489 6 × 2 = 1 + 0.312 191 966 556 979 2;
  • 52) 0.312 191 966 556 979 2 × 2 = 0 + 0.624 383 933 113 958 4;
  • 53) 0.624 383 933 113 958 4 × 2 = 1 + 0.248 767 866 227 916 8;

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:


0.777 777 777 777 779 15(10) =

0.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0010 0010 1(2)

5. Positive number before normalization:

24.777 777 777 777 779 15(10) =

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


24.777 777 777 777 779 15(10) =

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

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

1.1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0010 0010 1(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 0010 1


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 0101 =

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 24.777 777 777 777 779 15 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