24.777 777 777 777 778 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 778 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
24.777 777 777 777 778 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: 0.777 777 777 777 778 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) 0.777 777 777 777 778 8 × 2 = 1 + 0.555 555 555 555 557 6;
  • 2) 0.555 555 555 555 557 6 × 2 = 1 + 0.111 111 111 111 115 2;
  • 3) 0.111 111 111 111 115 2 × 2 = 0 + 0.222 222 222 222 230 4;
  • 4) 0.222 222 222 222 230 4 × 2 = 0 + 0.444 444 444 444 460 8;
  • 5) 0.444 444 444 444 460 8 × 2 = 0 + 0.888 888 888 888 921 6;
  • 6) 0.888 888 888 888 921 6 × 2 = 1 + 0.777 777 777 777 843 2;
  • 7) 0.777 777 777 777 843 2 × 2 = 1 + 0.555 555 555 555 686 4;
  • 8) 0.555 555 555 555 686 4 × 2 = 1 + 0.111 111 111 111 372 8;
  • 9) 0.111 111 111 111 372 8 × 2 = 0 + 0.222 222 222 222 745 6;
  • 10) 0.222 222 222 222 745 6 × 2 = 0 + 0.444 444 444 445 491 2;
  • 11) 0.444 444 444 445 491 2 × 2 = 0 + 0.888 888 888 890 982 4;
  • 12) 0.888 888 888 890 982 4 × 2 = 1 + 0.777 777 777 781 964 8;
  • 13) 0.777 777 777 781 964 8 × 2 = 1 + 0.555 555 555 563 929 6;
  • 14) 0.555 555 555 563 929 6 × 2 = 1 + 0.111 111 111 127 859 2;
  • 15) 0.111 111 111 127 859 2 × 2 = 0 + 0.222 222 222 255 718 4;
  • 16) 0.222 222 222 255 718 4 × 2 = 0 + 0.444 444 444 511 436 8;
  • 17) 0.444 444 444 511 436 8 × 2 = 0 + 0.888 888 889 022 873 6;
  • 18) 0.888 888 889 022 873 6 × 2 = 1 + 0.777 777 778 045 747 2;
  • 19) 0.777 777 778 045 747 2 × 2 = 1 + 0.555 555 556 091 494 4;
  • 20) 0.555 555 556 091 494 4 × 2 = 1 + 0.111 111 112 182 988 8;
  • 21) 0.111 111 112 182 988 8 × 2 = 0 + 0.222 222 224 365 977 6;
  • 22) 0.222 222 224 365 977 6 × 2 = 0 + 0.444 444 448 731 955 2;
  • 23) 0.444 444 448 731 955 2 × 2 = 0 + 0.888 888 897 463 910 4;
  • 24) 0.888 888 897 463 910 4 × 2 = 1 + 0.777 777 794 927 820 8;
  • 25) 0.777 777 794 927 820 8 × 2 = 1 + 0.555 555 589 855 641 6;
  • 26) 0.555 555 589 855 641 6 × 2 = 1 + 0.111 111 179 711 283 2;
  • 27) 0.111 111 179 711 283 2 × 2 = 0 + 0.222 222 359 422 566 4;
  • 28) 0.222 222 359 422 566 4 × 2 = 0 + 0.444 444 718 845 132 8;
  • 29) 0.444 444 718 845 132 8 × 2 = 0 + 0.888 889 437 690 265 6;
  • 30) 0.888 889 437 690 265 6 × 2 = 1 + 0.777 778 875 380 531 2;
  • 31) 0.777 778 875 380 531 2 × 2 = 1 + 0.555 557 750 761 062 4;
  • 32) 0.555 557 750 761 062 4 × 2 = 1 + 0.111 115 501 522 124 8;
  • 33) 0.111 115 501 522 124 8 × 2 = 0 + 0.222 231 003 044 249 6;
  • 34) 0.222 231 003 044 249 6 × 2 = 0 + 0.444 462 006 088 499 2;
  • 35) 0.444 462 006 088 499 2 × 2 = 0 + 0.888 924 012 176 998 4;
  • 36) 0.888 924 012 176 998 4 × 2 = 1 + 0.777 848 024 353 996 8;
  • 37) 0.777 848 024 353 996 8 × 2 = 1 + 0.555 696 048 707 993 6;
  • 38) 0.555 696 048 707 993 6 × 2 = 1 + 0.111 392 097 415 987 2;
  • 39) 0.111 392 097 415 987 2 × 2 = 0 + 0.222 784 194 831 974 4;
  • 40) 0.222 784 194 831 974 4 × 2 = 0 + 0.445 568 389 663 948 8;
  • 41) 0.445 568 389 663 948 8 × 2 = 0 + 0.891 136 779 327 897 6;
  • 42) 0.891 136 779 327 897 6 × 2 = 1 + 0.782 273 558 655 795 2;
  • 43) 0.782 273 558 655 795 2 × 2 = 1 + 0.564 547 117 311 590 4;
  • 44) 0.564 547 117 311 590 4 × 2 = 1 + 0.129 094 234 623 180 8;
  • 45) 0.129 094 234 623 180 8 × 2 = 0 + 0.258 188 469 246 361 6;
  • 46) 0.258 188 469 246 361 6 × 2 = 0 + 0.516 376 938 492 723 2;
  • 47) 0.516 376 938 492 723 2 × 2 = 1 + 0.032 753 876 985 446 4;
  • 48) 0.032 753 876 985 446 4 × 2 = 0 + 0.065 507 753 970 892 8;
  • 49) 0.065 507 753 970 892 8 × 2 = 0 + 0.131 015 507 941 785 6;
  • 50) 0.131 015 507 941 785 6 × 2 = 0 + 0.262 031 015 883 571 2;
  • 51) 0.262 031 015 883 571 2 × 2 = 0 + 0.524 062 031 767 142 4;
  • 52) 0.524 062 031 767 142 4 × 2 = 1 + 0.048 124 063 534 284 8;
  • 53) 0.048 124 063 534 284 8 × 2 = 0 + 0.096 248 127 068 569 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:


0.777 777 777 777 778 8(10) =

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

5. Positive number before normalization:

24.777 777 777 777 778 8(10) =

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


24.777 777 777 777 778 8(10) =

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

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

1.1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0010 0001 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 0001 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 0010 =

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 24.777 777 777 777 778 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 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