24.777 777 777 777 779 11 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 779 11(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 11(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 11.

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 11 × 2 = 1 + 0.555 555 555 555 558 22;
  • 2) 0.555 555 555 555 558 22 × 2 = 1 + 0.111 111 111 111 116 44;
  • 3) 0.111 111 111 111 116 44 × 2 = 0 + 0.222 222 222 222 232 88;
  • 4) 0.222 222 222 222 232 88 × 2 = 0 + 0.444 444 444 444 465 76;
  • 5) 0.444 444 444 444 465 76 × 2 = 0 + 0.888 888 888 888 931 52;
  • 6) 0.888 888 888 888 931 52 × 2 = 1 + 0.777 777 777 777 863 04;
  • 7) 0.777 777 777 777 863 04 × 2 = 1 + 0.555 555 555 555 726 08;
  • 8) 0.555 555 555 555 726 08 × 2 = 1 + 0.111 111 111 111 452 16;
  • 9) 0.111 111 111 111 452 16 × 2 = 0 + 0.222 222 222 222 904 32;
  • 10) 0.222 222 222 222 904 32 × 2 = 0 + 0.444 444 444 445 808 64;
  • 11) 0.444 444 444 445 808 64 × 2 = 0 + 0.888 888 888 891 617 28;
  • 12) 0.888 888 888 891 617 28 × 2 = 1 + 0.777 777 777 783 234 56;
  • 13) 0.777 777 777 783 234 56 × 2 = 1 + 0.555 555 555 566 469 12;
  • 14) 0.555 555 555 566 469 12 × 2 = 1 + 0.111 111 111 132 938 24;
  • 15) 0.111 111 111 132 938 24 × 2 = 0 + 0.222 222 222 265 876 48;
  • 16) 0.222 222 222 265 876 48 × 2 = 0 + 0.444 444 444 531 752 96;
  • 17) 0.444 444 444 531 752 96 × 2 = 0 + 0.888 888 889 063 505 92;
  • 18) 0.888 888 889 063 505 92 × 2 = 1 + 0.777 777 778 127 011 84;
  • 19) 0.777 777 778 127 011 84 × 2 = 1 + 0.555 555 556 254 023 68;
  • 20) 0.555 555 556 254 023 68 × 2 = 1 + 0.111 111 112 508 047 36;
  • 21) 0.111 111 112 508 047 36 × 2 = 0 + 0.222 222 225 016 094 72;
  • 22) 0.222 222 225 016 094 72 × 2 = 0 + 0.444 444 450 032 189 44;
  • 23) 0.444 444 450 032 189 44 × 2 = 0 + 0.888 888 900 064 378 88;
  • 24) 0.888 888 900 064 378 88 × 2 = 1 + 0.777 777 800 128 757 76;
  • 25) 0.777 777 800 128 757 76 × 2 = 1 + 0.555 555 600 257 515 52;
  • 26) 0.555 555 600 257 515 52 × 2 = 1 + 0.111 111 200 515 031 04;
  • 27) 0.111 111 200 515 031 04 × 2 = 0 + 0.222 222 401 030 062 08;
  • 28) 0.222 222 401 030 062 08 × 2 = 0 + 0.444 444 802 060 124 16;
  • 29) 0.444 444 802 060 124 16 × 2 = 0 + 0.888 889 604 120 248 32;
  • 30) 0.888 889 604 120 248 32 × 2 = 1 + 0.777 779 208 240 496 64;
  • 31) 0.777 779 208 240 496 64 × 2 = 1 + 0.555 558 416 480 993 28;
  • 32) 0.555 558 416 480 993 28 × 2 = 1 + 0.111 116 832 961 986 56;
  • 33) 0.111 116 832 961 986 56 × 2 = 0 + 0.222 233 665 923 973 12;
  • 34) 0.222 233 665 923 973 12 × 2 = 0 + 0.444 467 331 847 946 24;
  • 35) 0.444 467 331 847 946 24 × 2 = 0 + 0.888 934 663 695 892 48;
  • 36) 0.888 934 663 695 892 48 × 2 = 1 + 0.777 869 327 391 784 96;
  • 37) 0.777 869 327 391 784 96 × 2 = 1 + 0.555 738 654 783 569 92;
  • 38) 0.555 738 654 783 569 92 × 2 = 1 + 0.111 477 309 567 139 84;
  • 39) 0.111 477 309 567 139 84 × 2 = 0 + 0.222 954 619 134 279 68;
  • 40) 0.222 954 619 134 279 68 × 2 = 0 + 0.445 909 238 268 559 36;
  • 41) 0.445 909 238 268 559 36 × 2 = 0 + 0.891 818 476 537 118 72;
  • 42) 0.891 818 476 537 118 72 × 2 = 1 + 0.783 636 953 074 237 44;
  • 43) 0.783 636 953 074 237 44 × 2 = 1 + 0.567 273 906 148 474 88;
  • 44) 0.567 273 906 148 474 88 × 2 = 1 + 0.134 547 812 296 949 76;
  • 45) 0.134 547 812 296 949 76 × 2 = 0 + 0.269 095 624 593 899 52;
  • 46) 0.269 095 624 593 899 52 × 2 = 0 + 0.538 191 249 187 799 04;
  • 47) 0.538 191 249 187 799 04 × 2 = 1 + 0.076 382 498 375 598 08;
  • 48) 0.076 382 498 375 598 08 × 2 = 0 + 0.152 764 996 751 196 16;
  • 49) 0.152 764 996 751 196 16 × 2 = 0 + 0.305 529 993 502 392 32;
  • 50) 0.305 529 993 502 392 32 × 2 = 0 + 0.611 059 987 004 784 64;
  • 51) 0.611 059 987 004 784 64 × 2 = 1 + 0.222 119 974 009 569 28;
  • 52) 0.222 119 974 009 569 28 × 2 = 0 + 0.444 239 948 019 138 56;
  • 53) 0.444 239 948 019 138 56 × 2 = 0 + 0.888 479 896 038 277 12;

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 11(10) =

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

5. Positive number before normalization:

24.777 777 777 777 779 11(10) =

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


24.777 777 777 777 779 11(10) =

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

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

1.1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0010 0010 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 0010 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 0100 =

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 11 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