24.777 777 777 777 935 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 935(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
24.777 777 777 777 935(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: 0.777 777 777 777 935.

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) 0.777 777 777 777 935 × 2 = 1 + 0.555 555 555 555 87;
  • 2) 0.555 555 555 555 87 × 2 = 1 + 0.111 111 111 111 74;
  • 3) 0.111 111 111 111 74 × 2 = 0 + 0.222 222 222 223 48;
  • 4) 0.222 222 222 223 48 × 2 = 0 + 0.444 444 444 446 96;
  • 5) 0.444 444 444 446 96 × 2 = 0 + 0.888 888 888 893 92;
  • 6) 0.888 888 888 893 92 × 2 = 1 + 0.777 777 777 787 84;
  • 7) 0.777 777 777 787 84 × 2 = 1 + 0.555 555 555 575 68;
  • 8) 0.555 555 555 575 68 × 2 = 1 + 0.111 111 111 151 36;
  • 9) 0.111 111 111 151 36 × 2 = 0 + 0.222 222 222 302 72;
  • 10) 0.222 222 222 302 72 × 2 = 0 + 0.444 444 444 605 44;
  • 11) 0.444 444 444 605 44 × 2 = 0 + 0.888 888 889 210 88;
  • 12) 0.888 888 889 210 88 × 2 = 1 + 0.777 777 778 421 76;
  • 13) 0.777 777 778 421 76 × 2 = 1 + 0.555 555 556 843 52;
  • 14) 0.555 555 556 843 52 × 2 = 1 + 0.111 111 113 687 04;
  • 15) 0.111 111 113 687 04 × 2 = 0 + 0.222 222 227 374 08;
  • 16) 0.222 222 227 374 08 × 2 = 0 + 0.444 444 454 748 16;
  • 17) 0.444 444 454 748 16 × 2 = 0 + 0.888 888 909 496 32;
  • 18) 0.888 888 909 496 32 × 2 = 1 + 0.777 777 818 992 64;
  • 19) 0.777 777 818 992 64 × 2 = 1 + 0.555 555 637 985 28;
  • 20) 0.555 555 637 985 28 × 2 = 1 + 0.111 111 275 970 56;
  • 21) 0.111 111 275 970 56 × 2 = 0 + 0.222 222 551 941 12;
  • 22) 0.222 222 551 941 12 × 2 = 0 + 0.444 445 103 882 24;
  • 23) 0.444 445 103 882 24 × 2 = 0 + 0.888 890 207 764 48;
  • 24) 0.888 890 207 764 48 × 2 = 1 + 0.777 780 415 528 96;
  • 25) 0.777 780 415 528 96 × 2 = 1 + 0.555 560 831 057 92;
  • 26) 0.555 560 831 057 92 × 2 = 1 + 0.111 121 662 115 84;
  • 27) 0.111 121 662 115 84 × 2 = 0 + 0.222 243 324 231 68;
  • 28) 0.222 243 324 231 68 × 2 = 0 + 0.444 486 648 463 36;
  • 29) 0.444 486 648 463 36 × 2 = 0 + 0.888 973 296 926 72;
  • 30) 0.888 973 296 926 72 × 2 = 1 + 0.777 946 593 853 44;
  • 31) 0.777 946 593 853 44 × 2 = 1 + 0.555 893 187 706 88;
  • 32) 0.555 893 187 706 88 × 2 = 1 + 0.111 786 375 413 76;
  • 33) 0.111 786 375 413 76 × 2 = 0 + 0.223 572 750 827 52;
  • 34) 0.223 572 750 827 52 × 2 = 0 + 0.447 145 501 655 04;
  • 35) 0.447 145 501 655 04 × 2 = 0 + 0.894 291 003 310 08;
  • 36) 0.894 291 003 310 08 × 2 = 1 + 0.788 582 006 620 16;
  • 37) 0.788 582 006 620 16 × 2 = 1 + 0.577 164 013 240 32;
  • 38) 0.577 164 013 240 32 × 2 = 1 + 0.154 328 026 480 64;
  • 39) 0.154 328 026 480 64 × 2 = 0 + 0.308 656 052 961 28;
  • 40) 0.308 656 052 961 28 × 2 = 0 + 0.617 312 105 922 56;
  • 41) 0.617 312 105 922 56 × 2 = 1 + 0.234 624 211 845 12;
  • 42) 0.234 624 211 845 12 × 2 = 0 + 0.469 248 423 690 24;
  • 43) 0.469 248 423 690 24 × 2 = 0 + 0.938 496 847 380 48;
  • 44) 0.938 496 847 380 48 × 2 = 1 + 0.876 993 694 760 96;
  • 45) 0.876 993 694 760 96 × 2 = 1 + 0.753 987 389 521 92;
  • 46) 0.753 987 389 521 92 × 2 = 1 + 0.507 974 779 043 84;
  • 47) 0.507 974 779 043 84 × 2 = 1 + 0.015 949 558 087 68;
  • 48) 0.015 949 558 087 68 × 2 = 0 + 0.031 899 116 175 36;
  • 49) 0.031 899 116 175 36 × 2 = 0 + 0.063 798 232 350 72;
  • 50) 0.063 798 232 350 72 × 2 = 0 + 0.127 596 464 701 44;
  • 51) 0.127 596 464 701 44 × 2 = 0 + 0.255 192 929 402 88;
  • 52) 0.255 192 929 402 88 × 2 = 0 + 0.510 385 858 805 76;
  • 53) 0.510 385 858 805 76 × 2 = 1 + 0.020 771 717 611 52;

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:


0.777 777 777 777 935(10) =

0.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1001 1110 0000 1(2)

5. Positive number before normalization:

24.777 777 777 777 935(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1001 1110 0000 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:


24.777 777 777 777 935(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1001 1110 0000 1(2) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1001 1110 0000 1(2) × 20 =

1.1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1001 1110 0000 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 1001 1110 0000 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 1001 1110 0 0001 =

1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1001 1110


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 1001 1110


Decimal number 24.777 777 777 777 935 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 1001 1110

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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