24.777 777 777 777 777 780 15 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 780 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
24.777 777 777 777 777 780 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: 0.777 777 777 777 777 780 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) 0.777 777 777 777 777 780 15 × 2 = 1 + 0.555 555 555 555 555 560 3;
  • 2) 0.555 555 555 555 555 560 3 × 2 = 1 + 0.111 111 111 111 111 120 6;
  • 3) 0.111 111 111 111 111 120 6 × 2 = 0 + 0.222 222 222 222 222 241 2;
  • 4) 0.222 222 222 222 222 241 2 × 2 = 0 + 0.444 444 444 444 444 482 4;
  • 5) 0.444 444 444 444 444 482 4 × 2 = 0 + 0.888 888 888 888 888 964 8;
  • 6) 0.888 888 888 888 888 964 8 × 2 = 1 + 0.777 777 777 777 777 929 6;
  • 7) 0.777 777 777 777 777 929 6 × 2 = 1 + 0.555 555 555 555 555 859 2;
  • 8) 0.555 555 555 555 555 859 2 × 2 = 1 + 0.111 111 111 111 111 718 4;
  • 9) 0.111 111 111 111 111 718 4 × 2 = 0 + 0.222 222 222 222 223 436 8;
  • 10) 0.222 222 222 222 223 436 8 × 2 = 0 + 0.444 444 444 444 446 873 6;
  • 11) 0.444 444 444 444 446 873 6 × 2 = 0 + 0.888 888 888 888 893 747 2;
  • 12) 0.888 888 888 888 893 747 2 × 2 = 1 + 0.777 777 777 777 787 494 4;
  • 13) 0.777 777 777 777 787 494 4 × 2 = 1 + 0.555 555 555 555 574 988 8;
  • 14) 0.555 555 555 555 574 988 8 × 2 = 1 + 0.111 111 111 111 149 977 6;
  • 15) 0.111 111 111 111 149 977 6 × 2 = 0 + 0.222 222 222 222 299 955 2;
  • 16) 0.222 222 222 222 299 955 2 × 2 = 0 + 0.444 444 444 444 599 910 4;
  • 17) 0.444 444 444 444 599 910 4 × 2 = 0 + 0.888 888 888 889 199 820 8;
  • 18) 0.888 888 888 889 199 820 8 × 2 = 1 + 0.777 777 777 778 399 641 6;
  • 19) 0.777 777 777 778 399 641 6 × 2 = 1 + 0.555 555 555 556 799 283 2;
  • 20) 0.555 555 555 556 799 283 2 × 2 = 1 + 0.111 111 111 113 598 566 4;
  • 21) 0.111 111 111 113 598 566 4 × 2 = 0 + 0.222 222 222 227 197 132 8;
  • 22) 0.222 222 222 227 197 132 8 × 2 = 0 + 0.444 444 444 454 394 265 6;
  • 23) 0.444 444 444 454 394 265 6 × 2 = 0 + 0.888 888 888 908 788 531 2;
  • 24) 0.888 888 888 908 788 531 2 × 2 = 1 + 0.777 777 777 817 577 062 4;
  • 25) 0.777 777 777 817 577 062 4 × 2 = 1 + 0.555 555 555 635 154 124 8;
  • 26) 0.555 555 555 635 154 124 8 × 2 = 1 + 0.111 111 111 270 308 249 6;
  • 27) 0.111 111 111 270 308 249 6 × 2 = 0 + 0.222 222 222 540 616 499 2;
  • 28) 0.222 222 222 540 616 499 2 × 2 = 0 + 0.444 444 445 081 232 998 4;
  • 29) 0.444 444 445 081 232 998 4 × 2 = 0 + 0.888 888 890 162 465 996 8;
  • 30) 0.888 888 890 162 465 996 8 × 2 = 1 + 0.777 777 780 324 931 993 6;
  • 31) 0.777 777 780 324 931 993 6 × 2 = 1 + 0.555 555 560 649 863 987 2;
  • 32) 0.555 555 560 649 863 987 2 × 2 = 1 + 0.111 111 121 299 727 974 4;
  • 33) 0.111 111 121 299 727 974 4 × 2 = 0 + 0.222 222 242 599 455 948 8;
  • 34) 0.222 222 242 599 455 948 8 × 2 = 0 + 0.444 444 485 198 911 897 6;
  • 35) 0.444 444 485 198 911 897 6 × 2 = 0 + 0.888 888 970 397 823 795 2;
  • 36) 0.888 888 970 397 823 795 2 × 2 = 1 + 0.777 777 940 795 647 590 4;
  • 37) 0.777 777 940 795 647 590 4 × 2 = 1 + 0.555 555 881 591 295 180 8;
  • 38) 0.555 555 881 591 295 180 8 × 2 = 1 + 0.111 111 763 182 590 361 6;
  • 39) 0.111 111 763 182 590 361 6 × 2 = 0 + 0.222 223 526 365 180 723 2;
  • 40) 0.222 223 526 365 180 723 2 × 2 = 0 + 0.444 447 052 730 361 446 4;
  • 41) 0.444 447 052 730 361 446 4 × 2 = 0 + 0.888 894 105 460 722 892 8;
  • 42) 0.888 894 105 460 722 892 8 × 2 = 1 + 0.777 788 210 921 445 785 6;
  • 43) 0.777 788 210 921 445 785 6 × 2 = 1 + 0.555 576 421 842 891 571 2;
  • 44) 0.555 576 421 842 891 571 2 × 2 = 1 + 0.111 152 843 685 783 142 4;
  • 45) 0.111 152 843 685 783 142 4 × 2 = 0 + 0.222 305 687 371 566 284 8;
  • 46) 0.222 305 687 371 566 284 8 × 2 = 0 + 0.444 611 374 743 132 569 6;
  • 47) 0.444 611 374 743 132 569 6 × 2 = 0 + 0.889 222 749 486 265 139 2;
  • 48) 0.889 222 749 486 265 139 2 × 2 = 1 + 0.778 445 498 972 530 278 4;
  • 49) 0.778 445 498 972 530 278 4 × 2 = 1 + 0.556 890 997 945 060 556 8;
  • 50) 0.556 890 997 945 060 556 8 × 2 = 1 + 0.113 781 995 890 121 113 6;
  • 51) 0.113 781 995 890 121 113 6 × 2 = 0 + 0.227 563 991 780 242 227 2;
  • 52) 0.227 563 991 780 242 227 2 × 2 = 0 + 0.455 127 983 560 484 454 4;
  • 53) 0.455 127 983 560 484 454 4 × 2 = 0 + 0.910 255 967 120 968 908 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:


0.777 777 777 777 777 780 15(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 780 15(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:


24.777 777 777 777 777 780 15(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 24.777 777 777 777 777 780 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 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