24.777 777 777 777 777 784 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 784 3(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
24.777 777 777 777 777 784 3(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: 0.777 777 777 777 777 784 3.

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) 0.777 777 777 777 777 784 3 × 2 = 1 + 0.555 555 555 555 555 568 6;
  • 2) 0.555 555 555 555 555 568 6 × 2 = 1 + 0.111 111 111 111 111 137 2;
  • 3) 0.111 111 111 111 111 137 2 × 2 = 0 + 0.222 222 222 222 222 274 4;
  • 4) 0.222 222 222 222 222 274 4 × 2 = 0 + 0.444 444 444 444 444 548 8;
  • 5) 0.444 444 444 444 444 548 8 × 2 = 0 + 0.888 888 888 888 889 097 6;
  • 6) 0.888 888 888 888 889 097 6 × 2 = 1 + 0.777 777 777 777 778 195 2;
  • 7) 0.777 777 777 777 778 195 2 × 2 = 1 + 0.555 555 555 555 556 390 4;
  • 8) 0.555 555 555 555 556 390 4 × 2 = 1 + 0.111 111 111 111 112 780 8;
  • 9) 0.111 111 111 111 112 780 8 × 2 = 0 + 0.222 222 222 222 225 561 6;
  • 10) 0.222 222 222 222 225 561 6 × 2 = 0 + 0.444 444 444 444 451 123 2;
  • 11) 0.444 444 444 444 451 123 2 × 2 = 0 + 0.888 888 888 888 902 246 4;
  • 12) 0.888 888 888 888 902 246 4 × 2 = 1 + 0.777 777 777 777 804 492 8;
  • 13) 0.777 777 777 777 804 492 8 × 2 = 1 + 0.555 555 555 555 608 985 6;
  • 14) 0.555 555 555 555 608 985 6 × 2 = 1 + 0.111 111 111 111 217 971 2;
  • 15) 0.111 111 111 111 217 971 2 × 2 = 0 + 0.222 222 222 222 435 942 4;
  • 16) 0.222 222 222 222 435 942 4 × 2 = 0 + 0.444 444 444 444 871 884 8;
  • 17) 0.444 444 444 444 871 884 8 × 2 = 0 + 0.888 888 888 889 743 769 6;
  • 18) 0.888 888 888 889 743 769 6 × 2 = 1 + 0.777 777 777 779 487 539 2;
  • 19) 0.777 777 777 779 487 539 2 × 2 = 1 + 0.555 555 555 558 975 078 4;
  • 20) 0.555 555 555 558 975 078 4 × 2 = 1 + 0.111 111 111 117 950 156 8;
  • 21) 0.111 111 111 117 950 156 8 × 2 = 0 + 0.222 222 222 235 900 313 6;
  • 22) 0.222 222 222 235 900 313 6 × 2 = 0 + 0.444 444 444 471 800 627 2;
  • 23) 0.444 444 444 471 800 627 2 × 2 = 0 + 0.888 888 888 943 601 254 4;
  • 24) 0.888 888 888 943 601 254 4 × 2 = 1 + 0.777 777 777 887 202 508 8;
  • 25) 0.777 777 777 887 202 508 8 × 2 = 1 + 0.555 555 555 774 405 017 6;
  • 26) 0.555 555 555 774 405 017 6 × 2 = 1 + 0.111 111 111 548 810 035 2;
  • 27) 0.111 111 111 548 810 035 2 × 2 = 0 + 0.222 222 223 097 620 070 4;
  • 28) 0.222 222 223 097 620 070 4 × 2 = 0 + 0.444 444 446 195 240 140 8;
  • 29) 0.444 444 446 195 240 140 8 × 2 = 0 + 0.888 888 892 390 480 281 6;
  • 30) 0.888 888 892 390 480 281 6 × 2 = 1 + 0.777 777 784 780 960 563 2;
  • 31) 0.777 777 784 780 960 563 2 × 2 = 1 + 0.555 555 569 561 921 126 4;
  • 32) 0.555 555 569 561 921 126 4 × 2 = 1 + 0.111 111 139 123 842 252 8;
  • 33) 0.111 111 139 123 842 252 8 × 2 = 0 + 0.222 222 278 247 684 505 6;
  • 34) 0.222 222 278 247 684 505 6 × 2 = 0 + 0.444 444 556 495 369 011 2;
  • 35) 0.444 444 556 495 369 011 2 × 2 = 0 + 0.888 889 112 990 738 022 4;
  • 36) 0.888 889 112 990 738 022 4 × 2 = 1 + 0.777 778 225 981 476 044 8;
  • 37) 0.777 778 225 981 476 044 8 × 2 = 1 + 0.555 556 451 962 952 089 6;
  • 38) 0.555 556 451 962 952 089 6 × 2 = 1 + 0.111 112 903 925 904 179 2;
  • 39) 0.111 112 903 925 904 179 2 × 2 = 0 + 0.222 225 807 851 808 358 4;
  • 40) 0.222 225 807 851 808 358 4 × 2 = 0 + 0.444 451 615 703 616 716 8;
  • 41) 0.444 451 615 703 616 716 8 × 2 = 0 + 0.888 903 231 407 233 433 6;
  • 42) 0.888 903 231 407 233 433 6 × 2 = 1 + 0.777 806 462 814 466 867 2;
  • 43) 0.777 806 462 814 466 867 2 × 2 = 1 + 0.555 612 925 628 933 734 4;
  • 44) 0.555 612 925 628 933 734 4 × 2 = 1 + 0.111 225 851 257 867 468 8;
  • 45) 0.111 225 851 257 867 468 8 × 2 = 0 + 0.222 451 702 515 734 937 6;
  • 46) 0.222 451 702 515 734 937 6 × 2 = 0 + 0.444 903 405 031 469 875 2;
  • 47) 0.444 903 405 031 469 875 2 × 2 = 0 + 0.889 806 810 062 939 750 4;
  • 48) 0.889 806 810 062 939 750 4 × 2 = 1 + 0.779 613 620 125 879 500 8;
  • 49) 0.779 613 620 125 879 500 8 × 2 = 1 + 0.559 227 240 251 759 001 6;
  • 50) 0.559 227 240 251 759 001 6 × 2 = 1 + 0.118 454 480 503 518 003 2;
  • 51) 0.118 454 480 503 518 003 2 × 2 = 0 + 0.236 908 961 007 036 006 4;
  • 52) 0.236 908 961 007 036 006 4 × 2 = 0 + 0.473 817 922 014 072 012 8;
  • 53) 0.473 817 922 014 072 012 8 × 2 = 0 + 0.947 635 844 028 144 025 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:


0.777 777 777 777 777 784 3(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 784 3(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:


24.777 777 777 777 777 784 3(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 24.777 777 777 777 777 784 3 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