24.777 777 777 777 777 762 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 762 6(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
24.777 777 777 777 777 762 6(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: 0.777 777 777 777 777 762 6.

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) 0.777 777 777 777 777 762 6 × 2 = 1 + 0.555 555 555 555 555 525 2;
  • 2) 0.555 555 555 555 555 525 2 × 2 = 1 + 0.111 111 111 111 111 050 4;
  • 3) 0.111 111 111 111 111 050 4 × 2 = 0 + 0.222 222 222 222 222 100 8;
  • 4) 0.222 222 222 222 222 100 8 × 2 = 0 + 0.444 444 444 444 444 201 6;
  • 5) 0.444 444 444 444 444 201 6 × 2 = 0 + 0.888 888 888 888 888 403 2;
  • 6) 0.888 888 888 888 888 403 2 × 2 = 1 + 0.777 777 777 777 776 806 4;
  • 7) 0.777 777 777 777 776 806 4 × 2 = 1 + 0.555 555 555 555 553 612 8;
  • 8) 0.555 555 555 555 553 612 8 × 2 = 1 + 0.111 111 111 111 107 225 6;
  • 9) 0.111 111 111 111 107 225 6 × 2 = 0 + 0.222 222 222 222 214 451 2;
  • 10) 0.222 222 222 222 214 451 2 × 2 = 0 + 0.444 444 444 444 428 902 4;
  • 11) 0.444 444 444 444 428 902 4 × 2 = 0 + 0.888 888 888 888 857 804 8;
  • 12) 0.888 888 888 888 857 804 8 × 2 = 1 + 0.777 777 777 777 715 609 6;
  • 13) 0.777 777 777 777 715 609 6 × 2 = 1 + 0.555 555 555 555 431 219 2;
  • 14) 0.555 555 555 555 431 219 2 × 2 = 1 + 0.111 111 111 110 862 438 4;
  • 15) 0.111 111 111 110 862 438 4 × 2 = 0 + 0.222 222 222 221 724 876 8;
  • 16) 0.222 222 222 221 724 876 8 × 2 = 0 + 0.444 444 444 443 449 753 6;
  • 17) 0.444 444 444 443 449 753 6 × 2 = 0 + 0.888 888 888 886 899 507 2;
  • 18) 0.888 888 888 886 899 507 2 × 2 = 1 + 0.777 777 777 773 799 014 4;
  • 19) 0.777 777 777 773 799 014 4 × 2 = 1 + 0.555 555 555 547 598 028 8;
  • 20) 0.555 555 555 547 598 028 8 × 2 = 1 + 0.111 111 111 095 196 057 6;
  • 21) 0.111 111 111 095 196 057 6 × 2 = 0 + 0.222 222 222 190 392 115 2;
  • 22) 0.222 222 222 190 392 115 2 × 2 = 0 + 0.444 444 444 380 784 230 4;
  • 23) 0.444 444 444 380 784 230 4 × 2 = 0 + 0.888 888 888 761 568 460 8;
  • 24) 0.888 888 888 761 568 460 8 × 2 = 1 + 0.777 777 777 523 136 921 6;
  • 25) 0.777 777 777 523 136 921 6 × 2 = 1 + 0.555 555 555 046 273 843 2;
  • 26) 0.555 555 555 046 273 843 2 × 2 = 1 + 0.111 111 110 092 547 686 4;
  • 27) 0.111 111 110 092 547 686 4 × 2 = 0 + 0.222 222 220 185 095 372 8;
  • 28) 0.222 222 220 185 095 372 8 × 2 = 0 + 0.444 444 440 370 190 745 6;
  • 29) 0.444 444 440 370 190 745 6 × 2 = 0 + 0.888 888 880 740 381 491 2;
  • 30) 0.888 888 880 740 381 491 2 × 2 = 1 + 0.777 777 761 480 762 982 4;
  • 31) 0.777 777 761 480 762 982 4 × 2 = 1 + 0.555 555 522 961 525 964 8;
  • 32) 0.555 555 522 961 525 964 8 × 2 = 1 + 0.111 111 045 923 051 929 6;
  • 33) 0.111 111 045 923 051 929 6 × 2 = 0 + 0.222 222 091 846 103 859 2;
  • 34) 0.222 222 091 846 103 859 2 × 2 = 0 + 0.444 444 183 692 207 718 4;
  • 35) 0.444 444 183 692 207 718 4 × 2 = 0 + 0.888 888 367 384 415 436 8;
  • 36) 0.888 888 367 384 415 436 8 × 2 = 1 + 0.777 776 734 768 830 873 6;
  • 37) 0.777 776 734 768 830 873 6 × 2 = 1 + 0.555 553 469 537 661 747 2;
  • 38) 0.555 553 469 537 661 747 2 × 2 = 1 + 0.111 106 939 075 323 494 4;
  • 39) 0.111 106 939 075 323 494 4 × 2 = 0 + 0.222 213 878 150 646 988 8;
  • 40) 0.222 213 878 150 646 988 8 × 2 = 0 + 0.444 427 756 301 293 977 6;
  • 41) 0.444 427 756 301 293 977 6 × 2 = 0 + 0.888 855 512 602 587 955 2;
  • 42) 0.888 855 512 602 587 955 2 × 2 = 1 + 0.777 711 025 205 175 910 4;
  • 43) 0.777 711 025 205 175 910 4 × 2 = 1 + 0.555 422 050 410 351 820 8;
  • 44) 0.555 422 050 410 351 820 8 × 2 = 1 + 0.110 844 100 820 703 641 6;
  • 45) 0.110 844 100 820 703 641 6 × 2 = 0 + 0.221 688 201 641 407 283 2;
  • 46) 0.221 688 201 641 407 283 2 × 2 = 0 + 0.443 376 403 282 814 566 4;
  • 47) 0.443 376 403 282 814 566 4 × 2 = 0 + 0.886 752 806 565 629 132 8;
  • 48) 0.886 752 806 565 629 132 8 × 2 = 1 + 0.773 505 613 131 258 265 6;
  • 49) 0.773 505 613 131 258 265 6 × 2 = 1 + 0.547 011 226 262 516 531 2;
  • 50) 0.547 011 226 262 516 531 2 × 2 = 1 + 0.094 022 452 525 033 062 4;
  • 51) 0.094 022 452 525 033 062 4 × 2 = 0 + 0.188 044 905 050 066 124 8;
  • 52) 0.188 044 905 050 066 124 8 × 2 = 0 + 0.376 089 810 100 132 249 6;
  • 53) 0.376 089 810 100 132 249 6 × 2 = 0 + 0.752 179 620 200 264 499 2;

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:


0.777 777 777 777 777 762 6(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 762 6(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:


24.777 777 777 777 777 762 6(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 24.777 777 777 777 777 762 6 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