24.777 777 777 777 777 788 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 788 1(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
24.777 777 777 777 777 788 1(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: 0.777 777 777 777 777 788 1.

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) 0.777 777 777 777 777 788 1 × 2 = 1 + 0.555 555 555 555 555 576 2;
  • 2) 0.555 555 555 555 555 576 2 × 2 = 1 + 0.111 111 111 111 111 152 4;
  • 3) 0.111 111 111 111 111 152 4 × 2 = 0 + 0.222 222 222 222 222 304 8;
  • 4) 0.222 222 222 222 222 304 8 × 2 = 0 + 0.444 444 444 444 444 609 6;
  • 5) 0.444 444 444 444 444 609 6 × 2 = 0 + 0.888 888 888 888 889 219 2;
  • 6) 0.888 888 888 888 889 219 2 × 2 = 1 + 0.777 777 777 777 778 438 4;
  • 7) 0.777 777 777 777 778 438 4 × 2 = 1 + 0.555 555 555 555 556 876 8;
  • 8) 0.555 555 555 555 556 876 8 × 2 = 1 + 0.111 111 111 111 113 753 6;
  • 9) 0.111 111 111 111 113 753 6 × 2 = 0 + 0.222 222 222 222 227 507 2;
  • 10) 0.222 222 222 222 227 507 2 × 2 = 0 + 0.444 444 444 444 455 014 4;
  • 11) 0.444 444 444 444 455 014 4 × 2 = 0 + 0.888 888 888 888 910 028 8;
  • 12) 0.888 888 888 888 910 028 8 × 2 = 1 + 0.777 777 777 777 820 057 6;
  • 13) 0.777 777 777 777 820 057 6 × 2 = 1 + 0.555 555 555 555 640 115 2;
  • 14) 0.555 555 555 555 640 115 2 × 2 = 1 + 0.111 111 111 111 280 230 4;
  • 15) 0.111 111 111 111 280 230 4 × 2 = 0 + 0.222 222 222 222 560 460 8;
  • 16) 0.222 222 222 222 560 460 8 × 2 = 0 + 0.444 444 444 445 120 921 6;
  • 17) 0.444 444 444 445 120 921 6 × 2 = 0 + 0.888 888 888 890 241 843 2;
  • 18) 0.888 888 888 890 241 843 2 × 2 = 1 + 0.777 777 777 780 483 686 4;
  • 19) 0.777 777 777 780 483 686 4 × 2 = 1 + 0.555 555 555 560 967 372 8;
  • 20) 0.555 555 555 560 967 372 8 × 2 = 1 + 0.111 111 111 121 934 745 6;
  • 21) 0.111 111 111 121 934 745 6 × 2 = 0 + 0.222 222 222 243 869 491 2;
  • 22) 0.222 222 222 243 869 491 2 × 2 = 0 + 0.444 444 444 487 738 982 4;
  • 23) 0.444 444 444 487 738 982 4 × 2 = 0 + 0.888 888 888 975 477 964 8;
  • 24) 0.888 888 888 975 477 964 8 × 2 = 1 + 0.777 777 777 950 955 929 6;
  • 25) 0.777 777 777 950 955 929 6 × 2 = 1 + 0.555 555 555 901 911 859 2;
  • 26) 0.555 555 555 901 911 859 2 × 2 = 1 + 0.111 111 111 803 823 718 4;
  • 27) 0.111 111 111 803 823 718 4 × 2 = 0 + 0.222 222 223 607 647 436 8;
  • 28) 0.222 222 223 607 647 436 8 × 2 = 0 + 0.444 444 447 215 294 873 6;
  • 29) 0.444 444 447 215 294 873 6 × 2 = 0 + 0.888 888 894 430 589 747 2;
  • 30) 0.888 888 894 430 589 747 2 × 2 = 1 + 0.777 777 788 861 179 494 4;
  • 31) 0.777 777 788 861 179 494 4 × 2 = 1 + 0.555 555 577 722 358 988 8;
  • 32) 0.555 555 577 722 358 988 8 × 2 = 1 + 0.111 111 155 444 717 977 6;
  • 33) 0.111 111 155 444 717 977 6 × 2 = 0 + 0.222 222 310 889 435 955 2;
  • 34) 0.222 222 310 889 435 955 2 × 2 = 0 + 0.444 444 621 778 871 910 4;
  • 35) 0.444 444 621 778 871 910 4 × 2 = 0 + 0.888 889 243 557 743 820 8;
  • 36) 0.888 889 243 557 743 820 8 × 2 = 1 + 0.777 778 487 115 487 641 6;
  • 37) 0.777 778 487 115 487 641 6 × 2 = 1 + 0.555 556 974 230 975 283 2;
  • 38) 0.555 556 974 230 975 283 2 × 2 = 1 + 0.111 113 948 461 950 566 4;
  • 39) 0.111 113 948 461 950 566 4 × 2 = 0 + 0.222 227 896 923 901 132 8;
  • 40) 0.222 227 896 923 901 132 8 × 2 = 0 + 0.444 455 793 847 802 265 6;
  • 41) 0.444 455 793 847 802 265 6 × 2 = 0 + 0.888 911 587 695 604 531 2;
  • 42) 0.888 911 587 695 604 531 2 × 2 = 1 + 0.777 823 175 391 209 062 4;
  • 43) 0.777 823 175 391 209 062 4 × 2 = 1 + 0.555 646 350 782 418 124 8;
  • 44) 0.555 646 350 782 418 124 8 × 2 = 1 + 0.111 292 701 564 836 249 6;
  • 45) 0.111 292 701 564 836 249 6 × 2 = 0 + 0.222 585 403 129 672 499 2;
  • 46) 0.222 585 403 129 672 499 2 × 2 = 0 + 0.445 170 806 259 344 998 4;
  • 47) 0.445 170 806 259 344 998 4 × 2 = 0 + 0.890 341 612 518 689 996 8;
  • 48) 0.890 341 612 518 689 996 8 × 2 = 1 + 0.780 683 225 037 379 993 6;
  • 49) 0.780 683 225 037 379 993 6 × 2 = 1 + 0.561 366 450 074 759 987 2;
  • 50) 0.561 366 450 074 759 987 2 × 2 = 1 + 0.122 732 900 149 519 974 4;
  • 51) 0.122 732 900 149 519 974 4 × 2 = 0 + 0.245 465 800 299 039 948 8;
  • 52) 0.245 465 800 299 039 948 8 × 2 = 0 + 0.490 931 600 598 079 897 6;
  • 53) 0.490 931 600 598 079 897 6 × 2 = 0 + 0.981 863 201 196 159 795 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:


0.777 777 777 777 777 788 1(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 788 1(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:


24.777 777 777 777 777 788 1(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 24.777 777 777 777 777 788 1 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