24.777 777 777 777 952 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 952(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
24.777 777 777 777 952(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: 0.777 777 777 777 952.

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) 0.777 777 777 777 952 × 2 = 1 + 0.555 555 555 555 904;
  • 2) 0.555 555 555 555 904 × 2 = 1 + 0.111 111 111 111 808;
  • 3) 0.111 111 111 111 808 × 2 = 0 + 0.222 222 222 223 616;
  • 4) 0.222 222 222 223 616 × 2 = 0 + 0.444 444 444 447 232;
  • 5) 0.444 444 444 447 232 × 2 = 0 + 0.888 888 888 894 464;
  • 6) 0.888 888 888 894 464 × 2 = 1 + 0.777 777 777 788 928;
  • 7) 0.777 777 777 788 928 × 2 = 1 + 0.555 555 555 577 856;
  • 8) 0.555 555 555 577 856 × 2 = 1 + 0.111 111 111 155 712;
  • 9) 0.111 111 111 155 712 × 2 = 0 + 0.222 222 222 311 424;
  • 10) 0.222 222 222 311 424 × 2 = 0 + 0.444 444 444 622 848;
  • 11) 0.444 444 444 622 848 × 2 = 0 + 0.888 888 889 245 696;
  • 12) 0.888 888 889 245 696 × 2 = 1 + 0.777 777 778 491 392;
  • 13) 0.777 777 778 491 392 × 2 = 1 + 0.555 555 556 982 784;
  • 14) 0.555 555 556 982 784 × 2 = 1 + 0.111 111 113 965 568;
  • 15) 0.111 111 113 965 568 × 2 = 0 + 0.222 222 227 931 136;
  • 16) 0.222 222 227 931 136 × 2 = 0 + 0.444 444 455 862 272;
  • 17) 0.444 444 455 862 272 × 2 = 0 + 0.888 888 911 724 544;
  • 18) 0.888 888 911 724 544 × 2 = 1 + 0.777 777 823 449 088;
  • 19) 0.777 777 823 449 088 × 2 = 1 + 0.555 555 646 898 176;
  • 20) 0.555 555 646 898 176 × 2 = 1 + 0.111 111 293 796 352;
  • 21) 0.111 111 293 796 352 × 2 = 0 + 0.222 222 587 592 704;
  • 22) 0.222 222 587 592 704 × 2 = 0 + 0.444 445 175 185 408;
  • 23) 0.444 445 175 185 408 × 2 = 0 + 0.888 890 350 370 816;
  • 24) 0.888 890 350 370 816 × 2 = 1 + 0.777 780 700 741 632;
  • 25) 0.777 780 700 741 632 × 2 = 1 + 0.555 561 401 483 264;
  • 26) 0.555 561 401 483 264 × 2 = 1 + 0.111 122 802 966 528;
  • 27) 0.111 122 802 966 528 × 2 = 0 + 0.222 245 605 933 056;
  • 28) 0.222 245 605 933 056 × 2 = 0 + 0.444 491 211 866 112;
  • 29) 0.444 491 211 866 112 × 2 = 0 + 0.888 982 423 732 224;
  • 30) 0.888 982 423 732 224 × 2 = 1 + 0.777 964 847 464 448;
  • 31) 0.777 964 847 464 448 × 2 = 1 + 0.555 929 694 928 896;
  • 32) 0.555 929 694 928 896 × 2 = 1 + 0.111 859 389 857 792;
  • 33) 0.111 859 389 857 792 × 2 = 0 + 0.223 718 779 715 584;
  • 34) 0.223 718 779 715 584 × 2 = 0 + 0.447 437 559 431 168;
  • 35) 0.447 437 559 431 168 × 2 = 0 + 0.894 875 118 862 336;
  • 36) 0.894 875 118 862 336 × 2 = 1 + 0.789 750 237 724 672;
  • 37) 0.789 750 237 724 672 × 2 = 1 + 0.579 500 475 449 344;
  • 38) 0.579 500 475 449 344 × 2 = 1 + 0.159 000 950 898 688;
  • 39) 0.159 000 950 898 688 × 2 = 0 + 0.318 001 901 797 376;
  • 40) 0.318 001 901 797 376 × 2 = 0 + 0.636 003 803 594 752;
  • 41) 0.636 003 803 594 752 × 2 = 1 + 0.272 007 607 189 504;
  • 42) 0.272 007 607 189 504 × 2 = 0 + 0.544 015 214 379 008;
  • 43) 0.544 015 214 379 008 × 2 = 1 + 0.088 030 428 758 016;
  • 44) 0.088 030 428 758 016 × 2 = 0 + 0.176 060 857 516 032;
  • 45) 0.176 060 857 516 032 × 2 = 0 + 0.352 121 715 032 064;
  • 46) 0.352 121 715 032 064 × 2 = 0 + 0.704 243 430 064 128;
  • 47) 0.704 243 430 064 128 × 2 = 1 + 0.408 486 860 128 256;
  • 48) 0.408 486 860 128 256 × 2 = 0 + 0.816 973 720 256 512;
  • 49) 0.816 973 720 256 512 × 2 = 1 + 0.633 947 440 513 024;
  • 50) 0.633 947 440 513 024 × 2 = 1 + 0.267 894 881 026 048;
  • 51) 0.267 894 881 026 048 × 2 = 0 + 0.535 789 762 052 096;
  • 52) 0.535 789 762 052 096 × 2 = 1 + 0.071 579 524 104 192;
  • 53) 0.071 579 524 104 192 × 2 = 0 + 0.143 159 048 208 384;

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:


0.777 777 777 777 952(10) =

0.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1010 0010 1101 0(2)

5. Positive number before normalization:

24.777 777 777 777 952(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1010 0010 1101 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:


24.777 777 777 777 952(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1010 0010 1101 0(2) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1010 0010 1101 0(2) × 20 =

1.1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1010 0010 1101 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 1010 0010 1101 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 1010 0010 1 1010 =

1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1010 0010


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 1010 0010


Decimal number 24.777 777 777 777 952 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 1010 0010

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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