24.777 777 777 777 874 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 874(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
24.777 777 777 777 874(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: 0.777 777 777 777 874.

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) 0.777 777 777 777 874 × 2 = 1 + 0.555 555 555 555 748;
  • 2) 0.555 555 555 555 748 × 2 = 1 + 0.111 111 111 111 496;
  • 3) 0.111 111 111 111 496 × 2 = 0 + 0.222 222 222 222 992;
  • 4) 0.222 222 222 222 992 × 2 = 0 + 0.444 444 444 445 984;
  • 5) 0.444 444 444 445 984 × 2 = 0 + 0.888 888 888 891 968;
  • 6) 0.888 888 888 891 968 × 2 = 1 + 0.777 777 777 783 936;
  • 7) 0.777 777 777 783 936 × 2 = 1 + 0.555 555 555 567 872;
  • 8) 0.555 555 555 567 872 × 2 = 1 + 0.111 111 111 135 744;
  • 9) 0.111 111 111 135 744 × 2 = 0 + 0.222 222 222 271 488;
  • 10) 0.222 222 222 271 488 × 2 = 0 + 0.444 444 444 542 976;
  • 11) 0.444 444 444 542 976 × 2 = 0 + 0.888 888 889 085 952;
  • 12) 0.888 888 889 085 952 × 2 = 1 + 0.777 777 778 171 904;
  • 13) 0.777 777 778 171 904 × 2 = 1 + 0.555 555 556 343 808;
  • 14) 0.555 555 556 343 808 × 2 = 1 + 0.111 111 112 687 616;
  • 15) 0.111 111 112 687 616 × 2 = 0 + 0.222 222 225 375 232;
  • 16) 0.222 222 225 375 232 × 2 = 0 + 0.444 444 450 750 464;
  • 17) 0.444 444 450 750 464 × 2 = 0 + 0.888 888 901 500 928;
  • 18) 0.888 888 901 500 928 × 2 = 1 + 0.777 777 803 001 856;
  • 19) 0.777 777 803 001 856 × 2 = 1 + 0.555 555 606 003 712;
  • 20) 0.555 555 606 003 712 × 2 = 1 + 0.111 111 212 007 424;
  • 21) 0.111 111 212 007 424 × 2 = 0 + 0.222 222 424 014 848;
  • 22) 0.222 222 424 014 848 × 2 = 0 + 0.444 444 848 029 696;
  • 23) 0.444 444 848 029 696 × 2 = 0 + 0.888 889 696 059 392;
  • 24) 0.888 889 696 059 392 × 2 = 1 + 0.777 779 392 118 784;
  • 25) 0.777 779 392 118 784 × 2 = 1 + 0.555 558 784 237 568;
  • 26) 0.555 558 784 237 568 × 2 = 1 + 0.111 117 568 475 136;
  • 27) 0.111 117 568 475 136 × 2 = 0 + 0.222 235 136 950 272;
  • 28) 0.222 235 136 950 272 × 2 = 0 + 0.444 470 273 900 544;
  • 29) 0.444 470 273 900 544 × 2 = 0 + 0.888 940 547 801 088;
  • 30) 0.888 940 547 801 088 × 2 = 1 + 0.777 881 095 602 176;
  • 31) 0.777 881 095 602 176 × 2 = 1 + 0.555 762 191 204 352;
  • 32) 0.555 762 191 204 352 × 2 = 1 + 0.111 524 382 408 704;
  • 33) 0.111 524 382 408 704 × 2 = 0 + 0.223 048 764 817 408;
  • 34) 0.223 048 764 817 408 × 2 = 0 + 0.446 097 529 634 816;
  • 35) 0.446 097 529 634 816 × 2 = 0 + 0.892 195 059 269 632;
  • 36) 0.892 195 059 269 632 × 2 = 1 + 0.784 390 118 539 264;
  • 37) 0.784 390 118 539 264 × 2 = 1 + 0.568 780 237 078 528;
  • 38) 0.568 780 237 078 528 × 2 = 1 + 0.137 560 474 157 056;
  • 39) 0.137 560 474 157 056 × 2 = 0 + 0.275 120 948 314 112;
  • 40) 0.275 120 948 314 112 × 2 = 0 + 0.550 241 896 628 224;
  • 41) 0.550 241 896 628 224 × 2 = 1 + 0.100 483 793 256 448;
  • 42) 0.100 483 793 256 448 × 2 = 0 + 0.200 967 586 512 896;
  • 43) 0.200 967 586 512 896 × 2 = 0 + 0.401 935 173 025 792;
  • 44) 0.401 935 173 025 792 × 2 = 0 + 0.803 870 346 051 584;
  • 45) 0.803 870 346 051 584 × 2 = 1 + 0.607 740 692 103 168;
  • 46) 0.607 740 692 103 168 × 2 = 1 + 0.215 481 384 206 336;
  • 47) 0.215 481 384 206 336 × 2 = 0 + 0.430 962 768 412 672;
  • 48) 0.430 962 768 412 672 × 2 = 0 + 0.861 925 536 825 344;
  • 49) 0.861 925 536 825 344 × 2 = 1 + 0.723 851 073 650 688;
  • 50) 0.723 851 073 650 688 × 2 = 1 + 0.447 702 147 301 376;
  • 51) 0.447 702 147 301 376 × 2 = 0 + 0.895 404 294 602 752;
  • 52) 0.895 404 294 602 752 × 2 = 1 + 0.790 808 589 205 504;
  • 53) 0.790 808 589 205 504 × 2 = 1 + 0.581 617 178 411 008;

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:


0.777 777 777 777 874(10) =

0.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1000 1100 1101 1(2)

5. Positive number before normalization:

24.777 777 777 777 874(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1000 1100 1101 1(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:


24.777 777 777 777 874(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1000 1100 1101 1(2) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1000 1100 1101 1(2) × 20 =

1.1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1000 1100 1101 1(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 1000 1100 1101 1


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 1000 1100 1 1011 =

1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1000 1100


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 1000 1100


Decimal number 24.777 777 777 777 874 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 1000 1100

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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