24.777 777 777 777 777 777 779 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 777 779(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
24.777 777 777 777 777 777 779(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: 0.777 777 777 777 777 777 779.

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) 0.777 777 777 777 777 777 779 × 2 = 1 + 0.555 555 555 555 555 555 558;
  • 2) 0.555 555 555 555 555 555 558 × 2 = 1 + 0.111 111 111 111 111 111 116;
  • 3) 0.111 111 111 111 111 111 116 × 2 = 0 + 0.222 222 222 222 222 222 232;
  • 4) 0.222 222 222 222 222 222 232 × 2 = 0 + 0.444 444 444 444 444 444 464;
  • 5) 0.444 444 444 444 444 444 464 × 2 = 0 + 0.888 888 888 888 888 888 928;
  • 6) 0.888 888 888 888 888 888 928 × 2 = 1 + 0.777 777 777 777 777 777 856;
  • 7) 0.777 777 777 777 777 777 856 × 2 = 1 + 0.555 555 555 555 555 555 712;
  • 8) 0.555 555 555 555 555 555 712 × 2 = 1 + 0.111 111 111 111 111 111 424;
  • 9) 0.111 111 111 111 111 111 424 × 2 = 0 + 0.222 222 222 222 222 222 848;
  • 10) 0.222 222 222 222 222 222 848 × 2 = 0 + 0.444 444 444 444 444 445 696;
  • 11) 0.444 444 444 444 444 445 696 × 2 = 0 + 0.888 888 888 888 888 891 392;
  • 12) 0.888 888 888 888 888 891 392 × 2 = 1 + 0.777 777 777 777 777 782 784;
  • 13) 0.777 777 777 777 777 782 784 × 2 = 1 + 0.555 555 555 555 555 565 568;
  • 14) 0.555 555 555 555 555 565 568 × 2 = 1 + 0.111 111 111 111 111 131 136;
  • 15) 0.111 111 111 111 111 131 136 × 2 = 0 + 0.222 222 222 222 222 262 272;
  • 16) 0.222 222 222 222 222 262 272 × 2 = 0 + 0.444 444 444 444 444 524 544;
  • 17) 0.444 444 444 444 444 524 544 × 2 = 0 + 0.888 888 888 888 889 049 088;
  • 18) 0.888 888 888 888 889 049 088 × 2 = 1 + 0.777 777 777 777 778 098 176;
  • 19) 0.777 777 777 777 778 098 176 × 2 = 1 + 0.555 555 555 555 556 196 352;
  • 20) 0.555 555 555 555 556 196 352 × 2 = 1 + 0.111 111 111 111 112 392 704;
  • 21) 0.111 111 111 111 112 392 704 × 2 = 0 + 0.222 222 222 222 224 785 408;
  • 22) 0.222 222 222 222 224 785 408 × 2 = 0 + 0.444 444 444 444 449 570 816;
  • 23) 0.444 444 444 444 449 570 816 × 2 = 0 + 0.888 888 888 888 899 141 632;
  • 24) 0.888 888 888 888 899 141 632 × 2 = 1 + 0.777 777 777 777 798 283 264;
  • 25) 0.777 777 777 777 798 283 264 × 2 = 1 + 0.555 555 555 555 596 566 528;
  • 26) 0.555 555 555 555 596 566 528 × 2 = 1 + 0.111 111 111 111 193 133 056;
  • 27) 0.111 111 111 111 193 133 056 × 2 = 0 + 0.222 222 222 222 386 266 112;
  • 28) 0.222 222 222 222 386 266 112 × 2 = 0 + 0.444 444 444 444 772 532 224;
  • 29) 0.444 444 444 444 772 532 224 × 2 = 0 + 0.888 888 888 889 545 064 448;
  • 30) 0.888 888 888 889 545 064 448 × 2 = 1 + 0.777 777 777 779 090 128 896;
  • 31) 0.777 777 777 779 090 128 896 × 2 = 1 + 0.555 555 555 558 180 257 792;
  • 32) 0.555 555 555 558 180 257 792 × 2 = 1 + 0.111 111 111 116 360 515 584;
  • 33) 0.111 111 111 116 360 515 584 × 2 = 0 + 0.222 222 222 232 721 031 168;
  • 34) 0.222 222 222 232 721 031 168 × 2 = 0 + 0.444 444 444 465 442 062 336;
  • 35) 0.444 444 444 465 442 062 336 × 2 = 0 + 0.888 888 888 930 884 124 672;
  • 36) 0.888 888 888 930 884 124 672 × 2 = 1 + 0.777 777 777 861 768 249 344;
  • 37) 0.777 777 777 861 768 249 344 × 2 = 1 + 0.555 555 555 723 536 498 688;
  • 38) 0.555 555 555 723 536 498 688 × 2 = 1 + 0.111 111 111 447 072 997 376;
  • 39) 0.111 111 111 447 072 997 376 × 2 = 0 + 0.222 222 222 894 145 994 752;
  • 40) 0.222 222 222 894 145 994 752 × 2 = 0 + 0.444 444 445 788 291 989 504;
  • 41) 0.444 444 445 788 291 989 504 × 2 = 0 + 0.888 888 891 576 583 979 008;
  • 42) 0.888 888 891 576 583 979 008 × 2 = 1 + 0.777 777 783 153 167 958 016;
  • 43) 0.777 777 783 153 167 958 016 × 2 = 1 + 0.555 555 566 306 335 916 032;
  • 44) 0.555 555 566 306 335 916 032 × 2 = 1 + 0.111 111 132 612 671 832 064;
  • 45) 0.111 111 132 612 671 832 064 × 2 = 0 + 0.222 222 265 225 343 664 128;
  • 46) 0.222 222 265 225 343 664 128 × 2 = 0 + 0.444 444 530 450 687 328 256;
  • 47) 0.444 444 530 450 687 328 256 × 2 = 0 + 0.888 889 060 901 374 656 512;
  • 48) 0.888 889 060 901 374 656 512 × 2 = 1 + 0.777 778 121 802 749 313 024;
  • 49) 0.777 778 121 802 749 313 024 × 2 = 1 + 0.555 556 243 605 498 626 048;
  • 50) 0.555 556 243 605 498 626 048 × 2 = 1 + 0.111 112 487 210 997 252 096;
  • 51) 0.111 112 487 210 997 252 096 × 2 = 0 + 0.222 224 974 421 994 504 192;
  • 52) 0.222 224 974 421 994 504 192 × 2 = 0 + 0.444 449 948 843 989 008 384;
  • 53) 0.444 449 948 843 989 008 384 × 2 = 0 + 0.888 899 897 687 978 016 768;

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:


0.777 777 777 777 777 777 779(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 777 779(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:


24.777 777 777 777 777 777 779(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 24.777 777 777 777 777 777 779 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