24.777 777 777 777 777 777 778 089 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 777 778 089(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
24.777 777 777 777 777 777 778 089(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: 0.777 777 777 777 777 777 778 089.

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) 0.777 777 777 777 777 777 778 089 × 2 = 1 + 0.555 555 555 555 555 555 556 178;
  • 2) 0.555 555 555 555 555 555 556 178 × 2 = 1 + 0.111 111 111 111 111 111 112 356;
  • 3) 0.111 111 111 111 111 111 112 356 × 2 = 0 + 0.222 222 222 222 222 222 224 712;
  • 4) 0.222 222 222 222 222 222 224 712 × 2 = 0 + 0.444 444 444 444 444 444 449 424;
  • 5) 0.444 444 444 444 444 444 449 424 × 2 = 0 + 0.888 888 888 888 888 888 898 848;
  • 6) 0.888 888 888 888 888 888 898 848 × 2 = 1 + 0.777 777 777 777 777 777 797 696;
  • 7) 0.777 777 777 777 777 777 797 696 × 2 = 1 + 0.555 555 555 555 555 555 595 392;
  • 8) 0.555 555 555 555 555 555 595 392 × 2 = 1 + 0.111 111 111 111 111 111 190 784;
  • 9) 0.111 111 111 111 111 111 190 784 × 2 = 0 + 0.222 222 222 222 222 222 381 568;
  • 10) 0.222 222 222 222 222 222 381 568 × 2 = 0 + 0.444 444 444 444 444 444 763 136;
  • 11) 0.444 444 444 444 444 444 763 136 × 2 = 0 + 0.888 888 888 888 888 889 526 272;
  • 12) 0.888 888 888 888 888 889 526 272 × 2 = 1 + 0.777 777 777 777 777 779 052 544;
  • 13) 0.777 777 777 777 777 779 052 544 × 2 = 1 + 0.555 555 555 555 555 558 105 088;
  • 14) 0.555 555 555 555 555 558 105 088 × 2 = 1 + 0.111 111 111 111 111 116 210 176;
  • 15) 0.111 111 111 111 111 116 210 176 × 2 = 0 + 0.222 222 222 222 222 232 420 352;
  • 16) 0.222 222 222 222 222 232 420 352 × 2 = 0 + 0.444 444 444 444 444 464 840 704;
  • 17) 0.444 444 444 444 444 464 840 704 × 2 = 0 + 0.888 888 888 888 888 929 681 408;
  • 18) 0.888 888 888 888 888 929 681 408 × 2 = 1 + 0.777 777 777 777 777 859 362 816;
  • 19) 0.777 777 777 777 777 859 362 816 × 2 = 1 + 0.555 555 555 555 555 718 725 632;
  • 20) 0.555 555 555 555 555 718 725 632 × 2 = 1 + 0.111 111 111 111 111 437 451 264;
  • 21) 0.111 111 111 111 111 437 451 264 × 2 = 0 + 0.222 222 222 222 222 874 902 528;
  • 22) 0.222 222 222 222 222 874 902 528 × 2 = 0 + 0.444 444 444 444 445 749 805 056;
  • 23) 0.444 444 444 444 445 749 805 056 × 2 = 0 + 0.888 888 888 888 891 499 610 112;
  • 24) 0.888 888 888 888 891 499 610 112 × 2 = 1 + 0.777 777 777 777 782 999 220 224;
  • 25) 0.777 777 777 777 782 999 220 224 × 2 = 1 + 0.555 555 555 555 565 998 440 448;
  • 26) 0.555 555 555 555 565 998 440 448 × 2 = 1 + 0.111 111 111 111 131 996 880 896;
  • 27) 0.111 111 111 111 131 996 880 896 × 2 = 0 + 0.222 222 222 222 263 993 761 792;
  • 28) 0.222 222 222 222 263 993 761 792 × 2 = 0 + 0.444 444 444 444 527 987 523 584;
  • 29) 0.444 444 444 444 527 987 523 584 × 2 = 0 + 0.888 888 888 889 055 975 047 168;
  • 30) 0.888 888 888 889 055 975 047 168 × 2 = 1 + 0.777 777 777 778 111 950 094 336;
  • 31) 0.777 777 777 778 111 950 094 336 × 2 = 1 + 0.555 555 555 556 223 900 188 672;
  • 32) 0.555 555 555 556 223 900 188 672 × 2 = 1 + 0.111 111 111 112 447 800 377 344;
  • 33) 0.111 111 111 112 447 800 377 344 × 2 = 0 + 0.222 222 222 224 895 600 754 688;
  • 34) 0.222 222 222 224 895 600 754 688 × 2 = 0 + 0.444 444 444 449 791 201 509 376;
  • 35) 0.444 444 444 449 791 201 509 376 × 2 = 0 + 0.888 888 888 899 582 403 018 752;
  • 36) 0.888 888 888 899 582 403 018 752 × 2 = 1 + 0.777 777 777 799 164 806 037 504;
  • 37) 0.777 777 777 799 164 806 037 504 × 2 = 1 + 0.555 555 555 598 329 612 075 008;
  • 38) 0.555 555 555 598 329 612 075 008 × 2 = 1 + 0.111 111 111 196 659 224 150 016;
  • 39) 0.111 111 111 196 659 224 150 016 × 2 = 0 + 0.222 222 222 393 318 448 300 032;
  • 40) 0.222 222 222 393 318 448 300 032 × 2 = 0 + 0.444 444 444 786 636 896 600 064;
  • 41) 0.444 444 444 786 636 896 600 064 × 2 = 0 + 0.888 888 889 573 273 793 200 128;
  • 42) 0.888 888 889 573 273 793 200 128 × 2 = 1 + 0.777 777 779 146 547 586 400 256;
  • 43) 0.777 777 779 146 547 586 400 256 × 2 = 1 + 0.555 555 558 293 095 172 800 512;
  • 44) 0.555 555 558 293 095 172 800 512 × 2 = 1 + 0.111 111 116 586 190 345 601 024;
  • 45) 0.111 111 116 586 190 345 601 024 × 2 = 0 + 0.222 222 233 172 380 691 202 048;
  • 46) 0.222 222 233 172 380 691 202 048 × 2 = 0 + 0.444 444 466 344 761 382 404 096;
  • 47) 0.444 444 466 344 761 382 404 096 × 2 = 0 + 0.888 888 932 689 522 764 808 192;
  • 48) 0.888 888 932 689 522 764 808 192 × 2 = 1 + 0.777 777 865 379 045 529 616 384;
  • 49) 0.777 777 865 379 045 529 616 384 × 2 = 1 + 0.555 555 730 758 091 059 232 768;
  • 50) 0.555 555 730 758 091 059 232 768 × 2 = 1 + 0.111 111 461 516 182 118 465 536;
  • 51) 0.111 111 461 516 182 118 465 536 × 2 = 0 + 0.222 222 923 032 364 236 931 072;
  • 52) 0.222 222 923 032 364 236 931 072 × 2 = 0 + 0.444 445 846 064 728 473 862 144;
  • 53) 0.444 445 846 064 728 473 862 144 × 2 = 0 + 0.888 891 692 129 456 947 724 288;

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:


0.777 777 777 777 777 777 778 089(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 777 778 089(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:


24.777 777 777 777 777 777 778 089(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 24.777 777 777 777 777 777 778 089 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