24.777 777 777 777 666 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 666(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
24.777 777 777 777 666(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: 0.777 777 777 777 666.

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) 0.777 777 777 777 666 × 2 = 1 + 0.555 555 555 555 332;
  • 2) 0.555 555 555 555 332 × 2 = 1 + 0.111 111 111 110 664;
  • 3) 0.111 111 111 110 664 × 2 = 0 + 0.222 222 222 221 328;
  • 4) 0.222 222 222 221 328 × 2 = 0 + 0.444 444 444 442 656;
  • 5) 0.444 444 444 442 656 × 2 = 0 + 0.888 888 888 885 312;
  • 6) 0.888 888 888 885 312 × 2 = 1 + 0.777 777 777 770 624;
  • 7) 0.777 777 777 770 624 × 2 = 1 + 0.555 555 555 541 248;
  • 8) 0.555 555 555 541 248 × 2 = 1 + 0.111 111 111 082 496;
  • 9) 0.111 111 111 082 496 × 2 = 0 + 0.222 222 222 164 992;
  • 10) 0.222 222 222 164 992 × 2 = 0 + 0.444 444 444 329 984;
  • 11) 0.444 444 444 329 984 × 2 = 0 + 0.888 888 888 659 968;
  • 12) 0.888 888 888 659 968 × 2 = 1 + 0.777 777 777 319 936;
  • 13) 0.777 777 777 319 936 × 2 = 1 + 0.555 555 554 639 872;
  • 14) 0.555 555 554 639 872 × 2 = 1 + 0.111 111 109 279 744;
  • 15) 0.111 111 109 279 744 × 2 = 0 + 0.222 222 218 559 488;
  • 16) 0.222 222 218 559 488 × 2 = 0 + 0.444 444 437 118 976;
  • 17) 0.444 444 437 118 976 × 2 = 0 + 0.888 888 874 237 952;
  • 18) 0.888 888 874 237 952 × 2 = 1 + 0.777 777 748 475 904;
  • 19) 0.777 777 748 475 904 × 2 = 1 + 0.555 555 496 951 808;
  • 20) 0.555 555 496 951 808 × 2 = 1 + 0.111 110 993 903 616;
  • 21) 0.111 110 993 903 616 × 2 = 0 + 0.222 221 987 807 232;
  • 22) 0.222 221 987 807 232 × 2 = 0 + 0.444 443 975 614 464;
  • 23) 0.444 443 975 614 464 × 2 = 0 + 0.888 887 951 228 928;
  • 24) 0.888 887 951 228 928 × 2 = 1 + 0.777 775 902 457 856;
  • 25) 0.777 775 902 457 856 × 2 = 1 + 0.555 551 804 915 712;
  • 26) 0.555 551 804 915 712 × 2 = 1 + 0.111 103 609 831 424;
  • 27) 0.111 103 609 831 424 × 2 = 0 + 0.222 207 219 662 848;
  • 28) 0.222 207 219 662 848 × 2 = 0 + 0.444 414 439 325 696;
  • 29) 0.444 414 439 325 696 × 2 = 0 + 0.888 828 878 651 392;
  • 30) 0.888 828 878 651 392 × 2 = 1 + 0.777 657 757 302 784;
  • 31) 0.777 657 757 302 784 × 2 = 1 + 0.555 315 514 605 568;
  • 32) 0.555 315 514 605 568 × 2 = 1 + 0.110 631 029 211 136;
  • 33) 0.110 631 029 211 136 × 2 = 0 + 0.221 262 058 422 272;
  • 34) 0.221 262 058 422 272 × 2 = 0 + 0.442 524 116 844 544;
  • 35) 0.442 524 116 844 544 × 2 = 0 + 0.885 048 233 689 088;
  • 36) 0.885 048 233 689 088 × 2 = 1 + 0.770 096 467 378 176;
  • 37) 0.770 096 467 378 176 × 2 = 1 + 0.540 192 934 756 352;
  • 38) 0.540 192 934 756 352 × 2 = 1 + 0.080 385 869 512 704;
  • 39) 0.080 385 869 512 704 × 2 = 0 + 0.160 771 739 025 408;
  • 40) 0.160 771 739 025 408 × 2 = 0 + 0.321 543 478 050 816;
  • 41) 0.321 543 478 050 816 × 2 = 0 + 0.643 086 956 101 632;
  • 42) 0.643 086 956 101 632 × 2 = 1 + 0.286 173 912 203 264;
  • 43) 0.286 173 912 203 264 × 2 = 0 + 0.572 347 824 406 528;
  • 44) 0.572 347 824 406 528 × 2 = 1 + 0.144 695 648 813 056;
  • 45) 0.144 695 648 813 056 × 2 = 0 + 0.289 391 297 626 112;
  • 46) 0.289 391 297 626 112 × 2 = 0 + 0.578 782 595 252 224;
  • 47) 0.578 782 595 252 224 × 2 = 1 + 0.157 565 190 504 448;
  • 48) 0.157 565 190 504 448 × 2 = 0 + 0.315 130 381 008 896;
  • 49) 0.315 130 381 008 896 × 2 = 0 + 0.630 260 762 017 792;
  • 50) 0.630 260 762 017 792 × 2 = 1 + 0.260 521 524 035 584;
  • 51) 0.260 521 524 035 584 × 2 = 0 + 0.521 043 048 071 168;
  • 52) 0.521 043 048 071 168 × 2 = 1 + 0.042 086 096 142 336;
  • 53) 0.042 086 096 142 336 × 2 = 0 + 0.084 172 192 284 672;

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:


0.777 777 777 777 666(10) =

0.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0101 0010 0101 0(2)

5. Positive number before normalization:

24.777 777 777 777 666(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0101 0010 0101 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:


24.777 777 777 777 666(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0101 0010 0101 0(2) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0101 0010 0101 0(2) × 20 =

1.1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0101 0010 0101 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 0101 0010 0101 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 0101 0010 0 1010 =

1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0101 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 0101 0010


Decimal number 24.777 777 777 777 666 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 0101 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