24.777 777 777 777 777 777 567 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 777 567(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
24.777 777 777 777 777 777 567(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: 0.777 777 777 777 777 777 567.

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) 0.777 777 777 777 777 777 567 × 2 = 1 + 0.555 555 555 555 555 555 134;
  • 2) 0.555 555 555 555 555 555 134 × 2 = 1 + 0.111 111 111 111 111 110 268;
  • 3) 0.111 111 111 111 111 110 268 × 2 = 0 + 0.222 222 222 222 222 220 536;
  • 4) 0.222 222 222 222 222 220 536 × 2 = 0 + 0.444 444 444 444 444 441 072;
  • 5) 0.444 444 444 444 444 441 072 × 2 = 0 + 0.888 888 888 888 888 882 144;
  • 6) 0.888 888 888 888 888 882 144 × 2 = 1 + 0.777 777 777 777 777 764 288;
  • 7) 0.777 777 777 777 777 764 288 × 2 = 1 + 0.555 555 555 555 555 528 576;
  • 8) 0.555 555 555 555 555 528 576 × 2 = 1 + 0.111 111 111 111 111 057 152;
  • 9) 0.111 111 111 111 111 057 152 × 2 = 0 + 0.222 222 222 222 222 114 304;
  • 10) 0.222 222 222 222 222 114 304 × 2 = 0 + 0.444 444 444 444 444 228 608;
  • 11) 0.444 444 444 444 444 228 608 × 2 = 0 + 0.888 888 888 888 888 457 216;
  • 12) 0.888 888 888 888 888 457 216 × 2 = 1 + 0.777 777 777 777 776 914 432;
  • 13) 0.777 777 777 777 776 914 432 × 2 = 1 + 0.555 555 555 555 553 828 864;
  • 14) 0.555 555 555 555 553 828 864 × 2 = 1 + 0.111 111 111 111 107 657 728;
  • 15) 0.111 111 111 111 107 657 728 × 2 = 0 + 0.222 222 222 222 215 315 456;
  • 16) 0.222 222 222 222 215 315 456 × 2 = 0 + 0.444 444 444 444 430 630 912;
  • 17) 0.444 444 444 444 430 630 912 × 2 = 0 + 0.888 888 888 888 861 261 824;
  • 18) 0.888 888 888 888 861 261 824 × 2 = 1 + 0.777 777 777 777 722 523 648;
  • 19) 0.777 777 777 777 722 523 648 × 2 = 1 + 0.555 555 555 555 445 047 296;
  • 20) 0.555 555 555 555 445 047 296 × 2 = 1 + 0.111 111 111 110 890 094 592;
  • 21) 0.111 111 111 110 890 094 592 × 2 = 0 + 0.222 222 222 221 780 189 184;
  • 22) 0.222 222 222 221 780 189 184 × 2 = 0 + 0.444 444 444 443 560 378 368;
  • 23) 0.444 444 444 443 560 378 368 × 2 = 0 + 0.888 888 888 887 120 756 736;
  • 24) 0.888 888 888 887 120 756 736 × 2 = 1 + 0.777 777 777 774 241 513 472;
  • 25) 0.777 777 777 774 241 513 472 × 2 = 1 + 0.555 555 555 548 483 026 944;
  • 26) 0.555 555 555 548 483 026 944 × 2 = 1 + 0.111 111 111 096 966 053 888;
  • 27) 0.111 111 111 096 966 053 888 × 2 = 0 + 0.222 222 222 193 932 107 776;
  • 28) 0.222 222 222 193 932 107 776 × 2 = 0 + 0.444 444 444 387 864 215 552;
  • 29) 0.444 444 444 387 864 215 552 × 2 = 0 + 0.888 888 888 775 728 431 104;
  • 30) 0.888 888 888 775 728 431 104 × 2 = 1 + 0.777 777 777 551 456 862 208;
  • 31) 0.777 777 777 551 456 862 208 × 2 = 1 + 0.555 555 555 102 913 724 416;
  • 32) 0.555 555 555 102 913 724 416 × 2 = 1 + 0.111 111 110 205 827 448 832;
  • 33) 0.111 111 110 205 827 448 832 × 2 = 0 + 0.222 222 220 411 654 897 664;
  • 34) 0.222 222 220 411 654 897 664 × 2 = 0 + 0.444 444 440 823 309 795 328;
  • 35) 0.444 444 440 823 309 795 328 × 2 = 0 + 0.888 888 881 646 619 590 656;
  • 36) 0.888 888 881 646 619 590 656 × 2 = 1 + 0.777 777 763 293 239 181 312;
  • 37) 0.777 777 763 293 239 181 312 × 2 = 1 + 0.555 555 526 586 478 362 624;
  • 38) 0.555 555 526 586 478 362 624 × 2 = 1 + 0.111 111 053 172 956 725 248;
  • 39) 0.111 111 053 172 956 725 248 × 2 = 0 + 0.222 222 106 345 913 450 496;
  • 40) 0.222 222 106 345 913 450 496 × 2 = 0 + 0.444 444 212 691 826 900 992;
  • 41) 0.444 444 212 691 826 900 992 × 2 = 0 + 0.888 888 425 383 653 801 984;
  • 42) 0.888 888 425 383 653 801 984 × 2 = 1 + 0.777 776 850 767 307 603 968;
  • 43) 0.777 776 850 767 307 603 968 × 2 = 1 + 0.555 553 701 534 615 207 936;
  • 44) 0.555 553 701 534 615 207 936 × 2 = 1 + 0.111 107 403 069 230 415 872;
  • 45) 0.111 107 403 069 230 415 872 × 2 = 0 + 0.222 214 806 138 460 831 744;
  • 46) 0.222 214 806 138 460 831 744 × 2 = 0 + 0.444 429 612 276 921 663 488;
  • 47) 0.444 429 612 276 921 663 488 × 2 = 0 + 0.888 859 224 553 843 326 976;
  • 48) 0.888 859 224 553 843 326 976 × 2 = 1 + 0.777 718 449 107 686 653 952;
  • 49) 0.777 718 449 107 686 653 952 × 2 = 1 + 0.555 436 898 215 373 307 904;
  • 50) 0.555 436 898 215 373 307 904 × 2 = 1 + 0.110 873 796 430 746 615 808;
  • 51) 0.110 873 796 430 746 615 808 × 2 = 0 + 0.221 747 592 861 493 231 616;
  • 52) 0.221 747 592 861 493 231 616 × 2 = 0 + 0.443 495 185 722 986 463 232;
  • 53) 0.443 495 185 722 986 463 232 × 2 = 0 + 0.886 990 371 445 972 926 464;

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:


0.777 777 777 777 777 777 567(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 777 567(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:


24.777 777 777 777 777 777 567(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 24.777 777 777 777 777 777 567 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