24.777 777 777 777 777 777 841 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 777 841(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
24.777 777 777 777 777 777 841(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: 0.777 777 777 777 777 777 841.

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) 0.777 777 777 777 777 777 841 × 2 = 1 + 0.555 555 555 555 555 555 682;
  • 2) 0.555 555 555 555 555 555 682 × 2 = 1 + 0.111 111 111 111 111 111 364;
  • 3) 0.111 111 111 111 111 111 364 × 2 = 0 + 0.222 222 222 222 222 222 728;
  • 4) 0.222 222 222 222 222 222 728 × 2 = 0 + 0.444 444 444 444 444 445 456;
  • 5) 0.444 444 444 444 444 445 456 × 2 = 0 + 0.888 888 888 888 888 890 912;
  • 6) 0.888 888 888 888 888 890 912 × 2 = 1 + 0.777 777 777 777 777 781 824;
  • 7) 0.777 777 777 777 777 781 824 × 2 = 1 + 0.555 555 555 555 555 563 648;
  • 8) 0.555 555 555 555 555 563 648 × 2 = 1 + 0.111 111 111 111 111 127 296;
  • 9) 0.111 111 111 111 111 127 296 × 2 = 0 + 0.222 222 222 222 222 254 592;
  • 10) 0.222 222 222 222 222 254 592 × 2 = 0 + 0.444 444 444 444 444 509 184;
  • 11) 0.444 444 444 444 444 509 184 × 2 = 0 + 0.888 888 888 888 889 018 368;
  • 12) 0.888 888 888 888 889 018 368 × 2 = 1 + 0.777 777 777 777 778 036 736;
  • 13) 0.777 777 777 777 778 036 736 × 2 = 1 + 0.555 555 555 555 556 073 472;
  • 14) 0.555 555 555 555 556 073 472 × 2 = 1 + 0.111 111 111 111 112 146 944;
  • 15) 0.111 111 111 111 112 146 944 × 2 = 0 + 0.222 222 222 222 224 293 888;
  • 16) 0.222 222 222 222 224 293 888 × 2 = 0 + 0.444 444 444 444 448 587 776;
  • 17) 0.444 444 444 444 448 587 776 × 2 = 0 + 0.888 888 888 888 897 175 552;
  • 18) 0.888 888 888 888 897 175 552 × 2 = 1 + 0.777 777 777 777 794 351 104;
  • 19) 0.777 777 777 777 794 351 104 × 2 = 1 + 0.555 555 555 555 588 702 208;
  • 20) 0.555 555 555 555 588 702 208 × 2 = 1 + 0.111 111 111 111 177 404 416;
  • 21) 0.111 111 111 111 177 404 416 × 2 = 0 + 0.222 222 222 222 354 808 832;
  • 22) 0.222 222 222 222 354 808 832 × 2 = 0 + 0.444 444 444 444 709 617 664;
  • 23) 0.444 444 444 444 709 617 664 × 2 = 0 + 0.888 888 888 889 419 235 328;
  • 24) 0.888 888 888 889 419 235 328 × 2 = 1 + 0.777 777 777 778 838 470 656;
  • 25) 0.777 777 777 778 838 470 656 × 2 = 1 + 0.555 555 555 557 676 941 312;
  • 26) 0.555 555 555 557 676 941 312 × 2 = 1 + 0.111 111 111 115 353 882 624;
  • 27) 0.111 111 111 115 353 882 624 × 2 = 0 + 0.222 222 222 230 707 765 248;
  • 28) 0.222 222 222 230 707 765 248 × 2 = 0 + 0.444 444 444 461 415 530 496;
  • 29) 0.444 444 444 461 415 530 496 × 2 = 0 + 0.888 888 888 922 831 060 992;
  • 30) 0.888 888 888 922 831 060 992 × 2 = 1 + 0.777 777 777 845 662 121 984;
  • 31) 0.777 777 777 845 662 121 984 × 2 = 1 + 0.555 555 555 691 324 243 968;
  • 32) 0.555 555 555 691 324 243 968 × 2 = 1 + 0.111 111 111 382 648 487 936;
  • 33) 0.111 111 111 382 648 487 936 × 2 = 0 + 0.222 222 222 765 296 975 872;
  • 34) 0.222 222 222 765 296 975 872 × 2 = 0 + 0.444 444 445 530 593 951 744;
  • 35) 0.444 444 445 530 593 951 744 × 2 = 0 + 0.888 888 891 061 187 903 488;
  • 36) 0.888 888 891 061 187 903 488 × 2 = 1 + 0.777 777 782 122 375 806 976;
  • 37) 0.777 777 782 122 375 806 976 × 2 = 1 + 0.555 555 564 244 751 613 952;
  • 38) 0.555 555 564 244 751 613 952 × 2 = 1 + 0.111 111 128 489 503 227 904;
  • 39) 0.111 111 128 489 503 227 904 × 2 = 0 + 0.222 222 256 979 006 455 808;
  • 40) 0.222 222 256 979 006 455 808 × 2 = 0 + 0.444 444 513 958 012 911 616;
  • 41) 0.444 444 513 958 012 911 616 × 2 = 0 + 0.888 889 027 916 025 823 232;
  • 42) 0.888 889 027 916 025 823 232 × 2 = 1 + 0.777 778 055 832 051 646 464;
  • 43) 0.777 778 055 832 051 646 464 × 2 = 1 + 0.555 556 111 664 103 292 928;
  • 44) 0.555 556 111 664 103 292 928 × 2 = 1 + 0.111 112 223 328 206 585 856;
  • 45) 0.111 112 223 328 206 585 856 × 2 = 0 + 0.222 224 446 656 413 171 712;
  • 46) 0.222 224 446 656 413 171 712 × 2 = 0 + 0.444 448 893 312 826 343 424;
  • 47) 0.444 448 893 312 826 343 424 × 2 = 0 + 0.888 897 786 625 652 686 848;
  • 48) 0.888 897 786 625 652 686 848 × 2 = 1 + 0.777 795 573 251 305 373 696;
  • 49) 0.777 795 573 251 305 373 696 × 2 = 1 + 0.555 591 146 502 610 747 392;
  • 50) 0.555 591 146 502 610 747 392 × 2 = 1 + 0.111 182 293 005 221 494 784;
  • 51) 0.111 182 293 005 221 494 784 × 2 = 0 + 0.222 364 586 010 442 989 568;
  • 52) 0.222 364 586 010 442 989 568 × 2 = 0 + 0.444 729 172 020 885 979 136;
  • 53) 0.444 729 172 020 885 979 136 × 2 = 0 + 0.889 458 344 041 771 958 272;

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:


0.777 777 777 777 777 777 841(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 777 841(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:


24.777 777 777 777 777 777 841(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 24.777 777 777 777 777 777 841 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