24.777 777 777 777 777 777 806 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 777 806 1(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
24.777 777 777 777 777 777 806 1(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: 0.777 777 777 777 777 777 806 1.

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) 0.777 777 777 777 777 777 806 1 × 2 = 1 + 0.555 555 555 555 555 555 612 2;
  • 2) 0.555 555 555 555 555 555 612 2 × 2 = 1 + 0.111 111 111 111 111 111 224 4;
  • 3) 0.111 111 111 111 111 111 224 4 × 2 = 0 + 0.222 222 222 222 222 222 448 8;
  • 4) 0.222 222 222 222 222 222 448 8 × 2 = 0 + 0.444 444 444 444 444 444 897 6;
  • 5) 0.444 444 444 444 444 444 897 6 × 2 = 0 + 0.888 888 888 888 888 889 795 2;
  • 6) 0.888 888 888 888 888 889 795 2 × 2 = 1 + 0.777 777 777 777 777 779 590 4;
  • 7) 0.777 777 777 777 777 779 590 4 × 2 = 1 + 0.555 555 555 555 555 559 180 8;
  • 8) 0.555 555 555 555 555 559 180 8 × 2 = 1 + 0.111 111 111 111 111 118 361 6;
  • 9) 0.111 111 111 111 111 118 361 6 × 2 = 0 + 0.222 222 222 222 222 236 723 2;
  • 10) 0.222 222 222 222 222 236 723 2 × 2 = 0 + 0.444 444 444 444 444 473 446 4;
  • 11) 0.444 444 444 444 444 473 446 4 × 2 = 0 + 0.888 888 888 888 888 946 892 8;
  • 12) 0.888 888 888 888 888 946 892 8 × 2 = 1 + 0.777 777 777 777 777 893 785 6;
  • 13) 0.777 777 777 777 777 893 785 6 × 2 = 1 + 0.555 555 555 555 555 787 571 2;
  • 14) 0.555 555 555 555 555 787 571 2 × 2 = 1 + 0.111 111 111 111 111 575 142 4;
  • 15) 0.111 111 111 111 111 575 142 4 × 2 = 0 + 0.222 222 222 222 223 150 284 8;
  • 16) 0.222 222 222 222 223 150 284 8 × 2 = 0 + 0.444 444 444 444 446 300 569 6;
  • 17) 0.444 444 444 444 446 300 569 6 × 2 = 0 + 0.888 888 888 888 892 601 139 2;
  • 18) 0.888 888 888 888 892 601 139 2 × 2 = 1 + 0.777 777 777 777 785 202 278 4;
  • 19) 0.777 777 777 777 785 202 278 4 × 2 = 1 + 0.555 555 555 555 570 404 556 8;
  • 20) 0.555 555 555 555 570 404 556 8 × 2 = 1 + 0.111 111 111 111 140 809 113 6;
  • 21) 0.111 111 111 111 140 809 113 6 × 2 = 0 + 0.222 222 222 222 281 618 227 2;
  • 22) 0.222 222 222 222 281 618 227 2 × 2 = 0 + 0.444 444 444 444 563 236 454 4;
  • 23) 0.444 444 444 444 563 236 454 4 × 2 = 0 + 0.888 888 888 889 126 472 908 8;
  • 24) 0.888 888 888 889 126 472 908 8 × 2 = 1 + 0.777 777 777 778 252 945 817 6;
  • 25) 0.777 777 777 778 252 945 817 6 × 2 = 1 + 0.555 555 555 556 505 891 635 2;
  • 26) 0.555 555 555 556 505 891 635 2 × 2 = 1 + 0.111 111 111 113 011 783 270 4;
  • 27) 0.111 111 111 113 011 783 270 4 × 2 = 0 + 0.222 222 222 226 023 566 540 8;
  • 28) 0.222 222 222 226 023 566 540 8 × 2 = 0 + 0.444 444 444 452 047 133 081 6;
  • 29) 0.444 444 444 452 047 133 081 6 × 2 = 0 + 0.888 888 888 904 094 266 163 2;
  • 30) 0.888 888 888 904 094 266 163 2 × 2 = 1 + 0.777 777 777 808 188 532 326 4;
  • 31) 0.777 777 777 808 188 532 326 4 × 2 = 1 + 0.555 555 555 616 377 064 652 8;
  • 32) 0.555 555 555 616 377 064 652 8 × 2 = 1 + 0.111 111 111 232 754 129 305 6;
  • 33) 0.111 111 111 232 754 129 305 6 × 2 = 0 + 0.222 222 222 465 508 258 611 2;
  • 34) 0.222 222 222 465 508 258 611 2 × 2 = 0 + 0.444 444 444 931 016 517 222 4;
  • 35) 0.444 444 444 931 016 517 222 4 × 2 = 0 + 0.888 888 889 862 033 034 444 8;
  • 36) 0.888 888 889 862 033 034 444 8 × 2 = 1 + 0.777 777 779 724 066 068 889 6;
  • 37) 0.777 777 779 724 066 068 889 6 × 2 = 1 + 0.555 555 559 448 132 137 779 2;
  • 38) 0.555 555 559 448 132 137 779 2 × 2 = 1 + 0.111 111 118 896 264 275 558 4;
  • 39) 0.111 111 118 896 264 275 558 4 × 2 = 0 + 0.222 222 237 792 528 551 116 8;
  • 40) 0.222 222 237 792 528 551 116 8 × 2 = 0 + 0.444 444 475 585 057 102 233 6;
  • 41) 0.444 444 475 585 057 102 233 6 × 2 = 0 + 0.888 888 951 170 114 204 467 2;
  • 42) 0.888 888 951 170 114 204 467 2 × 2 = 1 + 0.777 777 902 340 228 408 934 4;
  • 43) 0.777 777 902 340 228 408 934 4 × 2 = 1 + 0.555 555 804 680 456 817 868 8;
  • 44) 0.555 555 804 680 456 817 868 8 × 2 = 1 + 0.111 111 609 360 913 635 737 6;
  • 45) 0.111 111 609 360 913 635 737 6 × 2 = 0 + 0.222 223 218 721 827 271 475 2;
  • 46) 0.222 223 218 721 827 271 475 2 × 2 = 0 + 0.444 446 437 443 654 542 950 4;
  • 47) 0.444 446 437 443 654 542 950 4 × 2 = 0 + 0.888 892 874 887 309 085 900 8;
  • 48) 0.888 892 874 887 309 085 900 8 × 2 = 1 + 0.777 785 749 774 618 171 801 6;
  • 49) 0.777 785 749 774 618 171 801 6 × 2 = 1 + 0.555 571 499 549 236 343 603 2;
  • 50) 0.555 571 499 549 236 343 603 2 × 2 = 1 + 0.111 142 999 098 472 687 206 4;
  • 51) 0.111 142 999 098 472 687 206 4 × 2 = 0 + 0.222 285 998 196 945 374 412 8;
  • 52) 0.222 285 998 196 945 374 412 8 × 2 = 0 + 0.444 571 996 393 890 748 825 6;
  • 53) 0.444 571 996 393 890 748 825 6 × 2 = 0 + 0.889 143 992 787 781 497 651 2;

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:


0.777 777 777 777 777 777 806 1(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 777 806 1(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:


24.777 777 777 777 777 777 806 1(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 24.777 777 777 777 777 777 806 1 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