Convert Decimal 24.777 777 777 777 777 777 777 777 777 705 9 to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 777 777 777 777 705 9(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
24.777 777 777 777 777 777 777 777 777 705 9(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: 0.777 777 777 777 777 777 777 777 777 705 9.

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) 0.777 777 777 777 777 777 777 777 777 705 9 × 2 = 1 + 0.555 555 555 555 555 555 555 555 555 411 8;
  • 2) 0.555 555 555 555 555 555 555 555 555 411 8 × 2 = 1 + 0.111 111 111 111 111 111 111 111 110 823 6;
  • 3) 0.111 111 111 111 111 111 111 111 110 823 6 × 2 = 0 + 0.222 222 222 222 222 222 222 222 221 647 2;
  • 4) 0.222 222 222 222 222 222 222 222 221 647 2 × 2 = 0 + 0.444 444 444 444 444 444 444 444 443 294 4;
  • 5) 0.444 444 444 444 444 444 444 444 443 294 4 × 2 = 0 + 0.888 888 888 888 888 888 888 888 886 588 8;
  • 6) 0.888 888 888 888 888 888 888 888 886 588 8 × 2 = 1 + 0.777 777 777 777 777 777 777 777 773 177 6;
  • 7) 0.777 777 777 777 777 777 777 777 773 177 6 × 2 = 1 + 0.555 555 555 555 555 555 555 555 546 355 2;
  • 8) 0.555 555 555 555 555 555 555 555 546 355 2 × 2 = 1 + 0.111 111 111 111 111 111 111 111 092 710 4;
  • 9) 0.111 111 111 111 111 111 111 111 092 710 4 × 2 = 0 + 0.222 222 222 222 222 222 222 222 185 420 8;
  • 10) 0.222 222 222 222 222 222 222 222 185 420 8 × 2 = 0 + 0.444 444 444 444 444 444 444 444 370 841 6;
  • 11) 0.444 444 444 444 444 444 444 444 370 841 6 × 2 = 0 + 0.888 888 888 888 888 888 888 888 741 683 2;
  • 12) 0.888 888 888 888 888 888 888 888 741 683 2 × 2 = 1 + 0.777 777 777 777 777 777 777 777 483 366 4;
  • 13) 0.777 777 777 777 777 777 777 777 483 366 4 × 2 = 1 + 0.555 555 555 555 555 555 555 554 966 732 8;
  • 14) 0.555 555 555 555 555 555 555 554 966 732 8 × 2 = 1 + 0.111 111 111 111 111 111 111 109 933 465 6;
  • 15) 0.111 111 111 111 111 111 111 109 933 465 6 × 2 = 0 + 0.222 222 222 222 222 222 222 219 866 931 2;
  • 16) 0.222 222 222 222 222 222 222 219 866 931 2 × 2 = 0 + 0.444 444 444 444 444 444 444 439 733 862 4;
  • 17) 0.444 444 444 444 444 444 444 439 733 862 4 × 2 = 0 + 0.888 888 888 888 888 888 888 879 467 724 8;
  • 18) 0.888 888 888 888 888 888 888 879 467 724 8 × 2 = 1 + 0.777 777 777 777 777 777 777 758 935 449 6;
  • 19) 0.777 777 777 777 777 777 777 758 935 449 6 × 2 = 1 + 0.555 555 555 555 555 555 555 517 870 899 2;
  • 20) 0.555 555 555 555 555 555 555 517 870 899 2 × 2 = 1 + 0.111 111 111 111 111 111 111 035 741 798 4;
  • 21) 0.111 111 111 111 111 111 111 035 741 798 4 × 2 = 0 + 0.222 222 222 222 222 222 222 071 483 596 8;
  • 22) 0.222 222 222 222 222 222 222 071 483 596 8 × 2 = 0 + 0.444 444 444 444 444 444 444 142 967 193 6;
  • 23) 0.444 444 444 444 444 444 444 142 967 193 6 × 2 = 0 + 0.888 888 888 888 888 888 888 285 934 387 2;
  • 24) 0.888 888 888 888 888 888 888 285 934 387 2 × 2 = 1 + 0.777 777 777 777 777 777 776 571 868 774 4;
  • 25) 0.777 777 777 777 777 777 776 571 868 774 4 × 2 = 1 + 0.555 555 555 555 555 555 553 143 737 548 8;
  • 26) 0.555 555 555 555 555 555 553 143 737 548 8 × 2 = 1 + 0.111 111 111 111 111 111 106 287 475 097 6;
  • 27) 0.111 111 111 111 111 111 106 287 475 097 6 × 2 = 0 + 0.222 222 222 222 222 222 212 574 950 195 2;
  • 28) 0.222 222 222 222 222 222 212 574 950 195 2 × 2 = 0 + 0.444 444 444 444 444 444 425 149 900 390 4;
  • 29) 0.444 444 444 444 444 444 425 149 900 390 4 × 2 = 0 + 0.888 888 888 888 888 888 850 299 800 780 8;
  • 30) 0.888 888 888 888 888 888 850 299 800 780 8 × 2 = 1 + 0.777 777 777 777 777 777 700 599 601 561 6;
  • 31) 0.777 777 777 777 777 777 700 599 601 561 6 × 2 = 1 + 0.555 555 555 555 555 555 401 199 203 123 2;
  • 32) 0.555 555 555 555 555 555 401 199 203 123 2 × 2 = 1 + 0.111 111 111 111 111 110 802 398 406 246 4;
  • 33) 0.111 111 111 111 111 110 802 398 406 246 4 × 2 = 0 + 0.222 222 222 222 222 221 604 796 812 492 8;
  • 34) 0.222 222 222 222 222 221 604 796 812 492 8 × 2 = 0 + 0.444 444 444 444 444 443 209 593 624 985 6;
  • 35) 0.444 444 444 444 444 443 209 593 624 985 6 × 2 = 0 + 0.888 888 888 888 888 886 419 187 249 971 2;
  • 36) 0.888 888 888 888 888 886 419 187 249 971 2 × 2 = 1 + 0.777 777 777 777 777 772 838 374 499 942 4;
  • 37) 0.777 777 777 777 777 772 838 374 499 942 4 × 2 = 1 + 0.555 555 555 555 555 545 676 748 999 884 8;
  • 38) 0.555 555 555 555 555 545 676 748 999 884 8 × 2 = 1 + 0.111 111 111 111 111 091 353 497 999 769 6;
  • 39) 0.111 111 111 111 111 091 353 497 999 769 6 × 2 = 0 + 0.222 222 222 222 222 182 706 995 999 539 2;
  • 40) 0.222 222 222 222 222 182 706 995 999 539 2 × 2 = 0 + 0.444 444 444 444 444 365 413 991 999 078 4;
  • 41) 0.444 444 444 444 444 365 413 991 999 078 4 × 2 = 0 + 0.888 888 888 888 888 730 827 983 998 156 8;
  • 42) 0.888 888 888 888 888 730 827 983 998 156 8 × 2 = 1 + 0.777 777 777 777 777 461 655 967 996 313 6;
  • 43) 0.777 777 777 777 777 461 655 967 996 313 6 × 2 = 1 + 0.555 555 555 555 554 923 311 935 992 627 2;
  • 44) 0.555 555 555 555 554 923 311 935 992 627 2 × 2 = 1 + 0.111 111 111 111 109 846 623 871 985 254 4;
  • 45) 0.111 111 111 111 109 846 623 871 985 254 4 × 2 = 0 + 0.222 222 222 222 219 693 247 743 970 508 8;
  • 46) 0.222 222 222 222 219 693 247 743 970 508 8 × 2 = 0 + 0.444 444 444 444 439 386 495 487 941 017 6;
  • 47) 0.444 444 444 444 439 386 495 487 941 017 6 × 2 = 0 + 0.888 888 888 888 878 772 990 975 882 035 2;
  • 48) 0.888 888 888 888 878 772 990 975 882 035 2 × 2 = 1 + 0.777 777 777 777 757 545 981 951 764 070 4;
  • 49) 0.777 777 777 777 757 545 981 951 764 070 4 × 2 = 1 + 0.555 555 555 555 515 091 963 903 528 140 8;
  • 50) 0.555 555 555 555 515 091 963 903 528 140 8 × 2 = 1 + 0.111 111 111 111 030 183 927 807 056 281 6;
  • 51) 0.111 111 111 111 030 183 927 807 056 281 6 × 2 = 0 + 0.222 222 222 222 060 367 855 614 112 563 2;
  • 52) 0.222 222 222 222 060 367 855 614 112 563 2 × 2 = 0 + 0.444 444 444 444 120 735 711 228 225 126 4;
  • 53) 0.444 444 444 444 120 735 711 228 225 126 4 × 2 = 0 + 0.888 888 888 888 241 471 422 456 450 252 8;

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:


0.777 777 777 777 777 777 777 777 777 705 9(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 777 777 777 777 705 9(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:


24.777 777 777 777 777 777 777 777 777 705 9(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 24.777 777 777 777 777 777 777 777 777 705 9 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