24.777 777 777 777 777 777 51 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 777 51(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 51(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 51.

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 51 × 2 = 1 + 0.555 555 555 555 555 555 02;
  • 2) 0.555 555 555 555 555 555 02 × 2 = 1 + 0.111 111 111 111 111 110 04;
  • 3) 0.111 111 111 111 111 110 04 × 2 = 0 + 0.222 222 222 222 222 220 08;
  • 4) 0.222 222 222 222 222 220 08 × 2 = 0 + 0.444 444 444 444 444 440 16;
  • 5) 0.444 444 444 444 444 440 16 × 2 = 0 + 0.888 888 888 888 888 880 32;
  • 6) 0.888 888 888 888 888 880 32 × 2 = 1 + 0.777 777 777 777 777 760 64;
  • 7) 0.777 777 777 777 777 760 64 × 2 = 1 + 0.555 555 555 555 555 521 28;
  • 8) 0.555 555 555 555 555 521 28 × 2 = 1 + 0.111 111 111 111 111 042 56;
  • 9) 0.111 111 111 111 111 042 56 × 2 = 0 + 0.222 222 222 222 222 085 12;
  • 10) 0.222 222 222 222 222 085 12 × 2 = 0 + 0.444 444 444 444 444 170 24;
  • 11) 0.444 444 444 444 444 170 24 × 2 = 0 + 0.888 888 888 888 888 340 48;
  • 12) 0.888 888 888 888 888 340 48 × 2 = 1 + 0.777 777 777 777 776 680 96;
  • 13) 0.777 777 777 777 776 680 96 × 2 = 1 + 0.555 555 555 555 553 361 92;
  • 14) 0.555 555 555 555 553 361 92 × 2 = 1 + 0.111 111 111 111 106 723 84;
  • 15) 0.111 111 111 111 106 723 84 × 2 = 0 + 0.222 222 222 222 213 447 68;
  • 16) 0.222 222 222 222 213 447 68 × 2 = 0 + 0.444 444 444 444 426 895 36;
  • 17) 0.444 444 444 444 426 895 36 × 2 = 0 + 0.888 888 888 888 853 790 72;
  • 18) 0.888 888 888 888 853 790 72 × 2 = 1 + 0.777 777 777 777 707 581 44;
  • 19) 0.777 777 777 777 707 581 44 × 2 = 1 + 0.555 555 555 555 415 162 88;
  • 20) 0.555 555 555 555 415 162 88 × 2 = 1 + 0.111 111 111 110 830 325 76;
  • 21) 0.111 111 111 110 830 325 76 × 2 = 0 + 0.222 222 222 221 660 651 52;
  • 22) 0.222 222 222 221 660 651 52 × 2 = 0 + 0.444 444 444 443 321 303 04;
  • 23) 0.444 444 444 443 321 303 04 × 2 = 0 + 0.888 888 888 886 642 606 08;
  • 24) 0.888 888 888 886 642 606 08 × 2 = 1 + 0.777 777 777 773 285 212 16;
  • 25) 0.777 777 777 773 285 212 16 × 2 = 1 + 0.555 555 555 546 570 424 32;
  • 26) 0.555 555 555 546 570 424 32 × 2 = 1 + 0.111 111 111 093 140 848 64;
  • 27) 0.111 111 111 093 140 848 64 × 2 = 0 + 0.222 222 222 186 281 697 28;
  • 28) 0.222 222 222 186 281 697 28 × 2 = 0 + 0.444 444 444 372 563 394 56;
  • 29) 0.444 444 444 372 563 394 56 × 2 = 0 + 0.888 888 888 745 126 789 12;
  • 30) 0.888 888 888 745 126 789 12 × 2 = 1 + 0.777 777 777 490 253 578 24;
  • 31) 0.777 777 777 490 253 578 24 × 2 = 1 + 0.555 555 554 980 507 156 48;
  • 32) 0.555 555 554 980 507 156 48 × 2 = 1 + 0.111 111 109 961 014 312 96;
  • 33) 0.111 111 109 961 014 312 96 × 2 = 0 + 0.222 222 219 922 028 625 92;
  • 34) 0.222 222 219 922 028 625 92 × 2 = 0 + 0.444 444 439 844 057 251 84;
  • 35) 0.444 444 439 844 057 251 84 × 2 = 0 + 0.888 888 879 688 114 503 68;
  • 36) 0.888 888 879 688 114 503 68 × 2 = 1 + 0.777 777 759 376 229 007 36;
  • 37) 0.777 777 759 376 229 007 36 × 2 = 1 + 0.555 555 518 752 458 014 72;
  • 38) 0.555 555 518 752 458 014 72 × 2 = 1 + 0.111 111 037 504 916 029 44;
  • 39) 0.111 111 037 504 916 029 44 × 2 = 0 + 0.222 222 075 009 832 058 88;
  • 40) 0.222 222 075 009 832 058 88 × 2 = 0 + 0.444 444 150 019 664 117 76;
  • 41) 0.444 444 150 019 664 117 76 × 2 = 0 + 0.888 888 300 039 328 235 52;
  • 42) 0.888 888 300 039 328 235 52 × 2 = 1 + 0.777 776 600 078 656 471 04;
  • 43) 0.777 776 600 078 656 471 04 × 2 = 1 + 0.555 553 200 157 312 942 08;
  • 44) 0.555 553 200 157 312 942 08 × 2 = 1 + 0.111 106 400 314 625 884 16;
  • 45) 0.111 106 400 314 625 884 16 × 2 = 0 + 0.222 212 800 629 251 768 32;
  • 46) 0.222 212 800 629 251 768 32 × 2 = 0 + 0.444 425 601 258 503 536 64;
  • 47) 0.444 425 601 258 503 536 64 × 2 = 0 + 0.888 851 202 517 007 073 28;
  • 48) 0.888 851 202 517 007 073 28 × 2 = 1 + 0.777 702 405 034 014 146 56;
  • 49) 0.777 702 405 034 014 146 56 × 2 = 1 + 0.555 404 810 068 028 293 12;
  • 50) 0.555 404 810 068 028 293 12 × 2 = 1 + 0.110 809 620 136 056 586 24;
  • 51) 0.110 809 620 136 056 586 24 × 2 = 0 + 0.221 619 240 272 113 172 48;
  • 52) 0.221 619 240 272 113 172 48 × 2 = 0 + 0.443 238 480 544 226 344 96;
  • 53) 0.443 238 480 544 226 344 96 × 2 = 0 + 0.886 476 961 088 452 689 92;

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 51(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 51(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 51(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 51 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