24.777 777 777 777 777 760 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 760 7(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
24.777 777 777 777 777 760 7(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: 0.777 777 777 777 777 760 7.

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) 0.777 777 777 777 777 760 7 × 2 = 1 + 0.555 555 555 555 555 521 4;
  • 2) 0.555 555 555 555 555 521 4 × 2 = 1 + 0.111 111 111 111 111 042 8;
  • 3) 0.111 111 111 111 111 042 8 × 2 = 0 + 0.222 222 222 222 222 085 6;
  • 4) 0.222 222 222 222 222 085 6 × 2 = 0 + 0.444 444 444 444 444 171 2;
  • 5) 0.444 444 444 444 444 171 2 × 2 = 0 + 0.888 888 888 888 888 342 4;
  • 6) 0.888 888 888 888 888 342 4 × 2 = 1 + 0.777 777 777 777 776 684 8;
  • 7) 0.777 777 777 777 776 684 8 × 2 = 1 + 0.555 555 555 555 553 369 6;
  • 8) 0.555 555 555 555 553 369 6 × 2 = 1 + 0.111 111 111 111 106 739 2;
  • 9) 0.111 111 111 111 106 739 2 × 2 = 0 + 0.222 222 222 222 213 478 4;
  • 10) 0.222 222 222 222 213 478 4 × 2 = 0 + 0.444 444 444 444 426 956 8;
  • 11) 0.444 444 444 444 426 956 8 × 2 = 0 + 0.888 888 888 888 853 913 6;
  • 12) 0.888 888 888 888 853 913 6 × 2 = 1 + 0.777 777 777 777 707 827 2;
  • 13) 0.777 777 777 777 707 827 2 × 2 = 1 + 0.555 555 555 555 415 654 4;
  • 14) 0.555 555 555 555 415 654 4 × 2 = 1 + 0.111 111 111 110 831 308 8;
  • 15) 0.111 111 111 110 831 308 8 × 2 = 0 + 0.222 222 222 221 662 617 6;
  • 16) 0.222 222 222 221 662 617 6 × 2 = 0 + 0.444 444 444 443 325 235 2;
  • 17) 0.444 444 444 443 325 235 2 × 2 = 0 + 0.888 888 888 886 650 470 4;
  • 18) 0.888 888 888 886 650 470 4 × 2 = 1 + 0.777 777 777 773 300 940 8;
  • 19) 0.777 777 777 773 300 940 8 × 2 = 1 + 0.555 555 555 546 601 881 6;
  • 20) 0.555 555 555 546 601 881 6 × 2 = 1 + 0.111 111 111 093 203 763 2;
  • 21) 0.111 111 111 093 203 763 2 × 2 = 0 + 0.222 222 222 186 407 526 4;
  • 22) 0.222 222 222 186 407 526 4 × 2 = 0 + 0.444 444 444 372 815 052 8;
  • 23) 0.444 444 444 372 815 052 8 × 2 = 0 + 0.888 888 888 745 630 105 6;
  • 24) 0.888 888 888 745 630 105 6 × 2 = 1 + 0.777 777 777 491 260 211 2;
  • 25) 0.777 777 777 491 260 211 2 × 2 = 1 + 0.555 555 554 982 520 422 4;
  • 26) 0.555 555 554 982 520 422 4 × 2 = 1 + 0.111 111 109 965 040 844 8;
  • 27) 0.111 111 109 965 040 844 8 × 2 = 0 + 0.222 222 219 930 081 689 6;
  • 28) 0.222 222 219 930 081 689 6 × 2 = 0 + 0.444 444 439 860 163 379 2;
  • 29) 0.444 444 439 860 163 379 2 × 2 = 0 + 0.888 888 879 720 326 758 4;
  • 30) 0.888 888 879 720 326 758 4 × 2 = 1 + 0.777 777 759 440 653 516 8;
  • 31) 0.777 777 759 440 653 516 8 × 2 = 1 + 0.555 555 518 881 307 033 6;
  • 32) 0.555 555 518 881 307 033 6 × 2 = 1 + 0.111 111 037 762 614 067 2;
  • 33) 0.111 111 037 762 614 067 2 × 2 = 0 + 0.222 222 075 525 228 134 4;
  • 34) 0.222 222 075 525 228 134 4 × 2 = 0 + 0.444 444 151 050 456 268 8;
  • 35) 0.444 444 151 050 456 268 8 × 2 = 0 + 0.888 888 302 100 912 537 6;
  • 36) 0.888 888 302 100 912 537 6 × 2 = 1 + 0.777 776 604 201 825 075 2;
  • 37) 0.777 776 604 201 825 075 2 × 2 = 1 + 0.555 553 208 403 650 150 4;
  • 38) 0.555 553 208 403 650 150 4 × 2 = 1 + 0.111 106 416 807 300 300 8;
  • 39) 0.111 106 416 807 300 300 8 × 2 = 0 + 0.222 212 833 614 600 601 6;
  • 40) 0.222 212 833 614 600 601 6 × 2 = 0 + 0.444 425 667 229 201 203 2;
  • 41) 0.444 425 667 229 201 203 2 × 2 = 0 + 0.888 851 334 458 402 406 4;
  • 42) 0.888 851 334 458 402 406 4 × 2 = 1 + 0.777 702 668 916 804 812 8;
  • 43) 0.777 702 668 916 804 812 8 × 2 = 1 + 0.555 405 337 833 609 625 6;
  • 44) 0.555 405 337 833 609 625 6 × 2 = 1 + 0.110 810 675 667 219 251 2;
  • 45) 0.110 810 675 667 219 251 2 × 2 = 0 + 0.221 621 351 334 438 502 4;
  • 46) 0.221 621 351 334 438 502 4 × 2 = 0 + 0.443 242 702 668 877 004 8;
  • 47) 0.443 242 702 668 877 004 8 × 2 = 0 + 0.886 485 405 337 754 009 6;
  • 48) 0.886 485 405 337 754 009 6 × 2 = 1 + 0.772 970 810 675 508 019 2;
  • 49) 0.772 970 810 675 508 019 2 × 2 = 1 + 0.545 941 621 351 016 038 4;
  • 50) 0.545 941 621 351 016 038 4 × 2 = 1 + 0.091 883 242 702 032 076 8;
  • 51) 0.091 883 242 702 032 076 8 × 2 = 0 + 0.183 766 485 404 064 153 6;
  • 52) 0.183 766 485 404 064 153 6 × 2 = 0 + 0.367 532 970 808 128 307 2;
  • 53) 0.367 532 970 808 128 307 2 × 2 = 0 + 0.735 065 941 616 256 614 4;

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:


0.777 777 777 777 777 760 7(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 760 7(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:


24.777 777 777 777 777 760 7(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 24.777 777 777 777 777 760 7 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