24.777 777 777 777 777 774 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 774 4(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
24.777 777 777 777 777 774 4(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: 0.777 777 777 777 777 774 4.

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) 0.777 777 777 777 777 774 4 × 2 = 1 + 0.555 555 555 555 555 548 8;
  • 2) 0.555 555 555 555 555 548 8 × 2 = 1 + 0.111 111 111 111 111 097 6;
  • 3) 0.111 111 111 111 111 097 6 × 2 = 0 + 0.222 222 222 222 222 195 2;
  • 4) 0.222 222 222 222 222 195 2 × 2 = 0 + 0.444 444 444 444 444 390 4;
  • 5) 0.444 444 444 444 444 390 4 × 2 = 0 + 0.888 888 888 888 888 780 8;
  • 6) 0.888 888 888 888 888 780 8 × 2 = 1 + 0.777 777 777 777 777 561 6;
  • 7) 0.777 777 777 777 777 561 6 × 2 = 1 + 0.555 555 555 555 555 123 2;
  • 8) 0.555 555 555 555 555 123 2 × 2 = 1 + 0.111 111 111 111 110 246 4;
  • 9) 0.111 111 111 111 110 246 4 × 2 = 0 + 0.222 222 222 222 220 492 8;
  • 10) 0.222 222 222 222 220 492 8 × 2 = 0 + 0.444 444 444 444 440 985 6;
  • 11) 0.444 444 444 444 440 985 6 × 2 = 0 + 0.888 888 888 888 881 971 2;
  • 12) 0.888 888 888 888 881 971 2 × 2 = 1 + 0.777 777 777 777 763 942 4;
  • 13) 0.777 777 777 777 763 942 4 × 2 = 1 + 0.555 555 555 555 527 884 8;
  • 14) 0.555 555 555 555 527 884 8 × 2 = 1 + 0.111 111 111 111 055 769 6;
  • 15) 0.111 111 111 111 055 769 6 × 2 = 0 + 0.222 222 222 222 111 539 2;
  • 16) 0.222 222 222 222 111 539 2 × 2 = 0 + 0.444 444 444 444 223 078 4;
  • 17) 0.444 444 444 444 223 078 4 × 2 = 0 + 0.888 888 888 888 446 156 8;
  • 18) 0.888 888 888 888 446 156 8 × 2 = 1 + 0.777 777 777 776 892 313 6;
  • 19) 0.777 777 777 776 892 313 6 × 2 = 1 + 0.555 555 555 553 784 627 2;
  • 20) 0.555 555 555 553 784 627 2 × 2 = 1 + 0.111 111 111 107 569 254 4;
  • 21) 0.111 111 111 107 569 254 4 × 2 = 0 + 0.222 222 222 215 138 508 8;
  • 22) 0.222 222 222 215 138 508 8 × 2 = 0 + 0.444 444 444 430 277 017 6;
  • 23) 0.444 444 444 430 277 017 6 × 2 = 0 + 0.888 888 888 860 554 035 2;
  • 24) 0.888 888 888 860 554 035 2 × 2 = 1 + 0.777 777 777 721 108 070 4;
  • 25) 0.777 777 777 721 108 070 4 × 2 = 1 + 0.555 555 555 442 216 140 8;
  • 26) 0.555 555 555 442 216 140 8 × 2 = 1 + 0.111 111 110 884 432 281 6;
  • 27) 0.111 111 110 884 432 281 6 × 2 = 0 + 0.222 222 221 768 864 563 2;
  • 28) 0.222 222 221 768 864 563 2 × 2 = 0 + 0.444 444 443 537 729 126 4;
  • 29) 0.444 444 443 537 729 126 4 × 2 = 0 + 0.888 888 887 075 458 252 8;
  • 30) 0.888 888 887 075 458 252 8 × 2 = 1 + 0.777 777 774 150 916 505 6;
  • 31) 0.777 777 774 150 916 505 6 × 2 = 1 + 0.555 555 548 301 833 011 2;
  • 32) 0.555 555 548 301 833 011 2 × 2 = 1 + 0.111 111 096 603 666 022 4;
  • 33) 0.111 111 096 603 666 022 4 × 2 = 0 + 0.222 222 193 207 332 044 8;
  • 34) 0.222 222 193 207 332 044 8 × 2 = 0 + 0.444 444 386 414 664 089 6;
  • 35) 0.444 444 386 414 664 089 6 × 2 = 0 + 0.888 888 772 829 328 179 2;
  • 36) 0.888 888 772 829 328 179 2 × 2 = 1 + 0.777 777 545 658 656 358 4;
  • 37) 0.777 777 545 658 656 358 4 × 2 = 1 + 0.555 555 091 317 312 716 8;
  • 38) 0.555 555 091 317 312 716 8 × 2 = 1 + 0.111 110 182 634 625 433 6;
  • 39) 0.111 110 182 634 625 433 6 × 2 = 0 + 0.222 220 365 269 250 867 2;
  • 40) 0.222 220 365 269 250 867 2 × 2 = 0 + 0.444 440 730 538 501 734 4;
  • 41) 0.444 440 730 538 501 734 4 × 2 = 0 + 0.888 881 461 077 003 468 8;
  • 42) 0.888 881 461 077 003 468 8 × 2 = 1 + 0.777 762 922 154 006 937 6;
  • 43) 0.777 762 922 154 006 937 6 × 2 = 1 + 0.555 525 844 308 013 875 2;
  • 44) 0.555 525 844 308 013 875 2 × 2 = 1 + 0.111 051 688 616 027 750 4;
  • 45) 0.111 051 688 616 027 750 4 × 2 = 0 + 0.222 103 377 232 055 500 8;
  • 46) 0.222 103 377 232 055 500 8 × 2 = 0 + 0.444 206 754 464 111 001 6;
  • 47) 0.444 206 754 464 111 001 6 × 2 = 0 + 0.888 413 508 928 222 003 2;
  • 48) 0.888 413 508 928 222 003 2 × 2 = 1 + 0.776 827 017 856 444 006 4;
  • 49) 0.776 827 017 856 444 006 4 × 2 = 1 + 0.553 654 035 712 888 012 8;
  • 50) 0.553 654 035 712 888 012 8 × 2 = 1 + 0.107 308 071 425 776 025 6;
  • 51) 0.107 308 071 425 776 025 6 × 2 = 0 + 0.214 616 142 851 552 051 2;
  • 52) 0.214 616 142 851 552 051 2 × 2 = 0 + 0.429 232 285 703 104 102 4;
  • 53) 0.429 232 285 703 104 102 4 × 2 = 0 + 0.858 464 571 406 208 204 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:


0.777 777 777 777 777 774 4(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 774 4(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:


24.777 777 777 777 777 774 4(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 24.777 777 777 777 777 774 4 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