24.777 777 777 777 777 789 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 789 8(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
24.777 777 777 777 777 789 8(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: 0.777 777 777 777 777 789 8.

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) 0.777 777 777 777 777 789 8 × 2 = 1 + 0.555 555 555 555 555 579 6;
  • 2) 0.555 555 555 555 555 579 6 × 2 = 1 + 0.111 111 111 111 111 159 2;
  • 3) 0.111 111 111 111 111 159 2 × 2 = 0 + 0.222 222 222 222 222 318 4;
  • 4) 0.222 222 222 222 222 318 4 × 2 = 0 + 0.444 444 444 444 444 636 8;
  • 5) 0.444 444 444 444 444 636 8 × 2 = 0 + 0.888 888 888 888 889 273 6;
  • 6) 0.888 888 888 888 889 273 6 × 2 = 1 + 0.777 777 777 777 778 547 2;
  • 7) 0.777 777 777 777 778 547 2 × 2 = 1 + 0.555 555 555 555 557 094 4;
  • 8) 0.555 555 555 555 557 094 4 × 2 = 1 + 0.111 111 111 111 114 188 8;
  • 9) 0.111 111 111 111 114 188 8 × 2 = 0 + 0.222 222 222 222 228 377 6;
  • 10) 0.222 222 222 222 228 377 6 × 2 = 0 + 0.444 444 444 444 456 755 2;
  • 11) 0.444 444 444 444 456 755 2 × 2 = 0 + 0.888 888 888 888 913 510 4;
  • 12) 0.888 888 888 888 913 510 4 × 2 = 1 + 0.777 777 777 777 827 020 8;
  • 13) 0.777 777 777 777 827 020 8 × 2 = 1 + 0.555 555 555 555 654 041 6;
  • 14) 0.555 555 555 555 654 041 6 × 2 = 1 + 0.111 111 111 111 308 083 2;
  • 15) 0.111 111 111 111 308 083 2 × 2 = 0 + 0.222 222 222 222 616 166 4;
  • 16) 0.222 222 222 222 616 166 4 × 2 = 0 + 0.444 444 444 445 232 332 8;
  • 17) 0.444 444 444 445 232 332 8 × 2 = 0 + 0.888 888 888 890 464 665 6;
  • 18) 0.888 888 888 890 464 665 6 × 2 = 1 + 0.777 777 777 780 929 331 2;
  • 19) 0.777 777 777 780 929 331 2 × 2 = 1 + 0.555 555 555 561 858 662 4;
  • 20) 0.555 555 555 561 858 662 4 × 2 = 1 + 0.111 111 111 123 717 324 8;
  • 21) 0.111 111 111 123 717 324 8 × 2 = 0 + 0.222 222 222 247 434 649 6;
  • 22) 0.222 222 222 247 434 649 6 × 2 = 0 + 0.444 444 444 494 869 299 2;
  • 23) 0.444 444 444 494 869 299 2 × 2 = 0 + 0.888 888 888 989 738 598 4;
  • 24) 0.888 888 888 989 738 598 4 × 2 = 1 + 0.777 777 777 979 477 196 8;
  • 25) 0.777 777 777 979 477 196 8 × 2 = 1 + 0.555 555 555 958 954 393 6;
  • 26) 0.555 555 555 958 954 393 6 × 2 = 1 + 0.111 111 111 917 908 787 2;
  • 27) 0.111 111 111 917 908 787 2 × 2 = 0 + 0.222 222 223 835 817 574 4;
  • 28) 0.222 222 223 835 817 574 4 × 2 = 0 + 0.444 444 447 671 635 148 8;
  • 29) 0.444 444 447 671 635 148 8 × 2 = 0 + 0.888 888 895 343 270 297 6;
  • 30) 0.888 888 895 343 270 297 6 × 2 = 1 + 0.777 777 790 686 540 595 2;
  • 31) 0.777 777 790 686 540 595 2 × 2 = 1 + 0.555 555 581 373 081 190 4;
  • 32) 0.555 555 581 373 081 190 4 × 2 = 1 + 0.111 111 162 746 162 380 8;
  • 33) 0.111 111 162 746 162 380 8 × 2 = 0 + 0.222 222 325 492 324 761 6;
  • 34) 0.222 222 325 492 324 761 6 × 2 = 0 + 0.444 444 650 984 649 523 2;
  • 35) 0.444 444 650 984 649 523 2 × 2 = 0 + 0.888 889 301 969 299 046 4;
  • 36) 0.888 889 301 969 299 046 4 × 2 = 1 + 0.777 778 603 938 598 092 8;
  • 37) 0.777 778 603 938 598 092 8 × 2 = 1 + 0.555 557 207 877 196 185 6;
  • 38) 0.555 557 207 877 196 185 6 × 2 = 1 + 0.111 114 415 754 392 371 2;
  • 39) 0.111 114 415 754 392 371 2 × 2 = 0 + 0.222 228 831 508 784 742 4;
  • 40) 0.222 228 831 508 784 742 4 × 2 = 0 + 0.444 457 663 017 569 484 8;
  • 41) 0.444 457 663 017 569 484 8 × 2 = 0 + 0.888 915 326 035 138 969 6;
  • 42) 0.888 915 326 035 138 969 6 × 2 = 1 + 0.777 830 652 070 277 939 2;
  • 43) 0.777 830 652 070 277 939 2 × 2 = 1 + 0.555 661 304 140 555 878 4;
  • 44) 0.555 661 304 140 555 878 4 × 2 = 1 + 0.111 322 608 281 111 756 8;
  • 45) 0.111 322 608 281 111 756 8 × 2 = 0 + 0.222 645 216 562 223 513 6;
  • 46) 0.222 645 216 562 223 513 6 × 2 = 0 + 0.445 290 433 124 447 027 2;
  • 47) 0.445 290 433 124 447 027 2 × 2 = 0 + 0.890 580 866 248 894 054 4;
  • 48) 0.890 580 866 248 894 054 4 × 2 = 1 + 0.781 161 732 497 788 108 8;
  • 49) 0.781 161 732 497 788 108 8 × 2 = 1 + 0.562 323 464 995 576 217 6;
  • 50) 0.562 323 464 995 576 217 6 × 2 = 1 + 0.124 646 929 991 152 435 2;
  • 51) 0.124 646 929 991 152 435 2 × 2 = 0 + 0.249 293 859 982 304 870 4;
  • 52) 0.249 293 859 982 304 870 4 × 2 = 0 + 0.498 587 719 964 609 740 8;
  • 53) 0.498 587 719 964 609 740 8 × 2 = 0 + 0.997 175 439 929 219 481 6;

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:


0.777 777 777 777 777 789 8(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 789 8(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:


24.777 777 777 777 777 789 8(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 24.777 777 777 777 777 789 8 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