24.777 777 777 777 777 767 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 767 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
24.777 777 777 777 777 767 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: 0.777 777 777 777 777 767 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) 0.777 777 777 777 777 767 4 × 2 = 1 + 0.555 555 555 555 555 534 8;
  • 2) 0.555 555 555 555 555 534 8 × 2 = 1 + 0.111 111 111 111 111 069 6;
  • 3) 0.111 111 111 111 111 069 6 × 2 = 0 + 0.222 222 222 222 222 139 2;
  • 4) 0.222 222 222 222 222 139 2 × 2 = 0 + 0.444 444 444 444 444 278 4;
  • 5) 0.444 444 444 444 444 278 4 × 2 = 0 + 0.888 888 888 888 888 556 8;
  • 6) 0.888 888 888 888 888 556 8 × 2 = 1 + 0.777 777 777 777 777 113 6;
  • 7) 0.777 777 777 777 777 113 6 × 2 = 1 + 0.555 555 555 555 554 227 2;
  • 8) 0.555 555 555 555 554 227 2 × 2 = 1 + 0.111 111 111 111 108 454 4;
  • 9) 0.111 111 111 111 108 454 4 × 2 = 0 + 0.222 222 222 222 216 908 8;
  • 10) 0.222 222 222 222 216 908 8 × 2 = 0 + 0.444 444 444 444 433 817 6;
  • 11) 0.444 444 444 444 433 817 6 × 2 = 0 + 0.888 888 888 888 867 635 2;
  • 12) 0.888 888 888 888 867 635 2 × 2 = 1 + 0.777 777 777 777 735 270 4;
  • 13) 0.777 777 777 777 735 270 4 × 2 = 1 + 0.555 555 555 555 470 540 8;
  • 14) 0.555 555 555 555 470 540 8 × 2 = 1 + 0.111 111 111 110 941 081 6;
  • 15) 0.111 111 111 110 941 081 6 × 2 = 0 + 0.222 222 222 221 882 163 2;
  • 16) 0.222 222 222 221 882 163 2 × 2 = 0 + 0.444 444 444 443 764 326 4;
  • 17) 0.444 444 444 443 764 326 4 × 2 = 0 + 0.888 888 888 887 528 652 8;
  • 18) 0.888 888 888 887 528 652 8 × 2 = 1 + 0.777 777 777 775 057 305 6;
  • 19) 0.777 777 777 775 057 305 6 × 2 = 1 + 0.555 555 555 550 114 611 2;
  • 20) 0.555 555 555 550 114 611 2 × 2 = 1 + 0.111 111 111 100 229 222 4;
  • 21) 0.111 111 111 100 229 222 4 × 2 = 0 + 0.222 222 222 200 458 444 8;
  • 22) 0.222 222 222 200 458 444 8 × 2 = 0 + 0.444 444 444 400 916 889 6;
  • 23) 0.444 444 444 400 916 889 6 × 2 = 0 + 0.888 888 888 801 833 779 2;
  • 24) 0.888 888 888 801 833 779 2 × 2 = 1 + 0.777 777 777 603 667 558 4;
  • 25) 0.777 777 777 603 667 558 4 × 2 = 1 + 0.555 555 555 207 335 116 8;
  • 26) 0.555 555 555 207 335 116 8 × 2 = 1 + 0.111 111 110 414 670 233 6;
  • 27) 0.111 111 110 414 670 233 6 × 2 = 0 + 0.222 222 220 829 340 467 2;
  • 28) 0.222 222 220 829 340 467 2 × 2 = 0 + 0.444 444 441 658 680 934 4;
  • 29) 0.444 444 441 658 680 934 4 × 2 = 0 + 0.888 888 883 317 361 868 8;
  • 30) 0.888 888 883 317 361 868 8 × 2 = 1 + 0.777 777 766 634 723 737 6;
  • 31) 0.777 777 766 634 723 737 6 × 2 = 1 + 0.555 555 533 269 447 475 2;
  • 32) 0.555 555 533 269 447 475 2 × 2 = 1 + 0.111 111 066 538 894 950 4;
  • 33) 0.111 111 066 538 894 950 4 × 2 = 0 + 0.222 222 133 077 789 900 8;
  • 34) 0.222 222 133 077 789 900 8 × 2 = 0 + 0.444 444 266 155 579 801 6;
  • 35) 0.444 444 266 155 579 801 6 × 2 = 0 + 0.888 888 532 311 159 603 2;
  • 36) 0.888 888 532 311 159 603 2 × 2 = 1 + 0.777 777 064 622 319 206 4;
  • 37) 0.777 777 064 622 319 206 4 × 2 = 1 + 0.555 554 129 244 638 412 8;
  • 38) 0.555 554 129 244 638 412 8 × 2 = 1 + 0.111 108 258 489 276 825 6;
  • 39) 0.111 108 258 489 276 825 6 × 2 = 0 + 0.222 216 516 978 553 651 2;
  • 40) 0.222 216 516 978 553 651 2 × 2 = 0 + 0.444 433 033 957 107 302 4;
  • 41) 0.444 433 033 957 107 302 4 × 2 = 0 + 0.888 866 067 914 214 604 8;
  • 42) 0.888 866 067 914 214 604 8 × 2 = 1 + 0.777 732 135 828 429 209 6;
  • 43) 0.777 732 135 828 429 209 6 × 2 = 1 + 0.555 464 271 656 858 419 2;
  • 44) 0.555 464 271 656 858 419 2 × 2 = 1 + 0.110 928 543 313 716 838 4;
  • 45) 0.110 928 543 313 716 838 4 × 2 = 0 + 0.221 857 086 627 433 676 8;
  • 46) 0.221 857 086 627 433 676 8 × 2 = 0 + 0.443 714 173 254 867 353 6;
  • 47) 0.443 714 173 254 867 353 6 × 2 = 0 + 0.887 428 346 509 734 707 2;
  • 48) 0.887 428 346 509 734 707 2 × 2 = 1 + 0.774 856 693 019 469 414 4;
  • 49) 0.774 856 693 019 469 414 4 × 2 = 1 + 0.549 713 386 038 938 828 8;
  • 50) 0.549 713 386 038 938 828 8 × 2 = 1 + 0.099 426 772 077 877 657 6;
  • 51) 0.099 426 772 077 877 657 6 × 2 = 0 + 0.198 853 544 155 755 315 2;
  • 52) 0.198 853 544 155 755 315 2 × 2 = 0 + 0.397 707 088 311 510 630 4;
  • 53) 0.397 707 088 311 510 630 4 × 2 = 0 + 0.795 414 176 623 021 260 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:


0.777 777 777 777 777 767 4(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 767 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:


24.777 777 777 777 777 767 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 24.777 777 777 777 777 767 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