24.777 777 777 777 777 741 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 741 1(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
24.777 777 777 777 777 741 1(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: 0.777 777 777 777 777 741 1.

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) 0.777 777 777 777 777 741 1 × 2 = 1 + 0.555 555 555 555 555 482 2;
  • 2) 0.555 555 555 555 555 482 2 × 2 = 1 + 0.111 111 111 111 110 964 4;
  • 3) 0.111 111 111 111 110 964 4 × 2 = 0 + 0.222 222 222 222 221 928 8;
  • 4) 0.222 222 222 222 221 928 8 × 2 = 0 + 0.444 444 444 444 443 857 6;
  • 5) 0.444 444 444 444 443 857 6 × 2 = 0 + 0.888 888 888 888 887 715 2;
  • 6) 0.888 888 888 888 887 715 2 × 2 = 1 + 0.777 777 777 777 775 430 4;
  • 7) 0.777 777 777 777 775 430 4 × 2 = 1 + 0.555 555 555 555 550 860 8;
  • 8) 0.555 555 555 555 550 860 8 × 2 = 1 + 0.111 111 111 111 101 721 6;
  • 9) 0.111 111 111 111 101 721 6 × 2 = 0 + 0.222 222 222 222 203 443 2;
  • 10) 0.222 222 222 222 203 443 2 × 2 = 0 + 0.444 444 444 444 406 886 4;
  • 11) 0.444 444 444 444 406 886 4 × 2 = 0 + 0.888 888 888 888 813 772 8;
  • 12) 0.888 888 888 888 813 772 8 × 2 = 1 + 0.777 777 777 777 627 545 6;
  • 13) 0.777 777 777 777 627 545 6 × 2 = 1 + 0.555 555 555 555 255 091 2;
  • 14) 0.555 555 555 555 255 091 2 × 2 = 1 + 0.111 111 111 110 510 182 4;
  • 15) 0.111 111 111 110 510 182 4 × 2 = 0 + 0.222 222 222 221 020 364 8;
  • 16) 0.222 222 222 221 020 364 8 × 2 = 0 + 0.444 444 444 442 040 729 6;
  • 17) 0.444 444 444 442 040 729 6 × 2 = 0 + 0.888 888 888 884 081 459 2;
  • 18) 0.888 888 888 884 081 459 2 × 2 = 1 + 0.777 777 777 768 162 918 4;
  • 19) 0.777 777 777 768 162 918 4 × 2 = 1 + 0.555 555 555 536 325 836 8;
  • 20) 0.555 555 555 536 325 836 8 × 2 = 1 + 0.111 111 111 072 651 673 6;
  • 21) 0.111 111 111 072 651 673 6 × 2 = 0 + 0.222 222 222 145 303 347 2;
  • 22) 0.222 222 222 145 303 347 2 × 2 = 0 + 0.444 444 444 290 606 694 4;
  • 23) 0.444 444 444 290 606 694 4 × 2 = 0 + 0.888 888 888 581 213 388 8;
  • 24) 0.888 888 888 581 213 388 8 × 2 = 1 + 0.777 777 777 162 426 777 6;
  • 25) 0.777 777 777 162 426 777 6 × 2 = 1 + 0.555 555 554 324 853 555 2;
  • 26) 0.555 555 554 324 853 555 2 × 2 = 1 + 0.111 111 108 649 707 110 4;
  • 27) 0.111 111 108 649 707 110 4 × 2 = 0 + 0.222 222 217 299 414 220 8;
  • 28) 0.222 222 217 299 414 220 8 × 2 = 0 + 0.444 444 434 598 828 441 6;
  • 29) 0.444 444 434 598 828 441 6 × 2 = 0 + 0.888 888 869 197 656 883 2;
  • 30) 0.888 888 869 197 656 883 2 × 2 = 1 + 0.777 777 738 395 313 766 4;
  • 31) 0.777 777 738 395 313 766 4 × 2 = 1 + 0.555 555 476 790 627 532 8;
  • 32) 0.555 555 476 790 627 532 8 × 2 = 1 + 0.111 110 953 581 255 065 6;
  • 33) 0.111 110 953 581 255 065 6 × 2 = 0 + 0.222 221 907 162 510 131 2;
  • 34) 0.222 221 907 162 510 131 2 × 2 = 0 + 0.444 443 814 325 020 262 4;
  • 35) 0.444 443 814 325 020 262 4 × 2 = 0 + 0.888 887 628 650 040 524 8;
  • 36) 0.888 887 628 650 040 524 8 × 2 = 1 + 0.777 775 257 300 081 049 6;
  • 37) 0.777 775 257 300 081 049 6 × 2 = 1 + 0.555 550 514 600 162 099 2;
  • 38) 0.555 550 514 600 162 099 2 × 2 = 1 + 0.111 101 029 200 324 198 4;
  • 39) 0.111 101 029 200 324 198 4 × 2 = 0 + 0.222 202 058 400 648 396 8;
  • 40) 0.222 202 058 400 648 396 8 × 2 = 0 + 0.444 404 116 801 296 793 6;
  • 41) 0.444 404 116 801 296 793 6 × 2 = 0 + 0.888 808 233 602 593 587 2;
  • 42) 0.888 808 233 602 593 587 2 × 2 = 1 + 0.777 616 467 205 187 174 4;
  • 43) 0.777 616 467 205 187 174 4 × 2 = 1 + 0.555 232 934 410 374 348 8;
  • 44) 0.555 232 934 410 374 348 8 × 2 = 1 + 0.110 465 868 820 748 697 6;
  • 45) 0.110 465 868 820 748 697 6 × 2 = 0 + 0.220 931 737 641 497 395 2;
  • 46) 0.220 931 737 641 497 395 2 × 2 = 0 + 0.441 863 475 282 994 790 4;
  • 47) 0.441 863 475 282 994 790 4 × 2 = 0 + 0.883 726 950 565 989 580 8;
  • 48) 0.883 726 950 565 989 580 8 × 2 = 1 + 0.767 453 901 131 979 161 6;
  • 49) 0.767 453 901 131 979 161 6 × 2 = 1 + 0.534 907 802 263 958 323 2;
  • 50) 0.534 907 802 263 958 323 2 × 2 = 1 + 0.069 815 604 527 916 646 4;
  • 51) 0.069 815 604 527 916 646 4 × 2 = 0 + 0.139 631 209 055 833 292 8;
  • 52) 0.139 631 209 055 833 292 8 × 2 = 0 + 0.279 262 418 111 666 585 6;
  • 53) 0.279 262 418 111 666 585 6 × 2 = 0 + 0.558 524 836 223 333 171 2;

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:


0.777 777 777 777 777 741 1(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 741 1(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:


24.777 777 777 777 777 741 1(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 24.777 777 777 777 777 741 1 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