24.777 777 777 777 777 784 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 784 6(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
24.777 777 777 777 777 784 6(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: 0.777 777 777 777 777 784 6.

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) 0.777 777 777 777 777 784 6 × 2 = 1 + 0.555 555 555 555 555 569 2;
  • 2) 0.555 555 555 555 555 569 2 × 2 = 1 + 0.111 111 111 111 111 138 4;
  • 3) 0.111 111 111 111 111 138 4 × 2 = 0 + 0.222 222 222 222 222 276 8;
  • 4) 0.222 222 222 222 222 276 8 × 2 = 0 + 0.444 444 444 444 444 553 6;
  • 5) 0.444 444 444 444 444 553 6 × 2 = 0 + 0.888 888 888 888 889 107 2;
  • 6) 0.888 888 888 888 889 107 2 × 2 = 1 + 0.777 777 777 777 778 214 4;
  • 7) 0.777 777 777 777 778 214 4 × 2 = 1 + 0.555 555 555 555 556 428 8;
  • 8) 0.555 555 555 555 556 428 8 × 2 = 1 + 0.111 111 111 111 112 857 6;
  • 9) 0.111 111 111 111 112 857 6 × 2 = 0 + 0.222 222 222 222 225 715 2;
  • 10) 0.222 222 222 222 225 715 2 × 2 = 0 + 0.444 444 444 444 451 430 4;
  • 11) 0.444 444 444 444 451 430 4 × 2 = 0 + 0.888 888 888 888 902 860 8;
  • 12) 0.888 888 888 888 902 860 8 × 2 = 1 + 0.777 777 777 777 805 721 6;
  • 13) 0.777 777 777 777 805 721 6 × 2 = 1 + 0.555 555 555 555 611 443 2;
  • 14) 0.555 555 555 555 611 443 2 × 2 = 1 + 0.111 111 111 111 222 886 4;
  • 15) 0.111 111 111 111 222 886 4 × 2 = 0 + 0.222 222 222 222 445 772 8;
  • 16) 0.222 222 222 222 445 772 8 × 2 = 0 + 0.444 444 444 444 891 545 6;
  • 17) 0.444 444 444 444 891 545 6 × 2 = 0 + 0.888 888 888 889 783 091 2;
  • 18) 0.888 888 888 889 783 091 2 × 2 = 1 + 0.777 777 777 779 566 182 4;
  • 19) 0.777 777 777 779 566 182 4 × 2 = 1 + 0.555 555 555 559 132 364 8;
  • 20) 0.555 555 555 559 132 364 8 × 2 = 1 + 0.111 111 111 118 264 729 6;
  • 21) 0.111 111 111 118 264 729 6 × 2 = 0 + 0.222 222 222 236 529 459 2;
  • 22) 0.222 222 222 236 529 459 2 × 2 = 0 + 0.444 444 444 473 058 918 4;
  • 23) 0.444 444 444 473 058 918 4 × 2 = 0 + 0.888 888 888 946 117 836 8;
  • 24) 0.888 888 888 946 117 836 8 × 2 = 1 + 0.777 777 777 892 235 673 6;
  • 25) 0.777 777 777 892 235 673 6 × 2 = 1 + 0.555 555 555 784 471 347 2;
  • 26) 0.555 555 555 784 471 347 2 × 2 = 1 + 0.111 111 111 568 942 694 4;
  • 27) 0.111 111 111 568 942 694 4 × 2 = 0 + 0.222 222 223 137 885 388 8;
  • 28) 0.222 222 223 137 885 388 8 × 2 = 0 + 0.444 444 446 275 770 777 6;
  • 29) 0.444 444 446 275 770 777 6 × 2 = 0 + 0.888 888 892 551 541 555 2;
  • 30) 0.888 888 892 551 541 555 2 × 2 = 1 + 0.777 777 785 103 083 110 4;
  • 31) 0.777 777 785 103 083 110 4 × 2 = 1 + 0.555 555 570 206 166 220 8;
  • 32) 0.555 555 570 206 166 220 8 × 2 = 1 + 0.111 111 140 412 332 441 6;
  • 33) 0.111 111 140 412 332 441 6 × 2 = 0 + 0.222 222 280 824 664 883 2;
  • 34) 0.222 222 280 824 664 883 2 × 2 = 0 + 0.444 444 561 649 329 766 4;
  • 35) 0.444 444 561 649 329 766 4 × 2 = 0 + 0.888 889 123 298 659 532 8;
  • 36) 0.888 889 123 298 659 532 8 × 2 = 1 + 0.777 778 246 597 319 065 6;
  • 37) 0.777 778 246 597 319 065 6 × 2 = 1 + 0.555 556 493 194 638 131 2;
  • 38) 0.555 556 493 194 638 131 2 × 2 = 1 + 0.111 112 986 389 276 262 4;
  • 39) 0.111 112 986 389 276 262 4 × 2 = 0 + 0.222 225 972 778 552 524 8;
  • 40) 0.222 225 972 778 552 524 8 × 2 = 0 + 0.444 451 945 557 105 049 6;
  • 41) 0.444 451 945 557 105 049 6 × 2 = 0 + 0.888 903 891 114 210 099 2;
  • 42) 0.888 903 891 114 210 099 2 × 2 = 1 + 0.777 807 782 228 420 198 4;
  • 43) 0.777 807 782 228 420 198 4 × 2 = 1 + 0.555 615 564 456 840 396 8;
  • 44) 0.555 615 564 456 840 396 8 × 2 = 1 + 0.111 231 128 913 680 793 6;
  • 45) 0.111 231 128 913 680 793 6 × 2 = 0 + 0.222 462 257 827 361 587 2;
  • 46) 0.222 462 257 827 361 587 2 × 2 = 0 + 0.444 924 515 654 723 174 4;
  • 47) 0.444 924 515 654 723 174 4 × 2 = 0 + 0.889 849 031 309 446 348 8;
  • 48) 0.889 849 031 309 446 348 8 × 2 = 1 + 0.779 698 062 618 892 697 6;
  • 49) 0.779 698 062 618 892 697 6 × 2 = 1 + 0.559 396 125 237 785 395 2;
  • 50) 0.559 396 125 237 785 395 2 × 2 = 1 + 0.118 792 250 475 570 790 4;
  • 51) 0.118 792 250 475 570 790 4 × 2 = 0 + 0.237 584 500 951 141 580 8;
  • 52) 0.237 584 500 951 141 580 8 × 2 = 0 + 0.475 169 001 902 283 161 6;
  • 53) 0.475 169 001 902 283 161 6 × 2 = 0 + 0.950 338 003 804 566 323 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:


0.777 777 777 777 777 784 6(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 784 6(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:


24.777 777 777 777 777 784 6(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 24.777 777 777 777 777 784 6 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