24.777 777 777 777 777 782 95 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 782 95(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
24.777 777 777 777 777 782 95(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: 0.777 777 777 777 777 782 95.

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) 0.777 777 777 777 777 782 95 × 2 = 1 + 0.555 555 555 555 555 565 9;
  • 2) 0.555 555 555 555 555 565 9 × 2 = 1 + 0.111 111 111 111 111 131 8;
  • 3) 0.111 111 111 111 111 131 8 × 2 = 0 + 0.222 222 222 222 222 263 6;
  • 4) 0.222 222 222 222 222 263 6 × 2 = 0 + 0.444 444 444 444 444 527 2;
  • 5) 0.444 444 444 444 444 527 2 × 2 = 0 + 0.888 888 888 888 889 054 4;
  • 6) 0.888 888 888 888 889 054 4 × 2 = 1 + 0.777 777 777 777 778 108 8;
  • 7) 0.777 777 777 777 778 108 8 × 2 = 1 + 0.555 555 555 555 556 217 6;
  • 8) 0.555 555 555 555 556 217 6 × 2 = 1 + 0.111 111 111 111 112 435 2;
  • 9) 0.111 111 111 111 112 435 2 × 2 = 0 + 0.222 222 222 222 224 870 4;
  • 10) 0.222 222 222 222 224 870 4 × 2 = 0 + 0.444 444 444 444 449 740 8;
  • 11) 0.444 444 444 444 449 740 8 × 2 = 0 + 0.888 888 888 888 899 481 6;
  • 12) 0.888 888 888 888 899 481 6 × 2 = 1 + 0.777 777 777 777 798 963 2;
  • 13) 0.777 777 777 777 798 963 2 × 2 = 1 + 0.555 555 555 555 597 926 4;
  • 14) 0.555 555 555 555 597 926 4 × 2 = 1 + 0.111 111 111 111 195 852 8;
  • 15) 0.111 111 111 111 195 852 8 × 2 = 0 + 0.222 222 222 222 391 705 6;
  • 16) 0.222 222 222 222 391 705 6 × 2 = 0 + 0.444 444 444 444 783 411 2;
  • 17) 0.444 444 444 444 783 411 2 × 2 = 0 + 0.888 888 888 889 566 822 4;
  • 18) 0.888 888 888 889 566 822 4 × 2 = 1 + 0.777 777 777 779 133 644 8;
  • 19) 0.777 777 777 779 133 644 8 × 2 = 1 + 0.555 555 555 558 267 289 6;
  • 20) 0.555 555 555 558 267 289 6 × 2 = 1 + 0.111 111 111 116 534 579 2;
  • 21) 0.111 111 111 116 534 579 2 × 2 = 0 + 0.222 222 222 233 069 158 4;
  • 22) 0.222 222 222 233 069 158 4 × 2 = 0 + 0.444 444 444 466 138 316 8;
  • 23) 0.444 444 444 466 138 316 8 × 2 = 0 + 0.888 888 888 932 276 633 6;
  • 24) 0.888 888 888 932 276 633 6 × 2 = 1 + 0.777 777 777 864 553 267 2;
  • 25) 0.777 777 777 864 553 267 2 × 2 = 1 + 0.555 555 555 729 106 534 4;
  • 26) 0.555 555 555 729 106 534 4 × 2 = 1 + 0.111 111 111 458 213 068 8;
  • 27) 0.111 111 111 458 213 068 8 × 2 = 0 + 0.222 222 222 916 426 137 6;
  • 28) 0.222 222 222 916 426 137 6 × 2 = 0 + 0.444 444 445 832 852 275 2;
  • 29) 0.444 444 445 832 852 275 2 × 2 = 0 + 0.888 888 891 665 704 550 4;
  • 30) 0.888 888 891 665 704 550 4 × 2 = 1 + 0.777 777 783 331 409 100 8;
  • 31) 0.777 777 783 331 409 100 8 × 2 = 1 + 0.555 555 566 662 818 201 6;
  • 32) 0.555 555 566 662 818 201 6 × 2 = 1 + 0.111 111 133 325 636 403 2;
  • 33) 0.111 111 133 325 636 403 2 × 2 = 0 + 0.222 222 266 651 272 806 4;
  • 34) 0.222 222 266 651 272 806 4 × 2 = 0 + 0.444 444 533 302 545 612 8;
  • 35) 0.444 444 533 302 545 612 8 × 2 = 0 + 0.888 889 066 605 091 225 6;
  • 36) 0.888 889 066 605 091 225 6 × 2 = 1 + 0.777 778 133 210 182 451 2;
  • 37) 0.777 778 133 210 182 451 2 × 2 = 1 + 0.555 556 266 420 364 902 4;
  • 38) 0.555 556 266 420 364 902 4 × 2 = 1 + 0.111 112 532 840 729 804 8;
  • 39) 0.111 112 532 840 729 804 8 × 2 = 0 + 0.222 225 065 681 459 609 6;
  • 40) 0.222 225 065 681 459 609 6 × 2 = 0 + 0.444 450 131 362 919 219 2;
  • 41) 0.444 450 131 362 919 219 2 × 2 = 0 + 0.888 900 262 725 838 438 4;
  • 42) 0.888 900 262 725 838 438 4 × 2 = 1 + 0.777 800 525 451 676 876 8;
  • 43) 0.777 800 525 451 676 876 8 × 2 = 1 + 0.555 601 050 903 353 753 6;
  • 44) 0.555 601 050 903 353 753 6 × 2 = 1 + 0.111 202 101 806 707 507 2;
  • 45) 0.111 202 101 806 707 507 2 × 2 = 0 + 0.222 404 203 613 415 014 4;
  • 46) 0.222 404 203 613 415 014 4 × 2 = 0 + 0.444 808 407 226 830 028 8;
  • 47) 0.444 808 407 226 830 028 8 × 2 = 0 + 0.889 616 814 453 660 057 6;
  • 48) 0.889 616 814 453 660 057 6 × 2 = 1 + 0.779 233 628 907 320 115 2;
  • 49) 0.779 233 628 907 320 115 2 × 2 = 1 + 0.558 467 257 814 640 230 4;
  • 50) 0.558 467 257 814 640 230 4 × 2 = 1 + 0.116 934 515 629 280 460 8;
  • 51) 0.116 934 515 629 280 460 8 × 2 = 0 + 0.233 869 031 258 560 921 6;
  • 52) 0.233 869 031 258 560 921 6 × 2 = 0 + 0.467 738 062 517 121 843 2;
  • 53) 0.467 738 062 517 121 843 2 × 2 = 0 + 0.935 476 125 034 243 686 4;

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:


0.777 777 777 777 777 782 95(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 782 95(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:


24.777 777 777 777 777 782 95(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 24.777 777 777 777 777 782 95 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