24.777 777 777 777 778 65 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 778 65(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
24.777 777 777 777 778 65(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: 0.777 777 777 777 778 65.

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) 0.777 777 777 777 778 65 × 2 = 1 + 0.555 555 555 555 557 3;
  • 2) 0.555 555 555 555 557 3 × 2 = 1 + 0.111 111 111 111 114 6;
  • 3) 0.111 111 111 111 114 6 × 2 = 0 + 0.222 222 222 222 229 2;
  • 4) 0.222 222 222 222 229 2 × 2 = 0 + 0.444 444 444 444 458 4;
  • 5) 0.444 444 444 444 458 4 × 2 = 0 + 0.888 888 888 888 916 8;
  • 6) 0.888 888 888 888 916 8 × 2 = 1 + 0.777 777 777 777 833 6;
  • 7) 0.777 777 777 777 833 6 × 2 = 1 + 0.555 555 555 555 667 2;
  • 8) 0.555 555 555 555 667 2 × 2 = 1 + 0.111 111 111 111 334 4;
  • 9) 0.111 111 111 111 334 4 × 2 = 0 + 0.222 222 222 222 668 8;
  • 10) 0.222 222 222 222 668 8 × 2 = 0 + 0.444 444 444 445 337 6;
  • 11) 0.444 444 444 445 337 6 × 2 = 0 + 0.888 888 888 890 675 2;
  • 12) 0.888 888 888 890 675 2 × 2 = 1 + 0.777 777 777 781 350 4;
  • 13) 0.777 777 777 781 350 4 × 2 = 1 + 0.555 555 555 562 700 8;
  • 14) 0.555 555 555 562 700 8 × 2 = 1 + 0.111 111 111 125 401 6;
  • 15) 0.111 111 111 125 401 6 × 2 = 0 + 0.222 222 222 250 803 2;
  • 16) 0.222 222 222 250 803 2 × 2 = 0 + 0.444 444 444 501 606 4;
  • 17) 0.444 444 444 501 606 4 × 2 = 0 + 0.888 888 889 003 212 8;
  • 18) 0.888 888 889 003 212 8 × 2 = 1 + 0.777 777 778 006 425 6;
  • 19) 0.777 777 778 006 425 6 × 2 = 1 + 0.555 555 556 012 851 2;
  • 20) 0.555 555 556 012 851 2 × 2 = 1 + 0.111 111 112 025 702 4;
  • 21) 0.111 111 112 025 702 4 × 2 = 0 + 0.222 222 224 051 404 8;
  • 22) 0.222 222 224 051 404 8 × 2 = 0 + 0.444 444 448 102 809 6;
  • 23) 0.444 444 448 102 809 6 × 2 = 0 + 0.888 888 896 205 619 2;
  • 24) 0.888 888 896 205 619 2 × 2 = 1 + 0.777 777 792 411 238 4;
  • 25) 0.777 777 792 411 238 4 × 2 = 1 + 0.555 555 584 822 476 8;
  • 26) 0.555 555 584 822 476 8 × 2 = 1 + 0.111 111 169 644 953 6;
  • 27) 0.111 111 169 644 953 6 × 2 = 0 + 0.222 222 339 289 907 2;
  • 28) 0.222 222 339 289 907 2 × 2 = 0 + 0.444 444 678 579 814 4;
  • 29) 0.444 444 678 579 814 4 × 2 = 0 + 0.888 889 357 159 628 8;
  • 30) 0.888 889 357 159 628 8 × 2 = 1 + 0.777 778 714 319 257 6;
  • 31) 0.777 778 714 319 257 6 × 2 = 1 + 0.555 557 428 638 515 2;
  • 32) 0.555 557 428 638 515 2 × 2 = 1 + 0.111 114 857 277 030 4;
  • 33) 0.111 114 857 277 030 4 × 2 = 0 + 0.222 229 714 554 060 8;
  • 34) 0.222 229 714 554 060 8 × 2 = 0 + 0.444 459 429 108 121 6;
  • 35) 0.444 459 429 108 121 6 × 2 = 0 + 0.888 918 858 216 243 2;
  • 36) 0.888 918 858 216 243 2 × 2 = 1 + 0.777 837 716 432 486 4;
  • 37) 0.777 837 716 432 486 4 × 2 = 1 + 0.555 675 432 864 972 8;
  • 38) 0.555 675 432 864 972 8 × 2 = 1 + 0.111 350 865 729 945 6;
  • 39) 0.111 350 865 729 945 6 × 2 = 0 + 0.222 701 731 459 891 2;
  • 40) 0.222 701 731 459 891 2 × 2 = 0 + 0.445 403 462 919 782 4;
  • 41) 0.445 403 462 919 782 4 × 2 = 0 + 0.890 806 925 839 564 8;
  • 42) 0.890 806 925 839 564 8 × 2 = 1 + 0.781 613 851 679 129 6;
  • 43) 0.781 613 851 679 129 6 × 2 = 1 + 0.563 227 703 358 259 2;
  • 44) 0.563 227 703 358 259 2 × 2 = 1 + 0.126 455 406 716 518 4;
  • 45) 0.126 455 406 716 518 4 × 2 = 0 + 0.252 910 813 433 036 8;
  • 46) 0.252 910 813 433 036 8 × 2 = 0 + 0.505 821 626 866 073 6;
  • 47) 0.505 821 626 866 073 6 × 2 = 1 + 0.011 643 253 732 147 2;
  • 48) 0.011 643 253 732 147 2 × 2 = 0 + 0.023 286 507 464 294 4;
  • 49) 0.023 286 507 464 294 4 × 2 = 0 + 0.046 573 014 928 588 8;
  • 50) 0.046 573 014 928 588 8 × 2 = 0 + 0.093 146 029 857 177 6;
  • 51) 0.093 146 029 857 177 6 × 2 = 0 + 0.186 292 059 714 355 2;
  • 52) 0.186 292 059 714 355 2 × 2 = 0 + 0.372 584 119 428 710 4;
  • 53) 0.372 584 119 428 710 4 × 2 = 0 + 0.745 168 238 857 420 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:


0.777 777 777 777 778 65(10) =

0.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0010 0000 0(2)

5. Positive number before normalization:

24.777 777 777 777 778 65(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0010 0000 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:


24.777 777 777 777 778 65(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0010 0000 0(2) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0010 0000 0(2) × 20 =

1.1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0010 0000 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 0010 0000 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 0010 0 0000 =

1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0010


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 0010


Decimal number 24.777 777 777 777 778 65 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 0010

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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