24.777 777 777 777 777 787 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 787 3(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
24.777 777 777 777 777 787 3(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: 0.777 777 777 777 777 787 3.

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) 0.777 777 777 777 777 787 3 × 2 = 1 + 0.555 555 555 555 555 574 6;
  • 2) 0.555 555 555 555 555 574 6 × 2 = 1 + 0.111 111 111 111 111 149 2;
  • 3) 0.111 111 111 111 111 149 2 × 2 = 0 + 0.222 222 222 222 222 298 4;
  • 4) 0.222 222 222 222 222 298 4 × 2 = 0 + 0.444 444 444 444 444 596 8;
  • 5) 0.444 444 444 444 444 596 8 × 2 = 0 + 0.888 888 888 888 889 193 6;
  • 6) 0.888 888 888 888 889 193 6 × 2 = 1 + 0.777 777 777 777 778 387 2;
  • 7) 0.777 777 777 777 778 387 2 × 2 = 1 + 0.555 555 555 555 556 774 4;
  • 8) 0.555 555 555 555 556 774 4 × 2 = 1 + 0.111 111 111 111 113 548 8;
  • 9) 0.111 111 111 111 113 548 8 × 2 = 0 + 0.222 222 222 222 227 097 6;
  • 10) 0.222 222 222 222 227 097 6 × 2 = 0 + 0.444 444 444 444 454 195 2;
  • 11) 0.444 444 444 444 454 195 2 × 2 = 0 + 0.888 888 888 888 908 390 4;
  • 12) 0.888 888 888 888 908 390 4 × 2 = 1 + 0.777 777 777 777 816 780 8;
  • 13) 0.777 777 777 777 816 780 8 × 2 = 1 + 0.555 555 555 555 633 561 6;
  • 14) 0.555 555 555 555 633 561 6 × 2 = 1 + 0.111 111 111 111 267 123 2;
  • 15) 0.111 111 111 111 267 123 2 × 2 = 0 + 0.222 222 222 222 534 246 4;
  • 16) 0.222 222 222 222 534 246 4 × 2 = 0 + 0.444 444 444 445 068 492 8;
  • 17) 0.444 444 444 445 068 492 8 × 2 = 0 + 0.888 888 888 890 136 985 6;
  • 18) 0.888 888 888 890 136 985 6 × 2 = 1 + 0.777 777 777 780 273 971 2;
  • 19) 0.777 777 777 780 273 971 2 × 2 = 1 + 0.555 555 555 560 547 942 4;
  • 20) 0.555 555 555 560 547 942 4 × 2 = 1 + 0.111 111 111 121 095 884 8;
  • 21) 0.111 111 111 121 095 884 8 × 2 = 0 + 0.222 222 222 242 191 769 6;
  • 22) 0.222 222 222 242 191 769 6 × 2 = 0 + 0.444 444 444 484 383 539 2;
  • 23) 0.444 444 444 484 383 539 2 × 2 = 0 + 0.888 888 888 968 767 078 4;
  • 24) 0.888 888 888 968 767 078 4 × 2 = 1 + 0.777 777 777 937 534 156 8;
  • 25) 0.777 777 777 937 534 156 8 × 2 = 1 + 0.555 555 555 875 068 313 6;
  • 26) 0.555 555 555 875 068 313 6 × 2 = 1 + 0.111 111 111 750 136 627 2;
  • 27) 0.111 111 111 750 136 627 2 × 2 = 0 + 0.222 222 223 500 273 254 4;
  • 28) 0.222 222 223 500 273 254 4 × 2 = 0 + 0.444 444 447 000 546 508 8;
  • 29) 0.444 444 447 000 546 508 8 × 2 = 0 + 0.888 888 894 001 093 017 6;
  • 30) 0.888 888 894 001 093 017 6 × 2 = 1 + 0.777 777 788 002 186 035 2;
  • 31) 0.777 777 788 002 186 035 2 × 2 = 1 + 0.555 555 576 004 372 070 4;
  • 32) 0.555 555 576 004 372 070 4 × 2 = 1 + 0.111 111 152 008 744 140 8;
  • 33) 0.111 111 152 008 744 140 8 × 2 = 0 + 0.222 222 304 017 488 281 6;
  • 34) 0.222 222 304 017 488 281 6 × 2 = 0 + 0.444 444 608 034 976 563 2;
  • 35) 0.444 444 608 034 976 563 2 × 2 = 0 + 0.888 889 216 069 953 126 4;
  • 36) 0.888 889 216 069 953 126 4 × 2 = 1 + 0.777 778 432 139 906 252 8;
  • 37) 0.777 778 432 139 906 252 8 × 2 = 1 + 0.555 556 864 279 812 505 6;
  • 38) 0.555 556 864 279 812 505 6 × 2 = 1 + 0.111 113 728 559 625 011 2;
  • 39) 0.111 113 728 559 625 011 2 × 2 = 0 + 0.222 227 457 119 250 022 4;
  • 40) 0.222 227 457 119 250 022 4 × 2 = 0 + 0.444 454 914 238 500 044 8;
  • 41) 0.444 454 914 238 500 044 8 × 2 = 0 + 0.888 909 828 477 000 089 6;
  • 42) 0.888 909 828 477 000 089 6 × 2 = 1 + 0.777 819 656 954 000 179 2;
  • 43) 0.777 819 656 954 000 179 2 × 2 = 1 + 0.555 639 313 908 000 358 4;
  • 44) 0.555 639 313 908 000 358 4 × 2 = 1 + 0.111 278 627 816 000 716 8;
  • 45) 0.111 278 627 816 000 716 8 × 2 = 0 + 0.222 557 255 632 001 433 6;
  • 46) 0.222 557 255 632 001 433 6 × 2 = 0 + 0.445 114 511 264 002 867 2;
  • 47) 0.445 114 511 264 002 867 2 × 2 = 0 + 0.890 229 022 528 005 734 4;
  • 48) 0.890 229 022 528 005 734 4 × 2 = 1 + 0.780 458 045 056 011 468 8;
  • 49) 0.780 458 045 056 011 468 8 × 2 = 1 + 0.560 916 090 112 022 937 6;
  • 50) 0.560 916 090 112 022 937 6 × 2 = 1 + 0.121 832 180 224 045 875 2;
  • 51) 0.121 832 180 224 045 875 2 × 2 = 0 + 0.243 664 360 448 091 750 4;
  • 52) 0.243 664 360 448 091 750 4 × 2 = 0 + 0.487 328 720 896 183 500 8;
  • 53) 0.487 328 720 896 183 500 8 × 2 = 0 + 0.974 657 441 792 367 001 6;

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:


0.777 777 777 777 777 787 3(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 787 3(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:


24.777 777 777 777 777 787 3(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 24.777 777 777 777 777 787 3 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