24.777 777 777 777 777 731 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 731 9(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
24.777 777 777 777 777 731 9(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: 0.777 777 777 777 777 731 9.

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) 0.777 777 777 777 777 731 9 × 2 = 1 + 0.555 555 555 555 555 463 8;
  • 2) 0.555 555 555 555 555 463 8 × 2 = 1 + 0.111 111 111 111 110 927 6;
  • 3) 0.111 111 111 111 110 927 6 × 2 = 0 + 0.222 222 222 222 221 855 2;
  • 4) 0.222 222 222 222 221 855 2 × 2 = 0 + 0.444 444 444 444 443 710 4;
  • 5) 0.444 444 444 444 443 710 4 × 2 = 0 + 0.888 888 888 888 887 420 8;
  • 6) 0.888 888 888 888 887 420 8 × 2 = 1 + 0.777 777 777 777 774 841 6;
  • 7) 0.777 777 777 777 774 841 6 × 2 = 1 + 0.555 555 555 555 549 683 2;
  • 8) 0.555 555 555 555 549 683 2 × 2 = 1 + 0.111 111 111 111 099 366 4;
  • 9) 0.111 111 111 111 099 366 4 × 2 = 0 + 0.222 222 222 222 198 732 8;
  • 10) 0.222 222 222 222 198 732 8 × 2 = 0 + 0.444 444 444 444 397 465 6;
  • 11) 0.444 444 444 444 397 465 6 × 2 = 0 + 0.888 888 888 888 794 931 2;
  • 12) 0.888 888 888 888 794 931 2 × 2 = 1 + 0.777 777 777 777 589 862 4;
  • 13) 0.777 777 777 777 589 862 4 × 2 = 1 + 0.555 555 555 555 179 724 8;
  • 14) 0.555 555 555 555 179 724 8 × 2 = 1 + 0.111 111 111 110 359 449 6;
  • 15) 0.111 111 111 110 359 449 6 × 2 = 0 + 0.222 222 222 220 718 899 2;
  • 16) 0.222 222 222 220 718 899 2 × 2 = 0 + 0.444 444 444 441 437 798 4;
  • 17) 0.444 444 444 441 437 798 4 × 2 = 0 + 0.888 888 888 882 875 596 8;
  • 18) 0.888 888 888 882 875 596 8 × 2 = 1 + 0.777 777 777 765 751 193 6;
  • 19) 0.777 777 777 765 751 193 6 × 2 = 1 + 0.555 555 555 531 502 387 2;
  • 20) 0.555 555 555 531 502 387 2 × 2 = 1 + 0.111 111 111 063 004 774 4;
  • 21) 0.111 111 111 063 004 774 4 × 2 = 0 + 0.222 222 222 126 009 548 8;
  • 22) 0.222 222 222 126 009 548 8 × 2 = 0 + 0.444 444 444 252 019 097 6;
  • 23) 0.444 444 444 252 019 097 6 × 2 = 0 + 0.888 888 888 504 038 195 2;
  • 24) 0.888 888 888 504 038 195 2 × 2 = 1 + 0.777 777 777 008 076 390 4;
  • 25) 0.777 777 777 008 076 390 4 × 2 = 1 + 0.555 555 554 016 152 780 8;
  • 26) 0.555 555 554 016 152 780 8 × 2 = 1 + 0.111 111 108 032 305 561 6;
  • 27) 0.111 111 108 032 305 561 6 × 2 = 0 + 0.222 222 216 064 611 123 2;
  • 28) 0.222 222 216 064 611 123 2 × 2 = 0 + 0.444 444 432 129 222 246 4;
  • 29) 0.444 444 432 129 222 246 4 × 2 = 0 + 0.888 888 864 258 444 492 8;
  • 30) 0.888 888 864 258 444 492 8 × 2 = 1 + 0.777 777 728 516 888 985 6;
  • 31) 0.777 777 728 516 888 985 6 × 2 = 1 + 0.555 555 457 033 777 971 2;
  • 32) 0.555 555 457 033 777 971 2 × 2 = 1 + 0.111 110 914 067 555 942 4;
  • 33) 0.111 110 914 067 555 942 4 × 2 = 0 + 0.222 221 828 135 111 884 8;
  • 34) 0.222 221 828 135 111 884 8 × 2 = 0 + 0.444 443 656 270 223 769 6;
  • 35) 0.444 443 656 270 223 769 6 × 2 = 0 + 0.888 887 312 540 447 539 2;
  • 36) 0.888 887 312 540 447 539 2 × 2 = 1 + 0.777 774 625 080 895 078 4;
  • 37) 0.777 774 625 080 895 078 4 × 2 = 1 + 0.555 549 250 161 790 156 8;
  • 38) 0.555 549 250 161 790 156 8 × 2 = 1 + 0.111 098 500 323 580 313 6;
  • 39) 0.111 098 500 323 580 313 6 × 2 = 0 + 0.222 197 000 647 160 627 2;
  • 40) 0.222 197 000 647 160 627 2 × 2 = 0 + 0.444 394 001 294 321 254 4;
  • 41) 0.444 394 001 294 321 254 4 × 2 = 0 + 0.888 788 002 588 642 508 8;
  • 42) 0.888 788 002 588 642 508 8 × 2 = 1 + 0.777 576 005 177 285 017 6;
  • 43) 0.777 576 005 177 285 017 6 × 2 = 1 + 0.555 152 010 354 570 035 2;
  • 44) 0.555 152 010 354 570 035 2 × 2 = 1 + 0.110 304 020 709 140 070 4;
  • 45) 0.110 304 020 709 140 070 4 × 2 = 0 + 0.220 608 041 418 280 140 8;
  • 46) 0.220 608 041 418 280 140 8 × 2 = 0 + 0.441 216 082 836 560 281 6;
  • 47) 0.441 216 082 836 560 281 6 × 2 = 0 + 0.882 432 165 673 120 563 2;
  • 48) 0.882 432 165 673 120 563 2 × 2 = 1 + 0.764 864 331 346 241 126 4;
  • 49) 0.764 864 331 346 241 126 4 × 2 = 1 + 0.529 728 662 692 482 252 8;
  • 50) 0.529 728 662 692 482 252 8 × 2 = 1 + 0.059 457 325 384 964 505 6;
  • 51) 0.059 457 325 384 964 505 6 × 2 = 0 + 0.118 914 650 769 929 011 2;
  • 52) 0.118 914 650 769 929 011 2 × 2 = 0 + 0.237 829 301 539 858 022 4;
  • 53) 0.237 829 301 539 858 022 4 × 2 = 0 + 0.475 658 603 079 716 044 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:


0.777 777 777 777 777 731 9(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 731 9(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:


24.777 777 777 777 777 731 9(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 24.777 777 777 777 777 731 9 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