24.777 777 777 777 777 780 58 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 780 58(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
24.777 777 777 777 777 780 58(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: 0.777 777 777 777 777 780 58.

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) 0.777 777 777 777 777 780 58 × 2 = 1 + 0.555 555 555 555 555 561 16;
  • 2) 0.555 555 555 555 555 561 16 × 2 = 1 + 0.111 111 111 111 111 122 32;
  • 3) 0.111 111 111 111 111 122 32 × 2 = 0 + 0.222 222 222 222 222 244 64;
  • 4) 0.222 222 222 222 222 244 64 × 2 = 0 + 0.444 444 444 444 444 489 28;
  • 5) 0.444 444 444 444 444 489 28 × 2 = 0 + 0.888 888 888 888 888 978 56;
  • 6) 0.888 888 888 888 888 978 56 × 2 = 1 + 0.777 777 777 777 777 957 12;
  • 7) 0.777 777 777 777 777 957 12 × 2 = 1 + 0.555 555 555 555 555 914 24;
  • 8) 0.555 555 555 555 555 914 24 × 2 = 1 + 0.111 111 111 111 111 828 48;
  • 9) 0.111 111 111 111 111 828 48 × 2 = 0 + 0.222 222 222 222 223 656 96;
  • 10) 0.222 222 222 222 223 656 96 × 2 = 0 + 0.444 444 444 444 447 313 92;
  • 11) 0.444 444 444 444 447 313 92 × 2 = 0 + 0.888 888 888 888 894 627 84;
  • 12) 0.888 888 888 888 894 627 84 × 2 = 1 + 0.777 777 777 777 789 255 68;
  • 13) 0.777 777 777 777 789 255 68 × 2 = 1 + 0.555 555 555 555 578 511 36;
  • 14) 0.555 555 555 555 578 511 36 × 2 = 1 + 0.111 111 111 111 157 022 72;
  • 15) 0.111 111 111 111 157 022 72 × 2 = 0 + 0.222 222 222 222 314 045 44;
  • 16) 0.222 222 222 222 314 045 44 × 2 = 0 + 0.444 444 444 444 628 090 88;
  • 17) 0.444 444 444 444 628 090 88 × 2 = 0 + 0.888 888 888 889 256 181 76;
  • 18) 0.888 888 888 889 256 181 76 × 2 = 1 + 0.777 777 777 778 512 363 52;
  • 19) 0.777 777 777 778 512 363 52 × 2 = 1 + 0.555 555 555 557 024 727 04;
  • 20) 0.555 555 555 557 024 727 04 × 2 = 1 + 0.111 111 111 114 049 454 08;
  • 21) 0.111 111 111 114 049 454 08 × 2 = 0 + 0.222 222 222 228 098 908 16;
  • 22) 0.222 222 222 228 098 908 16 × 2 = 0 + 0.444 444 444 456 197 816 32;
  • 23) 0.444 444 444 456 197 816 32 × 2 = 0 + 0.888 888 888 912 395 632 64;
  • 24) 0.888 888 888 912 395 632 64 × 2 = 1 + 0.777 777 777 824 791 265 28;
  • 25) 0.777 777 777 824 791 265 28 × 2 = 1 + 0.555 555 555 649 582 530 56;
  • 26) 0.555 555 555 649 582 530 56 × 2 = 1 + 0.111 111 111 299 165 061 12;
  • 27) 0.111 111 111 299 165 061 12 × 2 = 0 + 0.222 222 222 598 330 122 24;
  • 28) 0.222 222 222 598 330 122 24 × 2 = 0 + 0.444 444 445 196 660 244 48;
  • 29) 0.444 444 445 196 660 244 48 × 2 = 0 + 0.888 888 890 393 320 488 96;
  • 30) 0.888 888 890 393 320 488 96 × 2 = 1 + 0.777 777 780 786 640 977 92;
  • 31) 0.777 777 780 786 640 977 92 × 2 = 1 + 0.555 555 561 573 281 955 84;
  • 32) 0.555 555 561 573 281 955 84 × 2 = 1 + 0.111 111 123 146 563 911 68;
  • 33) 0.111 111 123 146 563 911 68 × 2 = 0 + 0.222 222 246 293 127 823 36;
  • 34) 0.222 222 246 293 127 823 36 × 2 = 0 + 0.444 444 492 586 255 646 72;
  • 35) 0.444 444 492 586 255 646 72 × 2 = 0 + 0.888 888 985 172 511 293 44;
  • 36) 0.888 888 985 172 511 293 44 × 2 = 1 + 0.777 777 970 345 022 586 88;
  • 37) 0.777 777 970 345 022 586 88 × 2 = 1 + 0.555 555 940 690 045 173 76;
  • 38) 0.555 555 940 690 045 173 76 × 2 = 1 + 0.111 111 881 380 090 347 52;
  • 39) 0.111 111 881 380 090 347 52 × 2 = 0 + 0.222 223 762 760 180 695 04;
  • 40) 0.222 223 762 760 180 695 04 × 2 = 0 + 0.444 447 525 520 361 390 08;
  • 41) 0.444 447 525 520 361 390 08 × 2 = 0 + 0.888 895 051 040 722 780 16;
  • 42) 0.888 895 051 040 722 780 16 × 2 = 1 + 0.777 790 102 081 445 560 32;
  • 43) 0.777 790 102 081 445 560 32 × 2 = 1 + 0.555 580 204 162 891 120 64;
  • 44) 0.555 580 204 162 891 120 64 × 2 = 1 + 0.111 160 408 325 782 241 28;
  • 45) 0.111 160 408 325 782 241 28 × 2 = 0 + 0.222 320 816 651 564 482 56;
  • 46) 0.222 320 816 651 564 482 56 × 2 = 0 + 0.444 641 633 303 128 965 12;
  • 47) 0.444 641 633 303 128 965 12 × 2 = 0 + 0.889 283 266 606 257 930 24;
  • 48) 0.889 283 266 606 257 930 24 × 2 = 1 + 0.778 566 533 212 515 860 48;
  • 49) 0.778 566 533 212 515 860 48 × 2 = 1 + 0.557 133 066 425 031 720 96;
  • 50) 0.557 133 066 425 031 720 96 × 2 = 1 + 0.114 266 132 850 063 441 92;
  • 51) 0.114 266 132 850 063 441 92 × 2 = 0 + 0.228 532 265 700 126 883 84;
  • 52) 0.228 532 265 700 126 883 84 × 2 = 0 + 0.457 064 531 400 253 767 68;
  • 53) 0.457 064 531 400 253 767 68 × 2 = 0 + 0.914 129 062 800 507 535 36;

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:


0.777 777 777 777 777 780 58(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 780 58(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:


24.777 777 777 777 777 780 58(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 24.777 777 777 777 777 780 58 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