24.777 777 777 777 778 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 778 5(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 5(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 5.

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 5 × 2 = 1 + 0.555 555 555 555 557;
  • 2) 0.555 555 555 555 557 × 2 = 1 + 0.111 111 111 111 114;
  • 3) 0.111 111 111 111 114 × 2 = 0 + 0.222 222 222 222 228;
  • 4) 0.222 222 222 222 228 × 2 = 0 + 0.444 444 444 444 456;
  • 5) 0.444 444 444 444 456 × 2 = 0 + 0.888 888 888 888 912;
  • 6) 0.888 888 888 888 912 × 2 = 1 + 0.777 777 777 777 824;
  • 7) 0.777 777 777 777 824 × 2 = 1 + 0.555 555 555 555 648;
  • 8) 0.555 555 555 555 648 × 2 = 1 + 0.111 111 111 111 296;
  • 9) 0.111 111 111 111 296 × 2 = 0 + 0.222 222 222 222 592;
  • 10) 0.222 222 222 222 592 × 2 = 0 + 0.444 444 444 445 184;
  • 11) 0.444 444 444 445 184 × 2 = 0 + 0.888 888 888 890 368;
  • 12) 0.888 888 888 890 368 × 2 = 1 + 0.777 777 777 780 736;
  • 13) 0.777 777 777 780 736 × 2 = 1 + 0.555 555 555 561 472;
  • 14) 0.555 555 555 561 472 × 2 = 1 + 0.111 111 111 122 944;
  • 15) 0.111 111 111 122 944 × 2 = 0 + 0.222 222 222 245 888;
  • 16) 0.222 222 222 245 888 × 2 = 0 + 0.444 444 444 491 776;
  • 17) 0.444 444 444 491 776 × 2 = 0 + 0.888 888 888 983 552;
  • 18) 0.888 888 888 983 552 × 2 = 1 + 0.777 777 777 967 104;
  • 19) 0.777 777 777 967 104 × 2 = 1 + 0.555 555 555 934 208;
  • 20) 0.555 555 555 934 208 × 2 = 1 + 0.111 111 111 868 416;
  • 21) 0.111 111 111 868 416 × 2 = 0 + 0.222 222 223 736 832;
  • 22) 0.222 222 223 736 832 × 2 = 0 + 0.444 444 447 473 664;
  • 23) 0.444 444 447 473 664 × 2 = 0 + 0.888 888 894 947 328;
  • 24) 0.888 888 894 947 328 × 2 = 1 + 0.777 777 789 894 656;
  • 25) 0.777 777 789 894 656 × 2 = 1 + 0.555 555 579 789 312;
  • 26) 0.555 555 579 789 312 × 2 = 1 + 0.111 111 159 578 624;
  • 27) 0.111 111 159 578 624 × 2 = 0 + 0.222 222 319 157 248;
  • 28) 0.222 222 319 157 248 × 2 = 0 + 0.444 444 638 314 496;
  • 29) 0.444 444 638 314 496 × 2 = 0 + 0.888 889 276 628 992;
  • 30) 0.888 889 276 628 992 × 2 = 1 + 0.777 778 553 257 984;
  • 31) 0.777 778 553 257 984 × 2 = 1 + 0.555 557 106 515 968;
  • 32) 0.555 557 106 515 968 × 2 = 1 + 0.111 114 213 031 936;
  • 33) 0.111 114 213 031 936 × 2 = 0 + 0.222 228 426 063 872;
  • 34) 0.222 228 426 063 872 × 2 = 0 + 0.444 456 852 127 744;
  • 35) 0.444 456 852 127 744 × 2 = 0 + 0.888 913 704 255 488;
  • 36) 0.888 913 704 255 488 × 2 = 1 + 0.777 827 408 510 976;
  • 37) 0.777 827 408 510 976 × 2 = 1 + 0.555 654 817 021 952;
  • 38) 0.555 654 817 021 952 × 2 = 1 + 0.111 309 634 043 904;
  • 39) 0.111 309 634 043 904 × 2 = 0 + 0.222 619 268 087 808;
  • 40) 0.222 619 268 087 808 × 2 = 0 + 0.445 238 536 175 616;
  • 41) 0.445 238 536 175 616 × 2 = 0 + 0.890 477 072 351 232;
  • 42) 0.890 477 072 351 232 × 2 = 1 + 0.780 954 144 702 464;
  • 43) 0.780 954 144 702 464 × 2 = 1 + 0.561 908 289 404 928;
  • 44) 0.561 908 289 404 928 × 2 = 1 + 0.123 816 578 809 856;
  • 45) 0.123 816 578 809 856 × 2 = 0 + 0.247 633 157 619 712;
  • 46) 0.247 633 157 619 712 × 2 = 0 + 0.495 266 315 239 424;
  • 47) 0.495 266 315 239 424 × 2 = 0 + 0.990 532 630 478 848;
  • 48) 0.990 532 630 478 848 × 2 = 1 + 0.981 065 260 957 696;
  • 49) 0.981 065 260 957 696 × 2 = 1 + 0.962 130 521 915 392;
  • 50) 0.962 130 521 915 392 × 2 = 1 + 0.924 261 043 830 784;
  • 51) 0.924 261 043 830 784 × 2 = 1 + 0.848 522 087 661 568;
  • 52) 0.848 522 087 661 568 × 2 = 1 + 0.697 044 175 323 136;
  • 53) 0.697 044 175 323 136 × 2 = 1 + 0.394 088 350 646 272;

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 5(10) =

0.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1111 1(2)

5. Positive number before normalization:

24.777 777 777 777 778 5(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1111 1(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 5(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1111 1(2) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1111 1(2) × 20 =

1.1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1111 1(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 1111 1


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 1111 =

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 24.777 777 777 777 778 5 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