24.777 777 777 777 718 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 718(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
24.777 777 777 777 718(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: 0.777 777 777 777 718.

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) 0.777 777 777 777 718 × 2 = 1 + 0.555 555 555 555 436;
  • 2) 0.555 555 555 555 436 × 2 = 1 + 0.111 111 111 110 872;
  • 3) 0.111 111 111 110 872 × 2 = 0 + 0.222 222 222 221 744;
  • 4) 0.222 222 222 221 744 × 2 = 0 + 0.444 444 444 443 488;
  • 5) 0.444 444 444 443 488 × 2 = 0 + 0.888 888 888 886 976;
  • 6) 0.888 888 888 886 976 × 2 = 1 + 0.777 777 777 773 952;
  • 7) 0.777 777 777 773 952 × 2 = 1 + 0.555 555 555 547 904;
  • 8) 0.555 555 555 547 904 × 2 = 1 + 0.111 111 111 095 808;
  • 9) 0.111 111 111 095 808 × 2 = 0 + 0.222 222 222 191 616;
  • 10) 0.222 222 222 191 616 × 2 = 0 + 0.444 444 444 383 232;
  • 11) 0.444 444 444 383 232 × 2 = 0 + 0.888 888 888 766 464;
  • 12) 0.888 888 888 766 464 × 2 = 1 + 0.777 777 777 532 928;
  • 13) 0.777 777 777 532 928 × 2 = 1 + 0.555 555 555 065 856;
  • 14) 0.555 555 555 065 856 × 2 = 1 + 0.111 111 110 131 712;
  • 15) 0.111 111 110 131 712 × 2 = 0 + 0.222 222 220 263 424;
  • 16) 0.222 222 220 263 424 × 2 = 0 + 0.444 444 440 526 848;
  • 17) 0.444 444 440 526 848 × 2 = 0 + 0.888 888 881 053 696;
  • 18) 0.888 888 881 053 696 × 2 = 1 + 0.777 777 762 107 392;
  • 19) 0.777 777 762 107 392 × 2 = 1 + 0.555 555 524 214 784;
  • 20) 0.555 555 524 214 784 × 2 = 1 + 0.111 111 048 429 568;
  • 21) 0.111 111 048 429 568 × 2 = 0 + 0.222 222 096 859 136;
  • 22) 0.222 222 096 859 136 × 2 = 0 + 0.444 444 193 718 272;
  • 23) 0.444 444 193 718 272 × 2 = 0 + 0.888 888 387 436 544;
  • 24) 0.888 888 387 436 544 × 2 = 1 + 0.777 776 774 873 088;
  • 25) 0.777 776 774 873 088 × 2 = 1 + 0.555 553 549 746 176;
  • 26) 0.555 553 549 746 176 × 2 = 1 + 0.111 107 099 492 352;
  • 27) 0.111 107 099 492 352 × 2 = 0 + 0.222 214 198 984 704;
  • 28) 0.222 214 198 984 704 × 2 = 0 + 0.444 428 397 969 408;
  • 29) 0.444 428 397 969 408 × 2 = 0 + 0.888 856 795 938 816;
  • 30) 0.888 856 795 938 816 × 2 = 1 + 0.777 713 591 877 632;
  • 31) 0.777 713 591 877 632 × 2 = 1 + 0.555 427 183 755 264;
  • 32) 0.555 427 183 755 264 × 2 = 1 + 0.110 854 367 510 528;
  • 33) 0.110 854 367 510 528 × 2 = 0 + 0.221 708 735 021 056;
  • 34) 0.221 708 735 021 056 × 2 = 0 + 0.443 417 470 042 112;
  • 35) 0.443 417 470 042 112 × 2 = 0 + 0.886 834 940 084 224;
  • 36) 0.886 834 940 084 224 × 2 = 1 + 0.773 669 880 168 448;
  • 37) 0.773 669 880 168 448 × 2 = 1 + 0.547 339 760 336 896;
  • 38) 0.547 339 760 336 896 × 2 = 1 + 0.094 679 520 673 792;
  • 39) 0.094 679 520 673 792 × 2 = 0 + 0.189 359 041 347 584;
  • 40) 0.189 359 041 347 584 × 2 = 0 + 0.378 718 082 695 168;
  • 41) 0.378 718 082 695 168 × 2 = 0 + 0.757 436 165 390 336;
  • 42) 0.757 436 165 390 336 × 2 = 1 + 0.514 872 330 780 672;
  • 43) 0.514 872 330 780 672 × 2 = 1 + 0.029 744 661 561 344;
  • 44) 0.029 744 661 561 344 × 2 = 0 + 0.059 489 323 122 688;
  • 45) 0.059 489 323 122 688 × 2 = 0 + 0.118 978 646 245 376;
  • 46) 0.118 978 646 245 376 × 2 = 0 + 0.237 957 292 490 752;
  • 47) 0.237 957 292 490 752 × 2 = 0 + 0.475 914 584 981 504;
  • 48) 0.475 914 584 981 504 × 2 = 0 + 0.951 829 169 963 008;
  • 49) 0.951 829 169 963 008 × 2 = 1 + 0.903 658 339 926 016;
  • 50) 0.903 658 339 926 016 × 2 = 1 + 0.807 316 679 852 032;
  • 51) 0.807 316 679 852 032 × 2 = 1 + 0.614 633 359 704 064;
  • 52) 0.614 633 359 704 064 × 2 = 1 + 0.229 266 719 408 128;
  • 53) 0.229 266 719 408 128 × 2 = 0 + 0.458 533 438 816 256;

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:


0.777 777 777 777 718(10) =

0.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0110 0000 1111 0(2)

5. Positive number before normalization:

24.777 777 777 777 718(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0110 0000 1111 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:


24.777 777 777 777 718(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0110 0000 1111 0(2) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0110 0000 1111 0(2) × 20 =

1.1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0110 0000 1111 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 0110 0000 1111 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 0110 0000 1 1110 =

1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0110 0000


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 0110 0000


Decimal number 24.777 777 777 777 718 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 0110 0000

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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