24.777 777 777 777 712 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 712(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
24.777 777 777 777 712(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: 0.777 777 777 777 712.

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) 0.777 777 777 777 712 × 2 = 1 + 0.555 555 555 555 424;
  • 2) 0.555 555 555 555 424 × 2 = 1 + 0.111 111 111 110 848;
  • 3) 0.111 111 111 110 848 × 2 = 0 + 0.222 222 222 221 696;
  • 4) 0.222 222 222 221 696 × 2 = 0 + 0.444 444 444 443 392;
  • 5) 0.444 444 444 443 392 × 2 = 0 + 0.888 888 888 886 784;
  • 6) 0.888 888 888 886 784 × 2 = 1 + 0.777 777 777 773 568;
  • 7) 0.777 777 777 773 568 × 2 = 1 + 0.555 555 555 547 136;
  • 8) 0.555 555 555 547 136 × 2 = 1 + 0.111 111 111 094 272;
  • 9) 0.111 111 111 094 272 × 2 = 0 + 0.222 222 222 188 544;
  • 10) 0.222 222 222 188 544 × 2 = 0 + 0.444 444 444 377 088;
  • 11) 0.444 444 444 377 088 × 2 = 0 + 0.888 888 888 754 176;
  • 12) 0.888 888 888 754 176 × 2 = 1 + 0.777 777 777 508 352;
  • 13) 0.777 777 777 508 352 × 2 = 1 + 0.555 555 555 016 704;
  • 14) 0.555 555 555 016 704 × 2 = 1 + 0.111 111 110 033 408;
  • 15) 0.111 111 110 033 408 × 2 = 0 + 0.222 222 220 066 816;
  • 16) 0.222 222 220 066 816 × 2 = 0 + 0.444 444 440 133 632;
  • 17) 0.444 444 440 133 632 × 2 = 0 + 0.888 888 880 267 264;
  • 18) 0.888 888 880 267 264 × 2 = 1 + 0.777 777 760 534 528;
  • 19) 0.777 777 760 534 528 × 2 = 1 + 0.555 555 521 069 056;
  • 20) 0.555 555 521 069 056 × 2 = 1 + 0.111 111 042 138 112;
  • 21) 0.111 111 042 138 112 × 2 = 0 + 0.222 222 084 276 224;
  • 22) 0.222 222 084 276 224 × 2 = 0 + 0.444 444 168 552 448;
  • 23) 0.444 444 168 552 448 × 2 = 0 + 0.888 888 337 104 896;
  • 24) 0.888 888 337 104 896 × 2 = 1 + 0.777 776 674 209 792;
  • 25) 0.777 776 674 209 792 × 2 = 1 + 0.555 553 348 419 584;
  • 26) 0.555 553 348 419 584 × 2 = 1 + 0.111 106 696 839 168;
  • 27) 0.111 106 696 839 168 × 2 = 0 + 0.222 213 393 678 336;
  • 28) 0.222 213 393 678 336 × 2 = 0 + 0.444 426 787 356 672;
  • 29) 0.444 426 787 356 672 × 2 = 0 + 0.888 853 574 713 344;
  • 30) 0.888 853 574 713 344 × 2 = 1 + 0.777 707 149 426 688;
  • 31) 0.777 707 149 426 688 × 2 = 1 + 0.555 414 298 853 376;
  • 32) 0.555 414 298 853 376 × 2 = 1 + 0.110 828 597 706 752;
  • 33) 0.110 828 597 706 752 × 2 = 0 + 0.221 657 195 413 504;
  • 34) 0.221 657 195 413 504 × 2 = 0 + 0.443 314 390 827 008;
  • 35) 0.443 314 390 827 008 × 2 = 0 + 0.886 628 781 654 016;
  • 36) 0.886 628 781 654 016 × 2 = 1 + 0.773 257 563 308 032;
  • 37) 0.773 257 563 308 032 × 2 = 1 + 0.546 515 126 616 064;
  • 38) 0.546 515 126 616 064 × 2 = 1 + 0.093 030 253 232 128;
  • 39) 0.093 030 253 232 128 × 2 = 0 + 0.186 060 506 464 256;
  • 40) 0.186 060 506 464 256 × 2 = 0 + 0.372 121 012 928 512;
  • 41) 0.372 121 012 928 512 × 2 = 0 + 0.744 242 025 857 024;
  • 42) 0.744 242 025 857 024 × 2 = 1 + 0.488 484 051 714 048;
  • 43) 0.488 484 051 714 048 × 2 = 0 + 0.976 968 103 428 096;
  • 44) 0.976 968 103 428 096 × 2 = 1 + 0.953 936 206 856 192;
  • 45) 0.953 936 206 856 192 × 2 = 1 + 0.907 872 413 712 384;
  • 46) 0.907 872 413 712 384 × 2 = 1 + 0.815 744 827 424 768;
  • 47) 0.815 744 827 424 768 × 2 = 1 + 0.631 489 654 849 536;
  • 48) 0.631 489 654 849 536 × 2 = 1 + 0.262 979 309 699 072;
  • 49) 0.262 979 309 699 072 × 2 = 0 + 0.525 958 619 398 144;
  • 50) 0.525 958 619 398 144 × 2 = 1 + 0.051 917 238 796 288;
  • 51) 0.051 917 238 796 288 × 2 = 0 + 0.103 834 477 592 576;
  • 52) 0.103 834 477 592 576 × 2 = 0 + 0.207 668 955 185 152;
  • 53) 0.207 668 955 185 152 × 2 = 0 + 0.415 337 910 370 304;

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:


0.777 777 777 777 712(10) =

0.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0101 1111 0100 0(2)

5. Positive number before normalization:

24.777 777 777 777 712(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0101 1111 0100 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:


24.777 777 777 777 712(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0101 1111 0100 0(2) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0101 1111 0100 0(2) × 20 =

1.1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0101 1111 0100 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 0101 1111 0100 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 0101 1111 0 1000 =

1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0101 1111


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 0101 1111


Decimal number 24.777 777 777 777 712 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 0101 1111

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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