24.777 777 777 777 777 777 777 742 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 777 777 742(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
24.777 777 777 777 777 777 777 742(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: 0.777 777 777 777 777 777 777 742.

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) 0.777 777 777 777 777 777 777 742 × 2 = 1 + 0.555 555 555 555 555 555 555 484;
  • 2) 0.555 555 555 555 555 555 555 484 × 2 = 1 + 0.111 111 111 111 111 111 110 968;
  • 3) 0.111 111 111 111 111 111 110 968 × 2 = 0 + 0.222 222 222 222 222 222 221 936;
  • 4) 0.222 222 222 222 222 222 221 936 × 2 = 0 + 0.444 444 444 444 444 444 443 872;
  • 5) 0.444 444 444 444 444 444 443 872 × 2 = 0 + 0.888 888 888 888 888 888 887 744;
  • 6) 0.888 888 888 888 888 888 887 744 × 2 = 1 + 0.777 777 777 777 777 777 775 488;
  • 7) 0.777 777 777 777 777 777 775 488 × 2 = 1 + 0.555 555 555 555 555 555 550 976;
  • 8) 0.555 555 555 555 555 555 550 976 × 2 = 1 + 0.111 111 111 111 111 111 101 952;
  • 9) 0.111 111 111 111 111 111 101 952 × 2 = 0 + 0.222 222 222 222 222 222 203 904;
  • 10) 0.222 222 222 222 222 222 203 904 × 2 = 0 + 0.444 444 444 444 444 444 407 808;
  • 11) 0.444 444 444 444 444 444 407 808 × 2 = 0 + 0.888 888 888 888 888 888 815 616;
  • 12) 0.888 888 888 888 888 888 815 616 × 2 = 1 + 0.777 777 777 777 777 777 631 232;
  • 13) 0.777 777 777 777 777 777 631 232 × 2 = 1 + 0.555 555 555 555 555 555 262 464;
  • 14) 0.555 555 555 555 555 555 262 464 × 2 = 1 + 0.111 111 111 111 111 110 524 928;
  • 15) 0.111 111 111 111 111 110 524 928 × 2 = 0 + 0.222 222 222 222 222 221 049 856;
  • 16) 0.222 222 222 222 222 221 049 856 × 2 = 0 + 0.444 444 444 444 444 442 099 712;
  • 17) 0.444 444 444 444 444 442 099 712 × 2 = 0 + 0.888 888 888 888 888 884 199 424;
  • 18) 0.888 888 888 888 888 884 199 424 × 2 = 1 + 0.777 777 777 777 777 768 398 848;
  • 19) 0.777 777 777 777 777 768 398 848 × 2 = 1 + 0.555 555 555 555 555 536 797 696;
  • 20) 0.555 555 555 555 555 536 797 696 × 2 = 1 + 0.111 111 111 111 111 073 595 392;
  • 21) 0.111 111 111 111 111 073 595 392 × 2 = 0 + 0.222 222 222 222 222 147 190 784;
  • 22) 0.222 222 222 222 222 147 190 784 × 2 = 0 + 0.444 444 444 444 444 294 381 568;
  • 23) 0.444 444 444 444 444 294 381 568 × 2 = 0 + 0.888 888 888 888 888 588 763 136;
  • 24) 0.888 888 888 888 888 588 763 136 × 2 = 1 + 0.777 777 777 777 777 177 526 272;
  • 25) 0.777 777 777 777 777 177 526 272 × 2 = 1 + 0.555 555 555 555 554 355 052 544;
  • 26) 0.555 555 555 555 554 355 052 544 × 2 = 1 + 0.111 111 111 111 108 710 105 088;
  • 27) 0.111 111 111 111 108 710 105 088 × 2 = 0 + 0.222 222 222 222 217 420 210 176;
  • 28) 0.222 222 222 222 217 420 210 176 × 2 = 0 + 0.444 444 444 444 434 840 420 352;
  • 29) 0.444 444 444 444 434 840 420 352 × 2 = 0 + 0.888 888 888 888 869 680 840 704;
  • 30) 0.888 888 888 888 869 680 840 704 × 2 = 1 + 0.777 777 777 777 739 361 681 408;
  • 31) 0.777 777 777 777 739 361 681 408 × 2 = 1 + 0.555 555 555 555 478 723 362 816;
  • 32) 0.555 555 555 555 478 723 362 816 × 2 = 1 + 0.111 111 111 110 957 446 725 632;
  • 33) 0.111 111 111 110 957 446 725 632 × 2 = 0 + 0.222 222 222 221 914 893 451 264;
  • 34) 0.222 222 222 221 914 893 451 264 × 2 = 0 + 0.444 444 444 443 829 786 902 528;
  • 35) 0.444 444 444 443 829 786 902 528 × 2 = 0 + 0.888 888 888 887 659 573 805 056;
  • 36) 0.888 888 888 887 659 573 805 056 × 2 = 1 + 0.777 777 777 775 319 147 610 112;
  • 37) 0.777 777 777 775 319 147 610 112 × 2 = 1 + 0.555 555 555 550 638 295 220 224;
  • 38) 0.555 555 555 550 638 295 220 224 × 2 = 1 + 0.111 111 111 101 276 590 440 448;
  • 39) 0.111 111 111 101 276 590 440 448 × 2 = 0 + 0.222 222 222 202 553 180 880 896;
  • 40) 0.222 222 222 202 553 180 880 896 × 2 = 0 + 0.444 444 444 405 106 361 761 792;
  • 41) 0.444 444 444 405 106 361 761 792 × 2 = 0 + 0.888 888 888 810 212 723 523 584;
  • 42) 0.888 888 888 810 212 723 523 584 × 2 = 1 + 0.777 777 777 620 425 447 047 168;
  • 43) 0.777 777 777 620 425 447 047 168 × 2 = 1 + 0.555 555 555 240 850 894 094 336;
  • 44) 0.555 555 555 240 850 894 094 336 × 2 = 1 + 0.111 111 110 481 701 788 188 672;
  • 45) 0.111 111 110 481 701 788 188 672 × 2 = 0 + 0.222 222 220 963 403 576 377 344;
  • 46) 0.222 222 220 963 403 576 377 344 × 2 = 0 + 0.444 444 441 926 807 152 754 688;
  • 47) 0.444 444 441 926 807 152 754 688 × 2 = 0 + 0.888 888 883 853 614 305 509 376;
  • 48) 0.888 888 883 853 614 305 509 376 × 2 = 1 + 0.777 777 767 707 228 611 018 752;
  • 49) 0.777 777 767 707 228 611 018 752 × 2 = 1 + 0.555 555 535 414 457 222 037 504;
  • 50) 0.555 555 535 414 457 222 037 504 × 2 = 1 + 0.111 111 070 828 914 444 075 008;
  • 51) 0.111 111 070 828 914 444 075 008 × 2 = 0 + 0.222 222 141 657 828 888 150 016;
  • 52) 0.222 222 141 657 828 888 150 016 × 2 = 0 + 0.444 444 283 315 657 776 300 032;
  • 53) 0.444 444 283 315 657 776 300 032 × 2 = 0 + 0.888 888 566 631 315 552 600 064;

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:


0.777 777 777 777 777 777 777 742(10) =

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

5. Positive number before normalization:

24.777 777 777 777 777 777 777 742(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:


24.777 777 777 777 777 777 777 742(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 24.777 777 777 777 777 777 777 742 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