24.777 777 777 777 882 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 882(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
24.777 777 777 777 882(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: 0.777 777 777 777 882.

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) 0.777 777 777 777 882 × 2 = 1 + 0.555 555 555 555 764;
  • 2) 0.555 555 555 555 764 × 2 = 1 + 0.111 111 111 111 528;
  • 3) 0.111 111 111 111 528 × 2 = 0 + 0.222 222 222 223 056;
  • 4) 0.222 222 222 223 056 × 2 = 0 + 0.444 444 444 446 112;
  • 5) 0.444 444 444 446 112 × 2 = 0 + 0.888 888 888 892 224;
  • 6) 0.888 888 888 892 224 × 2 = 1 + 0.777 777 777 784 448;
  • 7) 0.777 777 777 784 448 × 2 = 1 + 0.555 555 555 568 896;
  • 8) 0.555 555 555 568 896 × 2 = 1 + 0.111 111 111 137 792;
  • 9) 0.111 111 111 137 792 × 2 = 0 + 0.222 222 222 275 584;
  • 10) 0.222 222 222 275 584 × 2 = 0 + 0.444 444 444 551 168;
  • 11) 0.444 444 444 551 168 × 2 = 0 + 0.888 888 889 102 336;
  • 12) 0.888 888 889 102 336 × 2 = 1 + 0.777 777 778 204 672;
  • 13) 0.777 777 778 204 672 × 2 = 1 + 0.555 555 556 409 344;
  • 14) 0.555 555 556 409 344 × 2 = 1 + 0.111 111 112 818 688;
  • 15) 0.111 111 112 818 688 × 2 = 0 + 0.222 222 225 637 376;
  • 16) 0.222 222 225 637 376 × 2 = 0 + 0.444 444 451 274 752;
  • 17) 0.444 444 451 274 752 × 2 = 0 + 0.888 888 902 549 504;
  • 18) 0.888 888 902 549 504 × 2 = 1 + 0.777 777 805 099 008;
  • 19) 0.777 777 805 099 008 × 2 = 1 + 0.555 555 610 198 016;
  • 20) 0.555 555 610 198 016 × 2 = 1 + 0.111 111 220 396 032;
  • 21) 0.111 111 220 396 032 × 2 = 0 + 0.222 222 440 792 064;
  • 22) 0.222 222 440 792 064 × 2 = 0 + 0.444 444 881 584 128;
  • 23) 0.444 444 881 584 128 × 2 = 0 + 0.888 889 763 168 256;
  • 24) 0.888 889 763 168 256 × 2 = 1 + 0.777 779 526 336 512;
  • 25) 0.777 779 526 336 512 × 2 = 1 + 0.555 559 052 673 024;
  • 26) 0.555 559 052 673 024 × 2 = 1 + 0.111 118 105 346 048;
  • 27) 0.111 118 105 346 048 × 2 = 0 + 0.222 236 210 692 096;
  • 28) 0.222 236 210 692 096 × 2 = 0 + 0.444 472 421 384 192;
  • 29) 0.444 472 421 384 192 × 2 = 0 + 0.888 944 842 768 384;
  • 30) 0.888 944 842 768 384 × 2 = 1 + 0.777 889 685 536 768;
  • 31) 0.777 889 685 536 768 × 2 = 1 + 0.555 779 371 073 536;
  • 32) 0.555 779 371 073 536 × 2 = 1 + 0.111 558 742 147 072;
  • 33) 0.111 558 742 147 072 × 2 = 0 + 0.223 117 484 294 144;
  • 34) 0.223 117 484 294 144 × 2 = 0 + 0.446 234 968 588 288;
  • 35) 0.446 234 968 588 288 × 2 = 0 + 0.892 469 937 176 576;
  • 36) 0.892 469 937 176 576 × 2 = 1 + 0.784 939 874 353 152;
  • 37) 0.784 939 874 353 152 × 2 = 1 + 0.569 879 748 706 304;
  • 38) 0.569 879 748 706 304 × 2 = 1 + 0.139 759 497 412 608;
  • 39) 0.139 759 497 412 608 × 2 = 0 + 0.279 518 994 825 216;
  • 40) 0.279 518 994 825 216 × 2 = 0 + 0.559 037 989 650 432;
  • 41) 0.559 037 989 650 432 × 2 = 1 + 0.118 075 979 300 864;
  • 42) 0.118 075 979 300 864 × 2 = 0 + 0.236 151 958 601 728;
  • 43) 0.236 151 958 601 728 × 2 = 0 + 0.472 303 917 203 456;
  • 44) 0.472 303 917 203 456 × 2 = 0 + 0.944 607 834 406 912;
  • 45) 0.944 607 834 406 912 × 2 = 1 + 0.889 215 668 813 824;
  • 46) 0.889 215 668 813 824 × 2 = 1 + 0.778 431 337 627 648;
  • 47) 0.778 431 337 627 648 × 2 = 1 + 0.556 862 675 255 296;
  • 48) 0.556 862 675 255 296 × 2 = 1 + 0.113 725 350 510 592;
  • 49) 0.113 725 350 510 592 × 2 = 0 + 0.227 450 701 021 184;
  • 50) 0.227 450 701 021 184 × 2 = 0 + 0.454 901 402 042 368;
  • 51) 0.454 901 402 042 368 × 2 = 0 + 0.909 802 804 084 736;
  • 52) 0.909 802 804 084 736 × 2 = 1 + 0.819 605 608 169 472;
  • 53) 0.819 605 608 169 472 × 2 = 1 + 0.639 211 216 338 944;

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:


0.777 777 777 777 882(10) =

0.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1000 1111 0001 1(2)

5. Positive number before normalization:

24.777 777 777 777 882(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1000 1111 0001 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:


24.777 777 777 777 882(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1000 1111 0001 1(2) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1000 1111 0001 1(2) × 20 =

1.1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1000 1111 0001 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 1000 1111 0001 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 1000 1111 0 0011 =

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


Decimal number 24.777 777 777 777 882 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 1000 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