24.777 777 777 777 879 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 879(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
24.777 777 777 777 879(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: 0.777 777 777 777 879.

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) 0.777 777 777 777 879 × 2 = 1 + 0.555 555 555 555 758;
  • 2) 0.555 555 555 555 758 × 2 = 1 + 0.111 111 111 111 516;
  • 3) 0.111 111 111 111 516 × 2 = 0 + 0.222 222 222 223 032;
  • 4) 0.222 222 222 223 032 × 2 = 0 + 0.444 444 444 446 064;
  • 5) 0.444 444 444 446 064 × 2 = 0 + 0.888 888 888 892 128;
  • 6) 0.888 888 888 892 128 × 2 = 1 + 0.777 777 777 784 256;
  • 7) 0.777 777 777 784 256 × 2 = 1 + 0.555 555 555 568 512;
  • 8) 0.555 555 555 568 512 × 2 = 1 + 0.111 111 111 137 024;
  • 9) 0.111 111 111 137 024 × 2 = 0 + 0.222 222 222 274 048;
  • 10) 0.222 222 222 274 048 × 2 = 0 + 0.444 444 444 548 096;
  • 11) 0.444 444 444 548 096 × 2 = 0 + 0.888 888 889 096 192;
  • 12) 0.888 888 889 096 192 × 2 = 1 + 0.777 777 778 192 384;
  • 13) 0.777 777 778 192 384 × 2 = 1 + 0.555 555 556 384 768;
  • 14) 0.555 555 556 384 768 × 2 = 1 + 0.111 111 112 769 536;
  • 15) 0.111 111 112 769 536 × 2 = 0 + 0.222 222 225 539 072;
  • 16) 0.222 222 225 539 072 × 2 = 0 + 0.444 444 451 078 144;
  • 17) 0.444 444 451 078 144 × 2 = 0 + 0.888 888 902 156 288;
  • 18) 0.888 888 902 156 288 × 2 = 1 + 0.777 777 804 312 576;
  • 19) 0.777 777 804 312 576 × 2 = 1 + 0.555 555 608 625 152;
  • 20) 0.555 555 608 625 152 × 2 = 1 + 0.111 111 217 250 304;
  • 21) 0.111 111 217 250 304 × 2 = 0 + 0.222 222 434 500 608;
  • 22) 0.222 222 434 500 608 × 2 = 0 + 0.444 444 869 001 216;
  • 23) 0.444 444 869 001 216 × 2 = 0 + 0.888 889 738 002 432;
  • 24) 0.888 889 738 002 432 × 2 = 1 + 0.777 779 476 004 864;
  • 25) 0.777 779 476 004 864 × 2 = 1 + 0.555 558 952 009 728;
  • 26) 0.555 558 952 009 728 × 2 = 1 + 0.111 117 904 019 456;
  • 27) 0.111 117 904 019 456 × 2 = 0 + 0.222 235 808 038 912;
  • 28) 0.222 235 808 038 912 × 2 = 0 + 0.444 471 616 077 824;
  • 29) 0.444 471 616 077 824 × 2 = 0 + 0.888 943 232 155 648;
  • 30) 0.888 943 232 155 648 × 2 = 1 + 0.777 886 464 311 296;
  • 31) 0.777 886 464 311 296 × 2 = 1 + 0.555 772 928 622 592;
  • 32) 0.555 772 928 622 592 × 2 = 1 + 0.111 545 857 245 184;
  • 33) 0.111 545 857 245 184 × 2 = 0 + 0.223 091 714 490 368;
  • 34) 0.223 091 714 490 368 × 2 = 0 + 0.446 183 428 980 736;
  • 35) 0.446 183 428 980 736 × 2 = 0 + 0.892 366 857 961 472;
  • 36) 0.892 366 857 961 472 × 2 = 1 + 0.784 733 715 922 944;
  • 37) 0.784 733 715 922 944 × 2 = 1 + 0.569 467 431 845 888;
  • 38) 0.569 467 431 845 888 × 2 = 1 + 0.138 934 863 691 776;
  • 39) 0.138 934 863 691 776 × 2 = 0 + 0.277 869 727 383 552;
  • 40) 0.277 869 727 383 552 × 2 = 0 + 0.555 739 454 767 104;
  • 41) 0.555 739 454 767 104 × 2 = 1 + 0.111 478 909 534 208;
  • 42) 0.111 478 909 534 208 × 2 = 0 + 0.222 957 819 068 416;
  • 43) 0.222 957 819 068 416 × 2 = 0 + 0.445 915 638 136 832;
  • 44) 0.445 915 638 136 832 × 2 = 0 + 0.891 831 276 273 664;
  • 45) 0.891 831 276 273 664 × 2 = 1 + 0.783 662 552 547 328;
  • 46) 0.783 662 552 547 328 × 2 = 1 + 0.567 325 105 094 656;
  • 47) 0.567 325 105 094 656 × 2 = 1 + 0.134 650 210 189 312;
  • 48) 0.134 650 210 189 312 × 2 = 0 + 0.269 300 420 378 624;
  • 49) 0.269 300 420 378 624 × 2 = 0 + 0.538 600 840 757 248;
  • 50) 0.538 600 840 757 248 × 2 = 1 + 0.077 201 681 514 496;
  • 51) 0.077 201 681 514 496 × 2 = 0 + 0.154 403 363 028 992;
  • 52) 0.154 403 363 028 992 × 2 = 0 + 0.308 806 726 057 984;
  • 53) 0.308 806 726 057 984 × 2 = 0 + 0.617 613 452 115 968;

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:


0.777 777 777 777 879(10) =

0.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1000 1110 0100 0(2)

5. Positive number before normalization:

24.777 777 777 777 879(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1000 1110 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:


24.777 777 777 777 879(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1000 1110 0100 0(2) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1000 1110 0100 0(2) × 20 =

1.1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1000 1110 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 1000 1110 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 1000 1110 0 1000 =

1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1000 1110


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 1110


Decimal number 24.777 777 777 777 879 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 1110

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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