24.777 777 777 777 894 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 894(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
24.777 777 777 777 894(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: 0.777 777 777 777 894.

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) 0.777 777 777 777 894 × 2 = 1 + 0.555 555 555 555 788;
  • 2) 0.555 555 555 555 788 × 2 = 1 + 0.111 111 111 111 576;
  • 3) 0.111 111 111 111 576 × 2 = 0 + 0.222 222 222 223 152;
  • 4) 0.222 222 222 223 152 × 2 = 0 + 0.444 444 444 446 304;
  • 5) 0.444 444 444 446 304 × 2 = 0 + 0.888 888 888 892 608;
  • 6) 0.888 888 888 892 608 × 2 = 1 + 0.777 777 777 785 216;
  • 7) 0.777 777 777 785 216 × 2 = 1 + 0.555 555 555 570 432;
  • 8) 0.555 555 555 570 432 × 2 = 1 + 0.111 111 111 140 864;
  • 9) 0.111 111 111 140 864 × 2 = 0 + 0.222 222 222 281 728;
  • 10) 0.222 222 222 281 728 × 2 = 0 + 0.444 444 444 563 456;
  • 11) 0.444 444 444 563 456 × 2 = 0 + 0.888 888 889 126 912;
  • 12) 0.888 888 889 126 912 × 2 = 1 + 0.777 777 778 253 824;
  • 13) 0.777 777 778 253 824 × 2 = 1 + 0.555 555 556 507 648;
  • 14) 0.555 555 556 507 648 × 2 = 1 + 0.111 111 113 015 296;
  • 15) 0.111 111 113 015 296 × 2 = 0 + 0.222 222 226 030 592;
  • 16) 0.222 222 226 030 592 × 2 = 0 + 0.444 444 452 061 184;
  • 17) 0.444 444 452 061 184 × 2 = 0 + 0.888 888 904 122 368;
  • 18) 0.888 888 904 122 368 × 2 = 1 + 0.777 777 808 244 736;
  • 19) 0.777 777 808 244 736 × 2 = 1 + 0.555 555 616 489 472;
  • 20) 0.555 555 616 489 472 × 2 = 1 + 0.111 111 232 978 944;
  • 21) 0.111 111 232 978 944 × 2 = 0 + 0.222 222 465 957 888;
  • 22) 0.222 222 465 957 888 × 2 = 0 + 0.444 444 931 915 776;
  • 23) 0.444 444 931 915 776 × 2 = 0 + 0.888 889 863 831 552;
  • 24) 0.888 889 863 831 552 × 2 = 1 + 0.777 779 727 663 104;
  • 25) 0.777 779 727 663 104 × 2 = 1 + 0.555 559 455 326 208;
  • 26) 0.555 559 455 326 208 × 2 = 1 + 0.111 118 910 652 416;
  • 27) 0.111 118 910 652 416 × 2 = 0 + 0.222 237 821 304 832;
  • 28) 0.222 237 821 304 832 × 2 = 0 + 0.444 475 642 609 664;
  • 29) 0.444 475 642 609 664 × 2 = 0 + 0.888 951 285 219 328;
  • 30) 0.888 951 285 219 328 × 2 = 1 + 0.777 902 570 438 656;
  • 31) 0.777 902 570 438 656 × 2 = 1 + 0.555 805 140 877 312;
  • 32) 0.555 805 140 877 312 × 2 = 1 + 0.111 610 281 754 624;
  • 33) 0.111 610 281 754 624 × 2 = 0 + 0.223 220 563 509 248;
  • 34) 0.223 220 563 509 248 × 2 = 0 + 0.446 441 127 018 496;
  • 35) 0.446 441 127 018 496 × 2 = 0 + 0.892 882 254 036 992;
  • 36) 0.892 882 254 036 992 × 2 = 1 + 0.785 764 508 073 984;
  • 37) 0.785 764 508 073 984 × 2 = 1 + 0.571 529 016 147 968;
  • 38) 0.571 529 016 147 968 × 2 = 1 + 0.143 058 032 295 936;
  • 39) 0.143 058 032 295 936 × 2 = 0 + 0.286 116 064 591 872;
  • 40) 0.286 116 064 591 872 × 2 = 0 + 0.572 232 129 183 744;
  • 41) 0.572 232 129 183 744 × 2 = 1 + 0.144 464 258 367 488;
  • 42) 0.144 464 258 367 488 × 2 = 0 + 0.288 928 516 734 976;
  • 43) 0.288 928 516 734 976 × 2 = 0 + 0.577 857 033 469 952;
  • 44) 0.577 857 033 469 952 × 2 = 1 + 0.155 714 066 939 904;
  • 45) 0.155 714 066 939 904 × 2 = 0 + 0.311 428 133 879 808;
  • 46) 0.311 428 133 879 808 × 2 = 0 + 0.622 856 267 759 616;
  • 47) 0.622 856 267 759 616 × 2 = 1 + 0.245 712 535 519 232;
  • 48) 0.245 712 535 519 232 × 2 = 0 + 0.491 425 071 038 464;
  • 49) 0.491 425 071 038 464 × 2 = 0 + 0.982 850 142 076 928;
  • 50) 0.982 850 142 076 928 × 2 = 1 + 0.965 700 284 153 856;
  • 51) 0.965 700 284 153 856 × 2 = 1 + 0.931 400 568 307 712;
  • 52) 0.931 400 568 307 712 × 2 = 1 + 0.862 801 136 615 424;
  • 53) 0.862 801 136 615 424 × 2 = 1 + 0.725 602 273 230 848;

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:


0.777 777 777 777 894(10) =

0.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1001 0010 0111 1(2)

5. Positive number before normalization:

24.777 777 777 777 894(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1001 0010 0111 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:


24.777 777 777 777 894(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1001 0010 0111 1(2) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1001 0010 0111 1(2) × 20 =

1.1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1001 0010 0111 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 1001 0010 0111 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 1001 0010 0 1111 =

1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1001 0010


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 1001 0010


Decimal number 24.777 777 777 777 894 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 1001 0010

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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