24.777 777 777 777 941 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 941(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
24.777 777 777 777 941(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: 0.777 777 777 777 941.

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) 0.777 777 777 777 941 × 2 = 1 + 0.555 555 555 555 882;
  • 2) 0.555 555 555 555 882 × 2 = 1 + 0.111 111 111 111 764;
  • 3) 0.111 111 111 111 764 × 2 = 0 + 0.222 222 222 223 528;
  • 4) 0.222 222 222 223 528 × 2 = 0 + 0.444 444 444 447 056;
  • 5) 0.444 444 444 447 056 × 2 = 0 + 0.888 888 888 894 112;
  • 6) 0.888 888 888 894 112 × 2 = 1 + 0.777 777 777 788 224;
  • 7) 0.777 777 777 788 224 × 2 = 1 + 0.555 555 555 576 448;
  • 8) 0.555 555 555 576 448 × 2 = 1 + 0.111 111 111 152 896;
  • 9) 0.111 111 111 152 896 × 2 = 0 + 0.222 222 222 305 792;
  • 10) 0.222 222 222 305 792 × 2 = 0 + 0.444 444 444 611 584;
  • 11) 0.444 444 444 611 584 × 2 = 0 + 0.888 888 889 223 168;
  • 12) 0.888 888 889 223 168 × 2 = 1 + 0.777 777 778 446 336;
  • 13) 0.777 777 778 446 336 × 2 = 1 + 0.555 555 556 892 672;
  • 14) 0.555 555 556 892 672 × 2 = 1 + 0.111 111 113 785 344;
  • 15) 0.111 111 113 785 344 × 2 = 0 + 0.222 222 227 570 688;
  • 16) 0.222 222 227 570 688 × 2 = 0 + 0.444 444 455 141 376;
  • 17) 0.444 444 455 141 376 × 2 = 0 + 0.888 888 910 282 752;
  • 18) 0.888 888 910 282 752 × 2 = 1 + 0.777 777 820 565 504;
  • 19) 0.777 777 820 565 504 × 2 = 1 + 0.555 555 641 131 008;
  • 20) 0.555 555 641 131 008 × 2 = 1 + 0.111 111 282 262 016;
  • 21) 0.111 111 282 262 016 × 2 = 0 + 0.222 222 564 524 032;
  • 22) 0.222 222 564 524 032 × 2 = 0 + 0.444 445 129 048 064;
  • 23) 0.444 445 129 048 064 × 2 = 0 + 0.888 890 258 096 128;
  • 24) 0.888 890 258 096 128 × 2 = 1 + 0.777 780 516 192 256;
  • 25) 0.777 780 516 192 256 × 2 = 1 + 0.555 561 032 384 512;
  • 26) 0.555 561 032 384 512 × 2 = 1 + 0.111 122 064 769 024;
  • 27) 0.111 122 064 769 024 × 2 = 0 + 0.222 244 129 538 048;
  • 28) 0.222 244 129 538 048 × 2 = 0 + 0.444 488 259 076 096;
  • 29) 0.444 488 259 076 096 × 2 = 0 + 0.888 976 518 152 192;
  • 30) 0.888 976 518 152 192 × 2 = 1 + 0.777 953 036 304 384;
  • 31) 0.777 953 036 304 384 × 2 = 1 + 0.555 906 072 608 768;
  • 32) 0.555 906 072 608 768 × 2 = 1 + 0.111 812 145 217 536;
  • 33) 0.111 812 145 217 536 × 2 = 0 + 0.223 624 290 435 072;
  • 34) 0.223 624 290 435 072 × 2 = 0 + 0.447 248 580 870 144;
  • 35) 0.447 248 580 870 144 × 2 = 0 + 0.894 497 161 740 288;
  • 36) 0.894 497 161 740 288 × 2 = 1 + 0.788 994 323 480 576;
  • 37) 0.788 994 323 480 576 × 2 = 1 + 0.577 988 646 961 152;
  • 38) 0.577 988 646 961 152 × 2 = 1 + 0.155 977 293 922 304;
  • 39) 0.155 977 293 922 304 × 2 = 0 + 0.311 954 587 844 608;
  • 40) 0.311 954 587 844 608 × 2 = 0 + 0.623 909 175 689 216;
  • 41) 0.623 909 175 689 216 × 2 = 1 + 0.247 818 351 378 432;
  • 42) 0.247 818 351 378 432 × 2 = 0 + 0.495 636 702 756 864;
  • 43) 0.495 636 702 756 864 × 2 = 0 + 0.991 273 405 513 728;
  • 44) 0.991 273 405 513 728 × 2 = 1 + 0.982 546 811 027 456;
  • 45) 0.982 546 811 027 456 × 2 = 1 + 0.965 093 622 054 912;
  • 46) 0.965 093 622 054 912 × 2 = 1 + 0.930 187 244 109 824;
  • 47) 0.930 187 244 109 824 × 2 = 1 + 0.860 374 488 219 648;
  • 48) 0.860 374 488 219 648 × 2 = 1 + 0.720 748 976 439 296;
  • 49) 0.720 748 976 439 296 × 2 = 1 + 0.441 497 952 878 592;
  • 50) 0.441 497 952 878 592 × 2 = 0 + 0.882 995 905 757 184;
  • 51) 0.882 995 905 757 184 × 2 = 1 + 0.765 991 811 514 368;
  • 52) 0.765 991 811 514 368 × 2 = 1 + 0.531 983 623 028 736;
  • 53) 0.531 983 623 028 736 × 2 = 1 + 0.063 967 246 057 472;

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:


0.777 777 777 777 941(10) =

0.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1001 1111 1011 1(2)

5. Positive number before normalization:

24.777 777 777 777 941(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1001 1111 1011 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:


24.777 777 777 777 941(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1001 1111 1011 1(2) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1001 1111 1011 1(2) × 20 =

1.1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1001 1111 1011 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 1111 1011 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 1111 1 0111 =

1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 1001 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 1001 1111


Decimal number 24.777 777 777 777 941 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 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